MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements

MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements

MP Board Class 10th Science Chapter 5 Intext Questions

Intext Questions Page No. 81

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.
Answer:
In Newland’s Octaves, the properties of lithium and sodium were found to be the same. This arrangement is also found in Dobereiner triads.

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:
Dobereiner could identify only three ‘triads’ from the elements known at that time. Hence this system of classification into triads was not found to be useful.

MP Board Solutions

Question 3.
What were the limitations of Newlands’ Law of Octaves?
Answer:

The Limitations of Newlands’ Law of Oclaves is as follows:

  1. It was found that the Law of Octaves was applicable only upto calcium, as after calcium every eighth element did not possess properties similar to that of the first.
  2. It was assumed by Newlands’ that only 56 elements existed in nature and no more elements would be discovered in the future. But later on, several new elements were discovered, whose properties did not fit into the Law of Octaves.
  3. In order to fit elements into his Table, Newlands adjusted two elements in the same slot, but also put some unlike elements under the same note.

Intext Questions Page No. 85

Question 1.
Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements: K, C, Al, Si, Ba.
Answer:

  1. K is in I group. Its oxide is K2O
  2. C, is in IV group, its oxide is CO2
  3. Al, is in III group, its oxide is Al2O3
  4. Si, is IV group, its oxide is SiO2
  5. Ba, is in II group, its oxide is BaO

Question 2.
Besides gallium, which other elements have since been discovered that were left by Mendeleev in his Periodic Table? (any two)
Answer:
Scandium and germanium.

Question 3.
What were the criteria used by Mendeleev in creating his Periodic Table?
Answer:
Mendeleev used the relationship between the atomic masses of the elements and their physical and chemical properties. Among chemical properties, he examined the compound formed by elements with oxygen and hydrogen. He found that if the 63 elements known at that time were arranged in the increasing order of their atomic masses, the properties of elements and also formulae of their oxides and hydrides gradually changed from element to element and at a certain interval they suddenly started almost repealing relationship was expressed by Mendeleev’s periodic law. i,e the properties of examinants are the periodic functions of their atomic masses.

Question 4.
Why do you think the noble gases are placed in a separate group?
Answer:
All noble gases are inert elements. Their properties are different from other elements and are the least reactive. Therefore, the noble gases are placed in a separate group.

Intext Questions Page No. 90

Question 1.
How could the Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:

  1. In the Modern periodic Table atomic number of an elements is a more fundamental property than its atomic mass.
  2. The anomalous position of hydrogen can be discussed after we see what are the basis on which the position of an elements in the Modern Periodic Table depends.
  3. The elements present in any one group have the same number of valence electrons.
  4. Atoms of different elements with the same number of occupied shells are placed in the same period.
  5. In the Modern Periodic Table, a zig-zag line separated metals from non-metals.

MP Board Solutions

A valence electron calculator is a tool that helps you find the number of valence electrons in an atom.

Question 2.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer:
Calcium (Ca) and Strontium (Sr) is expected to show chemical reactions similar to magnesium (Mg). This is because the number of valence electrons (2) is the same in all of these three elements and since chemical properties are due to valence electrons, they show the same chemical reactions.

Question 3.
Name:

  1. Three elements that have a single electron in their outermost shells.
  2. Two elements that have two electrons in their outermost shells.
  3. Three elements with filled outermost shells.

Answer:

  1. Lithium (Li), Sodium (Na), and Potassium (K) has a single electron in their outermost shells.
  2. Magnesium (Mg) and Calcium (Ca) have two electrons in their outermost shells.
  3. Neon (Ne), Argon (Ar), and Xenon (Xe) have filled outermost shells.

Question 4.
Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
Answer:

Lithium, sodium and potassium – These three elements have one electron in their outermost orbit.

b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
Both helium (He) and neon (Ne) have filled outermost shells. Helium has a duplet in its K shells, while neon has an octet in its L shells.

Question 5.
In the Modern Periodic Table, which are the metals among the first ten elements?
Answer:
In the modem periodic table, Lithium and Beryllium are the metals among the first 10 elements.

Question 6.
By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic?
Ga Ge As Se Be
Answer:
Since, ‘Be’ lies to the extreme left-hand side of the periodic table, ‘Be’ is the most metallic among the given elements.

MP Board Class 10th Science Chapter 5 Ncert Textbook Exercises

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of the periodic table.
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl2, which is solid with a high melting point. X would most likely be in the same group of the Periodic table as:
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3.
Which element has:
(a) two shells, both of which are completely filled with electrons?
(b) Electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:
(a) Neon
(b) Magnesium
(c) Silicon
(d) Boron
(e) Carbon

Question 4.
(a) What property do all elements in the same column of the Periodic table as Boron have in common?
(b) What property do all elements in the same column of the Periodic table as Fluorine have in common?
Answer:
(a) Valency equal to 3.
(b) Valency equal to 1.

MP Board Solutions

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)
N(7) F(9) P(15) Ar(18)
Answer:
(a) The atomic number of this element is 17.
(b) It would be chemically similar to F(9) with configuration as 2, 7.

Question 6.
The position of three elements A, B and C in the Periodic table is shown below:
MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 1
(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
(a) A is a non-metal.
(b) C is less reactive than A because reactivity decreases down the group in halogens.
(c) C should be smaller in size than B as moving across a period, the nuclear charge increases and therefore, electrons come closer to the nucleus.
(d) A will form an anion as it will accept an electron to complete its octet.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer:
Nitrogen (7) : 2, 5
Phosphorus (15) : 2, 8, 5
Nitrogen is more electronegative because Metallic character decreases across a period and increased down a group.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modern Periodic table?
Answer:
In the modern periodic table, atoms with similar electronic configurations are placed in the same column. In a group, the number of valence electrons remains the same. Elements across a period show an increase in the number of valence electrons.

Question 9.
In the Modern Periodic table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21, and 38. Which of these have physical and chemical properties resembling calcium?
Answer:
The element with atomic number 12 has the same chemical properties as that of calcium. This is because both of them have same number of valence electrons (2).

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s Periodic table and the Modern Periodic table.
Answer:

Mendeleev s Periodic table:

    1. Elements are arranged in the increasing order of their atomic mass.
    2. This table has 8 groups and 6 periods. And each group is subdivided as an A and B.
    3. In this table, Hydrogen has no position.
    4. No position for isotopes, because in Mendeleev period these are not discovered.

Modern Periodic table:

    1. Elements are arranged in the increasing order of their atomic number.
    2. It has 18 groups and 7 periods.
    3. Inert gases are placed in separate groups.
    4. In this table, a zigzag line separates Metals from Non-metals.

(or)

Mendeleev’s Periodic table vs Modern Periodic table:

  1. Elements are arranged in the increasing order of their atomic masses, while in Modern Periodic table elements are arranged in the increasing order of their atomic numbers.
  2. There are a total of 7 groups (columns) and 6 periods (rows) while in Mendeleev’s’ Periodic Table, there are a total of 18 groups (columns) and 7 periods (rows).
  3. Elements having similar properties were placed directly under one another, while in Mendeleev’s’ Periodic Table elements having the same number of valence electrons are present in the same group.
  4. In Mendeleev’s Periodic Table the position of hydrogen could not be explained, while in Modern Periodic table hydrogen is placed above alkali metals.
  5. No distinguishing positions for metals and non-metals in Mendeleev’s Periodic Table while in Modern Periodic Table metals are present at the left-hand side of the periodic table whereas nonmetals are present at the right-hand side.

MP Board Class 10th Science Chapter 5 Additional Questions

MP Board Class 10th Science Chapter 5 Multiple Choice Questions

Question 1.
Which of the following statements is correct about the trends when going down in a group of the periodic table?
(a) Elements become less electropositive in nature.
(b) Element oxides become more acidic.
(c) Valence electrons increases.
(d) Elements lose their electrons more easily.
Answer:
(d) Elements lose their electrons more easily.

Question 2.
Element A forms a chloride with the formula ACl3, which is a stable compound. A would most likely be the same group of the Periodic Table as –
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(c) Al

Question 3.
Which of the following are coin metals?
(a) Ne, Ca, Na
(b) H2, N2, O2
(c) Li, Na, K
(d) Cu, Au, Ag
Answer:
(d) Cu, Au, Ag

Question 4.
Who gave the triad arrangement of elements?
(a) Mendeleev
(b) Newlands
(c) Dalton
(d) Dobereiner
Answer:
(d) Dobereiner

Question 5.
Newlands periodic table is based on the
(a) Atomic weight
(b) Atomic number
(c) Atomic radius
(d) Atomic volume
Answer:
(a) Atomic weight

Question 6.
Which of the following is not gas in normal atmospheric condition?
(a) Helium (He)
(b) Argon (Ar)
(c) Bromine (Br)
(d) Chlorine (Cl)
Answer:
(c) Bromine (Br)

Question 7.
While moving left to right across a period, the atomic radii –
(a) Remains the same
(b) Approaches zero
(c) Decreases
(d) Increases first then decreases
Answer:
(c) Decreases

Question 8.
Which element is a metalloid?
(a) Carbon
(b) Nitrogen
(c) Oxygen
(d) Silicon
Answer:
(d) Silicon

MP Board Solutions

Question 9.
Moseley’s periodic table is based on
(a) Atomic mass
(b) Mass number
(c) Atomic number
(d) Atomic volume
Answer:
(c) Atomic number

Question 10.
Which of the following is a group of highly electronegative elements?
(a) Cl, Br, I
(b) S, Se, Te
(c) Na, K, Rb
(d) Ca, Sr, Ba
Answer:
(a) Cl, Br, I

Question 11.
Which of the following elements is a non-metal?
(a) Aluminium
(b) Chlorine
(c) Sodium
(d) Silicon
Answer:
(b) Chlorine

Question 12.
As we move down in a group in Modern Periodic Table, the size of elements generally
(a) increases
(b) decreases
(c) remain the same
(d) first, increase then decrease
Answer:
(a) increases

Question 13.
As we move from top to bottom in a group in Modern Periodic Table, the electronegativity of elements
(a) Increases
(b) Decreases
(c) No change
(d) Not certain
Answer:
(b) Decreases

Question 14.
Which group of elements is considered most electropositive?
(a) Group 1
(b) Group 2
(c) Group 17
(d) Group 18
Answer:
(a) Group 1

Question 15.
Group 1 elements are also called as:
(a) Alkali metals
(b) Alkaline earth metals
(c) Halogens
(d) Noble gases
Answer:
(a) Alkali metals

Question 16.
Group 17 elements are also called as:
(a) Alkali Metals
(b) Alkaline Earth Metals
(c) Halogens
(d) Noble Gases
Answer:
(c) Halogens

Question 17.
How many elements were known when Mendeleev started his work?
(a) 100
(b) 215
(c) 65
(d) 80
Answer:
(c) 65

Question 18.
Why Mendeleev left spaces in his Periodic Table?
(a) A mistake
(b) For future elements
(c) For Isotopes
(d) For Isobars
Answer:
(b) For future elements

Question 19.
Why Lanthanoids and Actinoids are placed below in the Periodic Table?
(a) A mistake
(b) Better representation and view
(c) They were found very recently
(d) All of the above
Answer:
(c) They were found very recently

Question 20.
A period may have elements with –
(a) Variable atomic sizes
(b) Variable atomic number
(c) Variable valency
(d) All of the above
Answer:
(d) All of the above

Question 21.
Element A belongs to group 15. The formula of its hydride will:
(a) AH
(b) AH2
(c) AH3
(d) A3H
Answer:
(c) AH3

Question 22.
An electropositive element, A with 2 valence electron will form which type of oxide?
(a) AO
(b) A2O
(c) AO2
(d) AO3
Answer:
(a) AO

Question 23.
Most electronegative element in our Periodic Table:
(a) Iron
(b) Nitrogen
(c) Carbon
(d) Flourine
Answer:
(d) Flourine

Question 24.
Which of the following elements where not among metals/elements named to fill the gap of Mendeleev’s Periodic Tablespaces?
(a) Cobalt
(b) Scandium
(c) Gallium
(d) Germanium
Answer:
(a) Cobalt

Question 25.
In a group, all elements have similar ………..
(a) Electronic configuration
(b) Valence electron
(c) Electronegativity
(d) All of these
Answer:
(b) Valence electron

MP Board Solutions

Question 26.
Which among the following is a noble gas?
(a) C
(b) N
(c) O
(d) Ne
Answer:
(d) Ne

Question 27.
Which of the following elements has electronic configuration E = 2, 6?
(a) C
(b) N
(c) O
(d) Ne
Answer:
(c) O

Question 28.
Which of the following elements is a metalloid?
(a) B
(b) Al
(c) S
(d) P
Answer:
(a) B

Question 29.
Which of the following is not a halogen?
(a) Br
(b) I
(c) Te
(d) At
Answer:
(c) Te

Question 30.
Which element has a total of two shells, with four valence electrons?
(a) C
(b) N
(c) Br
(d) Co
Answer:
(a) C
(ii) Match column A’s description with column B’s Particulars.
MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 2
Answers:

  1. → 12
  2. → 3
  3. → 2
  4. → 4
  5. → 5
  6. → 13
  7. → 11
  8. → 14
  9. → 10
  10. → 10
  11. → 6
  12. → 1
  13. → 7
  14. → 8

MP Board Class 10th Science Chapter 5 Very Short Answer Type Questions

Question 1.
What would be the maximum number of electrons present in the outermost shell of atoms in the first period of Periodic Table?
Answer:
Two

Question 2.
What is Solder?
Answer:
It is an alloy of lead (Pb) and tin (Sn).

Question 3.
What is anode mud?
Answer:
During electrolytic refining, the soluble impurities go into the solution, Whereas, the insoluble impurities settle down at the bottom of anode and are known as anode mud.

Question 4.
Which metal is used with iron oxide to join railway tracks or cracked machine parts?
Answer:
Aluminium.

Question 5.
Give the thermit reaction.
Answer:
Fe2O3(s) + 2Al(s) → 2 Fe(l) + Al2O3(s) + Heat

Question 6.
Roasting is used for the extraction of which ore?
Answer:
Sulphide ore.

Question 7.
Name the metal lowest inactivity series (relative reactivities of metals).
Answer:
Au or Gold.

Question 8.
Which gas is evolved when a metal reacts with nitric acid?
Answer:
Hydrogen gas.

Question 9.
Name any two metal that does not react with water at all.
Answer:
Lead, copper, gold, silver. (any two)

Question 10.
Complete the following reaction: Metal oxide + water →.
Answer:
Metal hydroxide.

Question 11.
Which material is used to coat electric wires in homes?
Answer:
PVC or Polyvinylchloride.

MP Board Solutions

Question 12.
Name any two metals that are poor conductors of heat.
Answer:
Lead and mercury.

MP Board Class 10th Science Chapter 5 Short Answer Type Questions

Question 1.
What are the limitations of the Modern Periodic Table?
Answer:
The limitations of the Modern Periodic Table:
Position of hydrogen still dicey. It is not fixed till now. Position of lanthanides and actinides has not been given inside the main body of the Periodic Table. It does not reflect the exact distribution of electrons of some of the transition and inner transition elements.

Question 2.
Two elements X and Y have atomic numbers 12 and 16 respectively. Write the electronic configuration for these elements. To which period of the Modern Periodic Table do these two elements belong? What type of bond will be formed between them and why?
Answer:
Electronic configuration of X (Z= 12): 2, 8,2
Electronic configuration of Y (Z = 16): 2, 8,6
Both these elements belong to the third period. An ionic bond is formed between X and Y due to the transfer of two electrons from X to Y.

Question 3.
The present classification of elements is based on which fundamental property of elements?
Answer:
Atomic number.

Question 4.
Li, Na and K are the elements of a Dobereiner’s Triad. If the atomic mass of Li is 7 and that of K is 39, what would be the atomic mass of Na?
Answer:
According to of Dobereiner’s law of triads, the atomic mass of the middle element, in this case, Na, should be the arithmetic mean of Li and K. Thus, Arithmetic mean of Li and K = (7 + 39)/2 = 23.

Question 5.
Define Dobereiner’s law of triads.
Answer:
It states, “When elements are placed in order of the ascending order of atomic masses, groups of three elements having similar properties are obtained. The atomic mass of the middle element of the triad is equal to the mean of the atomic masses of the other two elements of the triad.”

Question 6.
Why did Dobereiner’s system of classification fail?
Answer:
The major drawback of Döbereiner’s classification was that it was valid only for a few groups of elements known during that time. He was able to identify three triads only. Also, more accurate measurements of atomic masses showed that the mid element of the triad did not really have the mean value of the sum of the other two elements of the triad. For elements of very low mass or very high mass, the law did not hold good. For example, Fluorine (F), Chlorine (Cl), Bromine (Br). The atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br.

Question 7.
Explain the position of metalloids in the Modern Periodic Table.
Answer:
In the Modern Periodic Table, a zig-zag line separates metals from non-metals. The borderline elements – boron, silicon, germanium, arsenic, antimony, tellurium and polonium – are intermediate in properties and are called metalloids or semi-metals.

Question 8.
Why silicon is classified as metalloid?
Answer:
Silicon is classified as a semi-metal or metalloid because it exhibits some properties of both metals and non-metals.

Question 9.
State Newlands law of octaves.
Answer;
Elements are arranged in increasing order of their atomic masses such that the properties of the eighth element are the repetition of the properties of the first element (similar to eighth note in an octave of music).

Question 10.
X and Y are the two elements having similar properties which obey Newlands law of octaves. How many elements are there in between X and Y?
Answer:
The law states there are eight elements in an octave (row). A number of elements between X and Y are six.

Question 11.
What are the drawbacks of Newlands law of octaves?
Answer:
Following are the major drawbacks:

  1. Worked well with lighter elements (upto calcium. After those elements in the eighth column did not possess properties similar to elements in the first column.
  2. Newland assumed only 56 elements existed so far. Later, new elements were discovered which did not fit into octaves table.
  3. Newland adjusted few elements in the same slot through their properties were quite different, e.g., Cobalt and nickel are in the same slot and these are placed in the same column as fluorine, chlorine and bromine which have very different properties than these elements. Iron, which resembles cobalt and nickel in properties, has been placed far away from these elements.

MP Board Class 10th Science Chapter 5 Long Answer Type Questions

Question 1.
What are the salient features of the Modern Periodic Table?
Answer:
In a period of the Periodic Table, the number of valence electrons increases as the atomic number increases. As a result, elements change from metal to metalloid to nonmetal to a noble gas. Atomic size is a periodic property. As atomic number increases in a period, the atomic radius decreases. As atomic number increases in a group, atomic radius increases.

Positive ions have smaller atomic radii than the neutral atoms from which they derive. Negative ions have larger atomic radii than their neutral atoms. Positive ions in the same group increase in size down the group. In a group, each element has the same number of valence electrons. As a result, the elements in a group show similar chemical behaviour.

Metallic character decreases from left to right in a period because of the increase in the effective nuclear charge. Non-metallic character increase from left to right in a period because of the increase in effective nuclear charge. Non-metallic character decreases down the group because of increase in the size of the atom.

MP Board Solutions

Question 2.
What periodic trends do we observe in terms of atomic radii or atomic sizes in Modern Periodic table?
Answer:
Following two trends are observed:
1. Within each column (group), atomic radius tends to increase from top to bottom. This trend results primarily from the increase in the number of the outer electrons. As we go down a column, the outer electrons have a greater probability of being farther from the nucleus, causing the atom to increase in size.

2. Within each row (period), the atomic radius tends to decrease from left to right. The major factor influencing this trend is the increase in the nuclear charge as we move across a row. The increasing effective nuclear charge steadily draws the valence electrons closer to the nucleus, causing the atomic radius to decrease.

Question 3.
An element A with atomic number 19 combines separately with NO3and (SO4)2,(PO4)3radicals:
(a) Give the electronic configuration of element A.
(b) Write the formulae of the three compounds so formed.
(c) To which group of the periodic table does the element ‘R’ belong?
(d) Does it form covalent or ionic compound? Why?
Answer:
(a) Electronic configuration of A: 2,8, 8, 1.
(b) Compounds formed are A(NO3), A2(SO4) and K3(PO4).
(c) A has one valence electron and hence, it belongs to the first group.
(d) It forms the ionic compound.

Question 4.
Describe types of periods, blocks and trends of periodic properties along periods associated with Modern Periodic Table.
Answer:
Periods:
First period (Atomic number 1 and 2): This is the shortest period. It contains only two elements (hydrogen and helium).

Second Period: (Atomic number 3 to 10): It contains eight elements (lithium to neon).

Third period (Atomic number 11 to 18): It contains eight elements (sodium to argon).

Fourth period (Atomic number 19 to 36): Row contains eighteen elements (potassium to krypton). i.e., 8 normal elements and 10 transition elements.

5th period (Atomic number 37 to 54): Contains 18 elements (rubidium to xenon) includes 8 normal elements and 10 transition elements.

Sixth period (Atomic number 55 to 86): The longest period. It contains 32 elements (caesium to radon) has 8 normal elements, 10 transition elements and 14 inner transition elements (lanthanides).

7th period (Atomic number 87 to 118): As like the sixth period, this period also can accommodate 32 elements. Till now 26 elements have been authenticated by IUPAC.

Blocks in Periodic Table:
The periodic table includes “blocks” defined in terms of which type of orbital is being filled via the Aufbau principle. This gives us the s-block, p-block, d-block, and f-block.

Blocks:
The s-, p-, d-, and f-blocks contain elements with outer electrons in the same type of orbital. Another key link between electron arrangement and position in the periodic table is that elements in any one main group have the same number of electrons in their highest energy level. The number of elements discovered so far is 118. The last element authenticated by IUPAC is Cn112 (Copernicium).

Properties of Periods: As you proceed to the left in a period or as you proceed down within a group:

  1. The metallic strength increases (Non-Metallic Strength decreases).
  2. The atomic radius increases.
  3. The ionization potential decreases.
  4. The electron affinity decreases.
  5. The electronegativity decreases.

MP Board Class 10th Science Chapter 5 Textbook Activities

Class 10 Science Activity 5.1 Page No. 84

  1. Looking at its resemblance to alkali metals and the halogen family, try to assign hydrogen a correct position in Mendeleev’s Period Table.
  2. To which group and period should hydrogen be assigned?

Observations:

  1. No position can be fixed for hydrogen in the Mendeleev’s Periodic Table.
  2. Properties of hydrogen fit with alkali metal as it combines with halogens, oxygen and sulphur to form compounds.
  3. Properties of hydrogen also fit or are similar to halogen as it exists in the form of diatomic molecules and combines with metals and non-metals forming covalent compounds.

Class 10 Science Activity 5.2 Page No. 85

  1. Consider the isotopes of chlorine, Cl-35 and CI-37.
  2. Would you place them in different slots because their atomic masses are different?
  3. Or would you place them in the same position because their chemical properties are the same?

Observations:
Two isotopes of chlorine are Cl-35 and Cl-37. Both isotopes have the same chemical properties and hence, both isotopes should be placed in the same position.

Class 10 Science Activity 5.3 Page No. 85

  1. How were the positions of cobalt and nickel resolved in the Modern Periodic Table?
  2. How were the positions of isotopes of various elements decided in the Modern Periodic Table?
  3. Is it possible to have an element with atomic number 1, 5 placed between hydrogen and helium?
  4. Where do you think should hydrogen be placed in the Modern Periodic Table?

Observations:
The position of Cobalt and Nickel were decided by placing them in the increasing order of atomic number in the Modern Periodic Table. Since isotopes are elements with the similar atomic number they are placed in the same position as its basic elements in the modern periodic table.

Class 10 Science Activity 5.4 Page No. 87

  1. Look at group 1 of the Modern Periodic Table, and name the elements present in it. Write down the electronic configuration of the first three elements of group 1.
  2. What similarity do you find in their electronic configurations?
  3. How many valence electrons are present in these three elements?

Observations:
Elements present in Group 1 are:
MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 3
Electronic configuration of the first three elements of Group I are as below:
MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 4

Class 10 Science Activity 5.6 Page No. 87

  1. If you look at the Modern Periodic Table, you will find that the elements Li, Be, B, C, N, O, F, and Ne are present in the second period. Write down their electronic configurations.
  2. Do these elements also contain the same number of valence electrons?
  3. Do they contain the same number of shells?

Observations:
No, these elements contain variable valence electrons as they belong to different groups:
MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 5

Class 10 Science Activity 5.6 Page No. 88

  1. How do you calculate the valency of an element from its electronic configuration?
  2. What is the valency of magnesium with atomic number 12 and sulphur with atomic number 16?
  3. Similarly, find out the valencies of the first twenty elements.
  4. How does the valency vary in a period on going from left to right?
  5. How does the valency vary in going down a group?

Observations:

  1. Valency of an element can be calculated by the numbers of valence electron present.
  2. Valency of Magnesium: 2
  3. Valency of Sulphur: 2 Variation of valency while moving left to right in a period.
    1 → 2 → 3 → 4 → 3 → 2 → 1 → 0
  4. Variation of valency while going down in a group does not change.

Class 10 Science Activity 5.7 Page No. 88

  1. Atomic radii of the elements of the second period are given below:
    MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 6
  2. Arrange them in decreasing order of their atomic radii.
  3. Are the elements now arranged in the pattern of a period in the Periodic Table?
  4. Which elements have the largest and the smallest atoms?
  5. How does the atomic radius change as you go from left to right in a period?

Observations:
Decreasing order of atomic radii of following elements:

  1. O < N < C < B < Be < Li
  2. No in pattern.
  3. Oxygen is smallest as per given data while Li is largest.
  4. Atomic radius reduces while moving right in a group a, nuclear charge increase.

Class 10 Science Activity 5.8 Page No. 89

  1. Study the variation in the atomic radii of first group elements given below and arrange them in increasing order.
    MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 7
  2. Name the elements which have the smallest and the largest atoms.
  3. How does the atomic size vary as you go down a group?

MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 8

  1. Sodium (Na) has the smallest atom and calcium (Ca) has the largest atom.
  2. Atomic size increases as we go down a group.

Class 10 Science Activity 5.9 Page No. 89

  1. Examine elements of the third period and classify them as metals and non-metals.
  2. On which side of the Periodic Table do you find the metals?
  3. On which side of the Periodic Table do you find the non-metals?

Observations:
Elements of the third period are:
MP Board Class 10th Science Solutions Chapter 5 Periodic Classification of Elements 9

Class 10 Science Activity 5.10 Page No. 89

  1. How do you think the tendency to lose electrons change in a group?
  2. How will this tendency change in a period?

Observations:
Metallic property reduces while moving right in a period.

Class 10 Science Activity 5.11 Page No. 90

  1. How would the tendency to gain electrons change as you go from left to right across a period? How
  2. would the tendency to gain electrons change as you go down a group?

Observations:

  1. The electrons increases as we go left to right in a period up to 17th group. It decreases in the 18th
  2. group. The tendency of gaining the electrons decreases as we go down a group.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts

In this article, we will share MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts

MP Board Class 10th Science Chapter 2 Intext Questions

Intext Questions Page No. 18

Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Answer:

  • Put the red litmus paper turn by turn in each of the three test tubes. The solution which turns the red litmus paper to blue will be a basic solution here, the blue litmus paper formed can now be used to test the acidic solution.
  • Put this blue litmus paper in the remaining two test tubes one by one. The solution which turns the blue litmus paper to red will be the acidic solution.
  • The solution which has no effect on any litmus paper will be neutral and hence it will be distilled water.

Intext Questions Page No. 22

Question 1.
Why should curd and sour substances not be kept in brass and copper vessels?
Answer:
Curd and other sour substances are acidic in nature. So, when they are kept in brass and copper vessels, harmful products along with hydrogen gas are produced which spoil the food.

MP Board Solutions

Question 2.
Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
Answer:
Hydrogen gas is usually liberated when an acid reacts with a metal. Let us illustrate it with the following examples:

  1. Add some pieces of zinc granules into 5ml of a dilute solution of sulphuric acid (H2SO4).
  2. Shake it well.
  3. Pass the produced gas into a soap solution.
  4. Now, soap bubbles are formed in the soap solution and these soap bubbles contain hydrogen.
  5. Bring a burning candle near a gas-filled bubble. A candle burns with a pop sound. So, the following reaction takes place:
    H2SO4(aq) + Zn(s) → ZnSO4(aq) + H2
    MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 1
  6. We can test the evolved hydrogen gas by its burning with a pop sound when a candle is brought near the soap bubbles.

pH Calculator is a free online tool that displays the pH value for the given chemical solution.

Question 3.
Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 2

Intext Questions Page No. 25

Question 1.
Why do HCl, HNO3 etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
HCl or HNO3 dissolve in water to form H+ or H3O+ ions in aqueous solutions which show their acidic character. The following reactions take place when HCl or HNO3 are mixed with water:
HCl(aq) → H+ + Cl
H+ + H2O → H3O+
On the other hand, when alcohol and glucose are mixed with water they do not dissolve to form ions due to the presence of hydrogen bonds and basic character. Hence, they do not show acidic character.

Question 2.
Why does an aqueous solution of acid conduct electricity?
Answer:
In the aqueous solution, acid forms ions and these ions are conductor of electricity.

Question 3.
Why does dry HCl gas not change the colour of the dry litmus paper?
Answer:
Dry HCl gas does not change the colour of the dry litmus paper because it has no hydrogen ions (H+) or hydronium (H3O+) ions in it.

MP Board Solutions

Question 4.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
Answer:
Mixing water to acid is an exothermic reaction. Hence while diluting an acid it is recommended that the acid should be added to water and not water to acid. If we mix water to acid explosion occurs and burning take place.

Question 5.
How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted?
Answer:
Concentration of hydronium ions (H3O+) decreases and becomes weak. In this way concentration of hydronium ion affects when a solution of acid is diluted.

Question 6.
How is the concentration of hydroxide ions (OH) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When excess base is dissolved in a solution of sodium hydroxide concentration of OH Hydroxide ion is more.
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 3

Intext Questions Page No. 28

Question 1.
You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:
Solution ‘A’ is acidic because pH of the solution A is 6 which is less than 7 while solution ‘B’ is basic because pH of the solution ‘B’ is 8 which is greater than 7. Solution ‘A’ has more hydrogen ion concentration in comparison to solution ‘B’ because solution ‘A’ is acidic.

If the pH value is less than 7, it represents an acidic solution.
If the pH value is more than 7, it represents a base.
It the pH – 6 is acidic it has more concentration of ions than pH-8 which is a base.

Question 2.
What effect does the concentration of H+(aq) ions have on the nature of the solution?
Answer:
If the concentration of H+(aq) ions is increased then the solution becomes acidic and if the concentration of H+(aq) ions is decreased then the solution becomes basic in nature.

Question 3.
Do basic solutions also have H+(aq) ions? If yes, then why are these basic?
Answer:
Yes. H+ ions are always present in basic solution. Concentration of Basic is more than OH ions.

Question 4.
Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
Answer:
If the soil is acidic and improper for cultivation, then to increase the basicity of soil, the farmer would treat the soil with quick lime or slaked lime or chalk.

Intext Questions Page No. 33

Question 1.
What is the common name of the compound CaOCl2?
Answer:
Bleaching powder.

Question 2.
Name the substance which on treatment with chlorine yields bleaching powder.
Answer:
Dry slaked lime or calcium hydroxide.

Question 3.
Name the sodium compound which is used for softening hard water.
Answer:
Sodium carbonate.
Na2CO3, 10 H2O is the compound of sodium to soften hard water.

MP Board Solutions

Question 4.
What will happen if a solution of sodium hydrocarbon is heated? Give the equation of the reaction involved.
Answer:
When sodium hydrocarbon is heated then sodium carbonate and water are formed along with the evolution of carbon dioxide gas.
2NaHCO3 → Na2CO3 + H2O + CO2

Question 5.
Write an equation to show the reaction between Plaster of Paris and water.
Answer:
Plaster of Paris reacts with water to form gypsum.
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 4

MP Board Class 10th Science Chapter 2 Ncert Textbook Exercises

Question 1.
A solution turns red litmus blue, its pH is likely to be
(a) 1
(b) 4
(c) 5
(d) 10
Answer:
(d) 10
Bases turn red litmus to blue. pH value of 7 is greater than 7. Hence this solution changes red litmus to blue.

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains
(a) NaCl
(b) HCl
(c) LiCl
(d) KCl
Answer:
b) the solution contains HCl.

Question 3.
10 ml of a solution of NaOH is found to be completely neutralised by 8 ml of a given solution of HCl. If we take 20 ml of the same solution of NaOH, the amount of HCl solution (the same solution as before) required to neutralise it will be
(a) 4 ml
(b) 8 ml
(c) 12 ml
(d) 16 ml
Answer:
d) 16 mL HCl solution is required.

Question 4.
Which one of the following types of medicines are used for treating indigestion?
(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic
Answer:
c) Antacid is used to treat indigestion.

Question 5.
Write word equations and then balanced equations for the reaction taking place when:
(a) dilute sulphuric acid reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron filings.
Answer:
a) Sulphuric acid + Zinc ➝ zinc Sulphate + Hydrogen
H2SO(aq) + Zn(s) ➝ ZnSO4(aq) + H2(g)
b) hydro Chloric acid + magnesium ➝ Magnisium Chloride + Hydrogen
2HCl(aq) + Mg(s) ➝ MgCl2(aq) + H2(g)
(c) Sulphuric Hydrigen Sulphate + Aluminium ➝ Aluminium + Hydrogen Chloride
3H2SO2(aq) + Mg(s) ➝ MgCl2(aq) + H2(g)
d) Hydrochloric acid  + Iron ➝ Ferric + Hydrogen
6HCl(aq) + 2Fe(s) ➝ 2FeCl2(aq) + 3H2(g)

Question 6.
Compounds such as alcohols and glucose also contain hydrogen but are not categorised as acids. Describe an activity to prove it.
Answer:
Experiment: Fix two nails on the cork and keep this in 100 ml beaker. Two nails are fixed to 6 volt battery, bulb and switch. Then pour some dilute HCl in the beaker and switch on the current. Repeat the experiment separately with glucose and alcohol solutions.
Observation: Bulb glows in HCl solution but do not glows in glucose solution.
Result: HCl ➝ H+ and Cl- ions.
These ions conduct electricity and bulb glows.
By this experiment we conclude that All acids contain Hydrogen.

Question 7.
Why does distilled water not conduct electricity, whereas rainwater does?
Answer:
Distilled water cannot conduct electricity because it does not contain ions while rainwater conducts electricity as it contains ions due to the presence of dissolved salts in it.

Question 8.
Why do acids not show acidic behavior in the absence of water?
Answer:
Acids do not show acidic property in the absence of water. Because Hydrogen ions dissociates in presence of water. Hydrogen ions are responsible for acidic nature.

Question 9.
Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9, respectively. Which solution is
(a) Neutral?
(b) Strongly alkaline?
(c) Strongly acidic?
(d) Weakly acidic?
(e) Weakly alkaline?
Arrange the pH in increasing order of hydrogen-ion concentration.
Answer:
a) Neutral ➝ solution D ➝ pH value of pH is 7.
b) strongly alkaline ➝ solution C ➝ pH is 11
c) strongly acidic ➝ solution B ➝ pH is 1
d) weakly acidic ➝ solution A ➝ pH is 4
e) weakly alkaline ➝ solution E —> pH is 9
We can arrange the pH in increasing order of hydrogen ion concentration as 11 < 9 <7 <4<1

MP Board Solutions

Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH3COOH) is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing
occur more vigorously and why?
Answer:
The fizzing will occur strongly in test tube A, in which hydrochloric acid (HCl) is added. This is because HCl is a stronger acid than CH3COOH and therefore, produces hydrogen gas at a faster speed due to which fizzing occurs.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.
Answer:
pH value of fresh milk is 6, but when it converts into curd value of pH decreases because curd is acidic. Hence this value is becoming less.

Question 12.
A milkman adds a very small amount of baking soda to fresh milk.

  1. Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
  2. Why does this milk take a long time to set as curd?

Answer:

  1. The milkman shifts the pH of the fresh milk from 6 to slightly alkaline because, in alkaline condition, milk does not set as curd easily. Hence, it does not get spoiled for longer period of time, in which he can sell it to make a profit.
  2. Since this milk is slightly basic than usual milk, acids produced to set the curd are neutralised by the base. Therefore, it takes a longer time for the curd to set which is usually acidic.

Question 13.
Plaster of Paris should be stored in a moisture-proof container. Explain why?
Answer:
The Plaster of Paris should be stored in a moisture-proof container as it absorbs water from moisture and turns into a hard substance (Gypsum) as shown in the following chemical equation:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 5

Question 14.
What is a neutralisation reaction? Give two examples.
Answer:
The reaction between an acid a base to give salt and water is known as a neutralisation reaction.
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 6

Question 15.
Give two important uses of washing soda and baking soda.
Answer:
1. Washing Soda:
(a) This is used in glass, soap and paper industries,
(b) It is used for removing permanent hardness of water.

2. Baking Soda:
(a) This is used cooking mixture of Baking soda and acid (tartaric acid – weak acid) is called Baking powder. When it is heated or combined with water. CO2 is evolved and soften the bread.
(b) It is also used in soda-acid fire extinguishers.

MP Board Class 10th Science Chapter 2 Additional Important Questions

MP Board Class 10th Science Chapter 2 Multiple Choice Questions

Question 1.
The range of a pH scale is:
(a) 1 – 10
(b) 1 – 100
(c) 0 – 14
(d) 1 – 14
Answer:
(c) 0 – 14

Question 2.
pH is defined as:
(a) The logarithm of hydrogen ion concentration
(b) The negative logarithm of hydrogen ion concentration
(c) Hydrogen ion concentration
(d) None of the above
Answer:
(a) The logarithm of hydrogen ion concentration

MP Board Solutions

Question 3.
Which of the following solution will have pH = 7?
(a) Tea
(b) The salt solution in distilled water
(c) Hydrochloric acid solution
(d) Water distilled with chlorine gas.
Answer:
(b) The salt solution in distilled water

Question 4.
Which colour indicate neutral solution on a pH paper?
(a) Brown
(b) Green
(c) Purple
(d) White or transparent
Answer:
(b) Green

Question 5.
A solution is acidic if:
(a) it releases H+ ions in the solution.
(b) it has a pH of less than 7.
(c) it has dark red, orange or greenish-yellow colour on a pH paper.
(d) all of the above.
Answer:
(d) all of the above.

Question 6.
Which of the following solutions will have pH < 7?
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 7
Choose correct combination:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) All of the above.
Answer:
(b) (ii) and (iii)

Question 7.
“p” in pH stands for:
(a) Phosphorus
(b) Potenz
(c) Potential
(d) Polarity
Answer:
(b) Potenz

Question 8.
Four different jars A, B, C, D contains hydrochloric acid, black coffee, ammonia and soap solution. Choose the order of decreasing acidic strength.
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 8
(a) A > B > C > D
(b) B > C > A > D
(c) D > C > B > A
(d) D = C > B > A
Answer:
(a) A > B > C > D

Question 9.
How can we find the pH of a solution?
(a) By dipping pH paper in it.
(b) By dropping some solution over pH paper.
(c) By heating pH paper in vapours of solution.
(d) By pouring all the solution over pH paper.
Answer:
(a) By dipping pH paper in it.

Question 10.
In a class, while doing practical on different solutions, four students give their observations. Which student/s has/ve given a correct explanation if their observation is as follows?
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 9
Choose the correct combination of students with wrong observations:
(a) A and B
(b) C and D
(c) A, B, C
(d) All students
Answer:
(b) C and D

Question 11.
Lemon juice gives orange colour over pH paper. What is its nature?
(a) Strong Acid
(b) Basic
(c) Neutral
(d) Moderate acid
Answer:
(a) Strong Acid

Question 12.
Pure water has a pH = 7, while distilled water has pH = 8 – 10, it represents that distilled water is:
(a) Slightly basic
(b) Strong base
(c) Mild acid
(d) Strong acid
Answer:
(a) Slightly basic

Question 13.
On a pH paper, pH values of 1 and 8 are represented by colours:
(a) Yellow and Orange
(b) Purple and Greenish
(c) Red and Bluish Green
(d) Green and Red
Answer:
(c) Red and Bluish Green

Question 14.
Which of the following acids gives a dark red colour?
(a) Lemon juice
(b) Hydrochloric acid
(c) Acetic acid
(d) Nitric acid
Answer:
(b) Hydrochloric acid

Question 15.
How pOH can by represented?
(a) -Log [H+] = pOH
(b) -Log [H] = pOH
(c) -Log [pH] = pOH
(d) -Log [OH] = pOH
Answer:
(d) -Log [OH] = pOH

Question 16.
If [H+] is 1.0 × 10-9 mole, what will be pH of solution?
(a) 10-9
(b) 1
(c) 9
(d) -9
Answer:
(c) 9

Question 17.
What is the nature of citric acid?
(a) Basic
(b) Acidic
(c) Neutral
(d) Both a and b
Answer:
(b) Acidic

Question 18.
What is the nature of Sodium Hydroxide?
(a) Basic
(b) Acidic
(c) Neutral
(d) Unpredictable
Answer:
(a) Basic

Question 19.
pH > 7 represents?
(a) Basic solutions
(b) Acidic solutions
(c) Neutral solution
(d) All
Answer:
(a) Basic solutions

Question 20.
pH < 7 represents?
(a) Basic solutions
(b) Acidic solutions
(c) Neutral solution
(d) All
Answer:
(b) Acidic solutions

Question 21.
Hydrogen ion concentration for pure water is:
(a) 7
(b) 10-7
(c) 10-7 mole/litre
(d) 107 moles/litre
Answer:
(c) 10-7 mole/litre

MP Board Class 10th Science Chapter 2 Very Short Answer Type Questions

Question 1.
How can pH be represented using log?
Answer:
pH = – log [H+].

Question 2.
What is the range of pH on pH paper?
Answer:
0 – 14.

Question 3.
What is the pH of a strong acid?
Answer:
0 – 2.

Question 4.
Why is water neutral?
Answer:
On dissociation, water has equal numbers of H+ and OH ions. So, it does not go through any change and remains neutral.

Question 5.
What is the pH of concentric HCl?
Answer:
1 – 2.

Question 6.
What will be the colour of pH paper when coffee is poured over it?
Answer:
Reddish as it is slightly acidic in nature.

Question 7.
Which is more acidic-lemon juice or baking powder?
Answer:
Lemon juice.

Question 8.
What is the best medium to check any chemical’s nature?
Answer:
Water.

Question 9.
What is the universal solvent?
Answer:
Aqua regia.

Question 10.
What kind of reactions are neutralisation reactions?
Answer:
Any reaction between an acid and a base to form salt and water is called neutralisation reaction.

Question 11.
Which solution is considered to be neutral?
Answer:
Solutions with no acidity or alkalinity are neutral. Acids and bases are present in equal amounts.

MP Board Solutions

Question 12.
Write the formula of brine and bleaching powder.
Answer:
NaCl and CaOCl2.

Question 13.
Name two products which we can be obtained by chemical processing of common salt.
Answer:
Baking soda and bleaching powder.

Question 14.
What is the common name of sodium hydrogen carbonate?
Answer:
Baking soda.

Question 15.
Write the formula for washing soda.
Answer:
Na2CO3.10H2O.

Question 16.
What is the source of naturally occurring acid lactic acid?
Answer:
Curd.

Question 17.
Which acid is present in tomato?
Answer:
Oxalic acid.

Question 18.
Name two olfactory indicators.
Answer:
Vanilla and clove.

Question 19.
What is the colour of methyl orange in acidic solution?
Answer:
Red.

Question 20.
What is the colour of phenolphthalein in basic medium?
Answer:
Pink.

MP Board Class 10th Science Chapter 2 Short Answer Type Questions

Question 1.
How acids are different from bases when dissolved in water?
Answer:
Acids on dissolving in water produce H+ ions while the base produces OHions.

Question 2.
Name two indicators and write their colour in different mediums.

Name of indicator Colour in acidic medium Colour in basic medium
Methyl Orange Red Yellow Pink
Phenolphthalein Colourless Pink

Question 3.
Name any three hydrated salts.
Answer:

  1. Barium chloride, BaCl2. 2H2O.
  2. Copper sulphate, CuSO4. 5H2O.
  3. Ferrous sulphate, FeSO4. 6H2O.

Question 4.
Give an equation of neutralisation reaction.
Answer:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 10

Question 5.
What causes acidity in our body? How can it be cured?
Answer:
Our stomach produces hydrochloric acid which helps in digestion of food. During indigestion, the stomach produces too much acid and this causes pain and irritation. This can be cured by using bases called antacids.

Question 6.
What is the result of the reaction between an acid and a metal?
Answer:
Corresponding salt is formed with the evolution of hydrogen gas when a metal reacts with acid.

Question 7.
Write two important uses of pH in everyday life. Also, give an example.
Answer:
pH balance and its particular range of maintenance are very important in nature because it affects animal and plant life very much.

For example:

  1. Curd formation: Atmospheric bacteria change the pH of milk which causes the curd formation.
  2. Aids in digestion: Slight acidic conditions in the stomach due to the presence of hydrochloric acid aids in the digestion of food.

Question 8.
Give examples of two acids and bases present in nature.
Answer:

  1. Acids: Citric acid, acetic acid.
  2. Bases: Calcium carbonate, sodium hydroxide.

Question 9.
Discuss the various types of salts.
Answer:
There are three types of salts:

  1. Neutral salts: Salts formed by the mixing of strong acid and strong base, e.g., NaCl, K2SO4 etc.
  2. Acidic salts: Salts formed by the mixing of a strong acid and weak base e.g., NH4Cl, CaSO4.
  3. Basic salts: Salts formed by the mixing of a strong base and weak acid e.g., Na2CO3, CH3COONa etc.

Question 10.
Common salt acts as raw material for many important daily use chemicals. Name some of them and also write their chemical formula.
Answer:
Sodium hydroxide (NaOH), Baking Soda (NaHCO3), Washing soda (Na2CO3. 10H2O) etc.

Question 11.
What are the products of Chlor-alkali process?
Answer:
The products of Chlor-alkali process are chlorine and sodium hydroxide.

Question 12.
Name two uses of each of the given salts:

  1. Bleaching powder
  2. Baking soda

Answer:
Use of given salts are:
1. Bleaching powder:

  • It is used as an oxidising agent in chemical industries.
  • It is used for disinfecting water to make it free of germs.

2. Baking soda:

  • It is used in soda acid fire extinguishers.
  • It is an ingredient in antacids.

Question 13.
What is the water of crystallisation?
Answer:
The water of crystallisation is the fixed number of water molecules present in one formula unit of salt.

Question 14.
Give reaction to show the formation of sodium zincate?
Answer:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 25

Question 15.
Name the products of electrolysis of brine and also give on use of each.
Answer:
Chlorine gas, H2 gas, and sodium hydroxide are the products of electrolysis of brine:

  1. Use of chlorine gas: It is used as a disinfectant.
  2. Use of H2 gas: It is used in the manufacture of ammonia.
  3. Use of sodium hydroxide: It is used for the manufacture of soaps and detergents.

MP Board Class 10th Science Chapter 2 Long Answer Type Questions

Question 1.
Discuss the nature of the solution and its type in brief and also explain the strength of a solution.
Answer:
Nature of the solution:
When a solute is dissolved solvent (generally water), it shows different kind of nature with regard to its reactivity and solubility. On the basis of removal of H+ ion or OH the solution is formed. The types of solutions are divided as follows:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 11
Strength of solution is determined by the,

  1. Speed of reactivity i.e., how fast the ions are found dissociated.
  2. Amount of ions (H+ or OH ) released or their ion concentration.

Question 2.
Give one example in each case:
(a) a weak mineral acid.
(b) a base which is not an alkali.
(c) a hydrogen-containing compound which is not an acid.
(d) a basic oxide soluble in water.
(e) a basic oxide insoluble in water.
Answer:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 12

MP Board Class 10th Science Chapter 2 Textbook Activities

Class 10 Science Activity 2.1 Page No. 18

  1. Collect the following solutions from the science laboratory – hydrochloric acid (HCl), sulphuric acid (H2SO4), nitric acid (HNO3), acetic acid (CH3COOH), sodium hydroxide (NaOH), calcium hydroxide (Ca(OH)2], potassium hydroxide (KOH), magnesium hydroxide [Mg(OH)2] and ammonium hydroxide (NH4OH).
  2. Put a drop of each of the above solutions on a watch-glass one by one and test with a drop of the indicators shown in the table.
  3. What change in colour did you observe with red litmus, blue litmus, phenolphthalein and methyl orange solutions for each of the solutions taken?
  4. Tabulate your observations in the table.

Result:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 13

Class 10 Science Activity 2.2 Page No. 18,19

  1. Take some finely chopped onions in a plastic bag along with some strips of clean cloth. Tie up the bag tightly and leave overnight in the fridge. The cloth strips can now be used to test for acids and bases.
  2. Take two of these cloth strips and check their odour.
  3. Keep them on a clean surface and put a few drops of dilute HCl solution on one strip and a few drops of dilute NaOH solution on the other.
  4. Rinse both cloth strips with water and again check their odour.
  5. Note your observations.
  6. Now take some dilute vanilla essence and clove oil and check their odour.
  7. Take some dilute HCl solution in one test tube and dilute NaOH solution in another. Add a few drops of dilute vanilla essence to both test tubes and shake well. Check the odour once again and record changes in odour, if any.
  8. Similarly, test the change in the odour of clove oil with dilute HCl and dilute NaOH solution and record your observations.

Observations:

  1. On putting the cloth strip into dilute.HCl solution, the red colour of cloth strip changes to pale red.
  2. On putting the cloth strip into NaOH solution, the red colour of strip changes to green in colour.

Odour test:

  1. Add vanilla essence in dilute NaOH.
  2. Add vanilla essence in dilute HCl.
  3. Add clove oil in to dilute HCl.
  4. Add clove oil in dilute NaOH.

Result:

  1. No smell found.
  2. The smell of vanilla exists.
  3. The smell of clove present.
  4. The smell of clove exists.

Class 10 Science Activity 2.3 Page No. 19,20

Caution:

  1. This activity needs the teacher’s assistance.
  2. Set the apparatus as shown in Figure.
  3. Take about 5 ml of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
  4. What do you observe on the surface of zinc granules?
  5. Pass the gas being evolved through the soap solution.
  6. Why are bubbles formed in the soap solution?
  7. Take a burning candle near a gas-filled bubble.
  8. What do you observe?
  9. Repeat this Activity with some more acids like HCl, HNO3 and CH3COOH.
  10. Are the observations in all the cases the same or different?

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 14

Observations:

  1. 5 ml of dilute sulphuric acid + zinc granules → On the surface of zinc granules + bubbles of hydrogen gas are formed.
  2. On passing the gas evolved through a soap solution → bubbles are formed due to low surface tension of soap solution. On taking a burning candle near a gas-filled bubbles → gas burns with a pop sound.
  3. All other acids HCl, HNO3 and CH3COOH show the same observation.

Class 10 Science Activity 2.4 Page No. 20

  1. Place a few pieces of granulated zinc metal in a test tube.
  2. Add 2 ml of sodium hydroxide solution and warm the contents of the test tube.
  3. This activity needs the teacher’s assistance.
  4. Set the apparatus as shown in Figure.
  5. Take about 5 ml of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
  6. What do you observe on the surface of zinc granules?
  7. Pass the gas being evolved through the soap solution.
  8. Why are bubbles formed in the soap solution?
  9. Take a burning candle near a gas-filled bubble.
  10. What do you observe?
  11. Repeat this Activity with some more acids like HCl, HNO3 and CH3COOH.
  12. Are the observations in all the cases the same or different?

Observations:
Reaction:
2NaOH + Zn → Na2ZnO3 + H2

Conclusion:
Bubbles of hydrogen gas are formed.

Class 10 Science Activity 2.5 Page No. 20

  1. Take two test tubes, label them as A and B.
  2. Take about 0.5 g of sodium carbonate (Na2CO3) in test tube A and about 0.5 g of sodium hydrogen carbonate (NaHCO3) in test tube B.
  3. Add about 2 ml of dilute HCl to both the test tubes.
  4. What do you observe?
  5. Pass the gas produced in each case through lime water (calcium hydroxide solution) as shown below and record your observations.

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 15

Observations:
Test Tube A:
Reaction:
Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) +CO2(g)

Test Tube B:
Reaction:
NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
On passing CO2 gas through lime water it turns milky because insoluble white precipitate of CaCO3 is formed as shown below:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 16
On passing excess gas through lime water, it becomes colourless.
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 17

Class 10 Science Activity 2.6 Page No. 21

  1. Take about 2 ml of dilute NaOH solution in a test tube and add two drops of phenolphthalein solution.
  2. What is the colour of the solution?
  3. Add dilute HCl solution to the above solution drop by drop.
  4. Is there any colour change for the reaction mixture?
  5. Why did the colour of phenolphthalein change after the addition of an acid?
  6. Now add a few drops of NaOH to the above mixture.
  7. Does the pink colour of phenolphthalein reappear?
  8. Why do you think this has happened?

Observations:

  1. Add 2 drops of phenolphthalein solution into NaOH solution → colour of the solution is pink.
  2. Now add dilute HCl solution drop by drop.
  3. Now, reaction mixture changes to colourless.
  4. This is because neutralisation of HCl and NaOH takes place.
  5. Now, add NaOH (few drops) → Pink colour reappears.

Class 10 Science Activity 2.7 Page No. 21

  1. Take a small amount of copper oxide in a beaker and add dilute hydrochloric acid slowly while stirring.
  2. Note the colour of the solution. What has happened to the copper oxide?

Observations:

  1. Add dilute HCl to copper oxide solution:
  2. Colour of the solution turns into green and CuCl2 dissolves.
  3. The blue-green colour formed due to formation of copper (II) chloride,
    CuO + 2HCl → CuCl + H2O

Class 10 Science Activity 2.8 Page No. 22

  1. Take solutions of glucose, alcohol, hydrochloric acid, sulphuric acid etc.
  2. Fix two nails on a cork and place the cork in a 100 ml beaker.
  3. Connect the nails to the two terminals of a 6-volt battery through a bulb and a switch, as shown in Figure.

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 18

  1. Now pour some dilute HCl in the beaker and switch on the current. Repeat with dilute sulphuric acid.
  2. What do you observe?
  3. Repeat the experiment separately with glucose and alcohol solutions.
  4. What do you observe now?
  5. Does the bulb glow in all cases?

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 19

Class 10 Science Activity 2.9 Page No. 23

  1. Take about 1g solid NaCl in a clean and dry test tube and set up the apparatus as shown in Figure.
  2. Add some concentrated sulphuric acid to the test tube.
  3. What do you observe? Is there a gas coming out of the delivery tube?
  4. Test the gas evolved successively with dry and wet blue litmus paper.
  5. In which case does the litmus paper change colour?
  6. On the basis of the above Activity, what do you infer about the acidic character of:
    • Dry HCl gas?
    • HCl solution?

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 20

Note to teachers:
If the climate is very humid, you will have to pass the gas produced through a guard tube (drying tube) containing calcium chloride to dry the gas.

Observations:

  1. Adding concentrated H2SO4 to test tube containing NaCl leads to the production of HCl gas. Now, testing this gas with litmus paper following were recorded,
  2. HCl gas passes → Colour change
  3. Dry litmus paper → No change
  4. Wet litmus paper → Blue litmus turns red

Thus, only HCl solution release H+ ions and acidic property exist due to H+ ions.

Class 10 Science Activity 2.10 Page No. 24

  1. Take 10 ml water in a beaker.
  2. Add a few drops of concentrated H2SO4 to it and swirl the beaker slowly.
  3. Touch the base of the beaker.
  4. Is there a change in temperature?
  5. Is this an exothermic or endothermic process?
  6. Repeat the above Activity with sodium hydroxide pellets and record your observations.

Observations:

  1. Add a few drops of concentrated H2SO4 to the water in a beaker → it becomes hot as the reaction is highly exothermic.
  2. Now, add NaOH pellets to water → the beaker becomes hot, the reaction is exothermic.

Class 10 Science Activity 2.11 Page No. 26

  1. Test the pH values of solutions given in the table.
  2. Record your observations.
  3. What is the nature of each substance on the basis of your observations?

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 21

Class 10 Science Activity 2.12 Page No. 27

  1. Put about 2 g soil in a test tube and add 5 mL water to it.
  2. Shake the contents of the test tube.
  3. Filter the contents and collect the filtrate in a test tube.
  4. Check the pH of this filtrate with the help of universal indicator paper.
  5. What can you conclude about the ideal soil pH for the growth of plants in your region?

Observations:
MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 22
Class 10 Science Activity 2.13 Page No. 28,29

  1. Write the chemical formulae of the salts given below.
    Potassium sulphate, sodium sulphate, calcium sulphate. magnesium sulphate, copper sulphate, sodium chloride, sodium nitrate, sodium carbonate and ammonium chloride.
  2. Identify the acids and bases from which the above salts may be obtained.
  3. Salts having the same positive or negative radicals are said to belong to a family. For example. NaCl and Na2SO4 belong to the family of sodium salts. Similarly, NaCl and KCl belong to the family of chloride salts. How many families can you identify among the salts given in this Activity?

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 23

Class 10 Science Activity 2.14 Page No. 28

  1. Collect the following salt samples – sodium chloride, potassium nitrate, aluminium chloride, zinc sulphate, copper sulphate, sodium acetate, sodium carbonate and sodium hydrogen carbonate (some other salts available can also be taken).
  2. Check their solubility in water (use distilled water only).
  3. Check the action of these solutions on litmus and find the pH using a pH paper.
  4. Which of the salts are acidic, basic or neutral?
  5. Identify the acid or base used to form the salt.
  6. Report your observations in the table.

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts 24

Class 10 Science Activity 2.15 Page No. 32

  1. Heat a few crystals of copper sulphate in a dry boiling tube.
  2. What is the colour of the copper sulphate after heating?
  3. Do you notice water droplets in the boiling tube? Where have these come from?
  4. Add 2-3 drops of water on the sample of copper sulphate obtained after heating
  5. What do you observe? Is the blue colour of copper sulphate restored?

Observations:

  1. On heating blue crystals of copper sulphate, it becomes colourless or white and few drops of water are seen on test tube due to condensation of water of crystallisation.
  2. On adding few drops of water to heated anhydrous copper sulphate, the blue colour of copper sulphate reappear.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals

In this article, we will share MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals

MP Board Class 10th Science Chapter 3 Intext Questions

Intext Questions Page No. 40

Question 1.
Give an example of a metal which:

  1. Is a liquid at room temperature.
  2. can be easily cut with a knife.
  3. Is the best conductor of heat.
  4. Is a poor conductor of heat.

Answer:

  1. Metal that exists in a liquid state at room temperature → Mercury.
  2. Metal that can be easily cut with a knife → Sodium.
  3. Metal that is the best conductor of heat → Silver.
  4. Metals that are poor conductors of heat → Lead.

MP Board Solutions

Question 2.
Explain the meanings of malleable and ductile.
Answer:

  1. Malleable: Materials that can be beaten into thin sheets are called malleable.
  2. Ductile: Materials that can be drawn into thin wires are called ductile.

Metals can be hammered into thin sheets. This property of a metal is called malleability and the metals showing this property are called malleable. Gold, Silver, Copper, aluminium etc are malleable metals. Metals can be drawn into wires. The ability of metals to be drawn into thin wires is called ductility. Gold is the most ductile metal. It is interesting to know that a wire of about 2 km length can be drawn from one gram of gold.

Intext Questions Page No. 46

Question 1.
Why is sodium kept immersed in kerosene oil?
Answer:
Sodium is highly reactive metal. It catches fire if kept in the open. Hence, to protect this to prevent accidental fires, it is kept immersed in kerosene oil.

Question 2.
Write equations for the reactions of:
Iron with steam.
Calcium and potassium with water.
Answer:

  1. 3Fe(s) + 4H2O(g) ➝ Fe3O4(aq) + 4H2(g)
  2. Ca(s) + 2H2O(l) ➝ Ca(OH)2(aq)+ H2(g)+ Heat
    2K(s) + 2H2O(l) ➝ 2KOH(aq) + H2(g) + Heat

Question 3.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 1
Use the table above to answer the following questions about metals A, B, C and D.

  1. Which is the most reactive metal?
  2. What would you observe if B is added to a solution of Copper(II) sulphate?
  3. Arrange the metals A, B, C and D in the order of decreasing reactivity.

Answer:

  1. B is most reactive metal.
  2. If B is added to a solution of copper sulphate it displaces copper from copper sulphate.
  3. If metals are written in the order of decreasing reactivity it is B > A > C > D.

Question 4.
Which gas is produced when diluting hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H2SO4.
Answer:
Hydrogen gas is evolved when diluting hydrochloric acid is added to a reactive metal. When iron reacts with dilute H2SO4, Iron(II) sulphate with the evolution of hydrogen gas is formed.
Fe(s) + H2SO4aq) → FeSO4(aq) + H2(g)

MP Board Solutions

Question 5.
What would you observe when zinc is added to a solution of iron(II) sulphate? Write the chemical reaction that takes place.
Answer:
Zinc is more reactive than iron. When zinc is added to iron (II) sulphate, then it will displace the iron from iron sulphate solution as shown in the following chemical reaction,
Zn(s) + FeSO4(aq) → ZnSO4(aq) + Fe(s)

Intext Questions Page No. 49

Question 1.
(i) Write the electron-dot structures for sodium, oxygen and magnesium.
(ii) Show the formation of Na2O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds?
Answer:
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 2
(iii) The ions present in Na2O are Na+ and O2- ions and MgO are Mg2+
and O2- ions.

Question 2.
Why do ionic compounds have high melting points?
Answer:
Ionic compounds have high melting points because there is electrostic forces of attraction between their charges.

Intext Questions Page No. 53

Question 1.
Define the following terms:

  1. Mineral
  2. Ore
  3. Gangue

Answer:

  1. Mineral: Compounds which occur Naturally are called minerals.
  2. Ore: Metals can be obtained from minerals. These are called ores.
  3. Gangue: Ores mined from the earth are usually contaminated with large amounts of impurities such as soil, sand etc., called gangue.

Question 2.
Name two metals which are found in nature in the free state.
Answer:
The metals at the bottom of the reactivity series are mostly found in a free state. For example gold, silver, and platinum.

Question 3.
What chemical process is used for obtaining a metal from its oxide?
Answer:
Metal can be extracted from its oxide by the process of reduction.

Intext Questions Page No. 55

Question 1.
Metallic oxides of zinc, magnesium and copper were heated with the following metals.

Metal Zinc Magnesium Copper
Zinc oxide
Magnesium oxide
Copper oxide

In which cases will you find displacement reactions taking place?
Answer:

Metal Zinc Magnesium Copper
Zinc oxide No reaction Displacement No reaction
Magnesium oxide No reaction Displacement No reaction
Copper oxide Displacement Displacement No reaction

Question 2.
Which metals do not corrode easily?
Answer:
Silver and Gold are not corrode easily.

Question 3.
What are alloys?
Answer:
An alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal. For example, brass is an alloy of copper and zinc.

MP Board Class 10th Science Chapter 3 Ncert Textbook Exercises

Question 1.
Which of the following pairs will give displacement reactions?
(a) NaCl solution and copper metal.
(b) MgCl2 solution and aluminium metal.
(c) FeSO4 solution and silver metal.
(d) AgNO3 solution and copper metal.
Answer:
(d) AgNO3 solution and copper metal.

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) all of the above.
Answer:
(c) Applying a coating of zinc

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be:
(a) Calcium
(b) Carbon
(c) Silicon
(d) Iron
Answer:
(a) Calcium

Question 4.
Food cans are coated with tin and not with zinc because:
(a) Zinc is costlier than tin.
(b) Zinc has a higher melting point than tin.
(c) Zinc is more reactive than tin.
(d) Zinc is less reactive than tin.
Answer:
(c) Zinc is more reactive than tin.

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch.

  1. How could you use them to distinguish between samples of metals and non-metals?
  2. Assess the usefulness of these tests in distinguishing between metals and non-metals.

Answer:

  1. Metals can be spread into sheets with the help of a hammer while non metals give powder. When metals are connected into circuit using battery, bulb, wires and a switch current passes through the circuit and the bulb glows.
  2. Hammer is a reliable method because no non metal can be spread into sheet.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
Metal oxides which react with both acids as well as bases to produce salt and water are known as amphoteric oxides.
Eg: Al2O3 – Aluminium oxide
ZnO – Zinc Oxide.

MP Board Solutions

Question 7.
Name two metals which will displace hydrogen from dilute acids, and two metals which will not.
Answer:
Iron and Aluminium are more reactive than Hydrogen and displace hydrogen from dilute acids. Mercury and copper re less reactive and these do not displace hydrogen from dilute acids.

Question 8.
In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?
Answer:
In the electrolytic refining of a metal, M:

  1. Anode → Impure metal, M.
  2. Cathode → Thin strip of pure metal, M.
  3. Electrolyte → Aqueous solution of a salt of the metal, M.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in the figure below.
(a) What will be the action of gas on:

  1. Dry litmus paper?
  2. Moist litmus paper?

(b) Write a balanced chemical equation for the reaction taking place.
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 3
Answer:
(a)

  1. There will be no action on dry litmus paper.
  2. The colour of litmus paper will turn red because sulphur is a non-metal and the oxides of non-metal are acidic in nature.

(b) MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 4

Question 10.
State two ways to prevent the rusting of iron.
Answer:
Two ways to prevent the rusting or iron are
(a) Applying oil, paint and grease we can prevent rusting.
(b) By coating with zinc to iron, we can prevent using. This is called Galvanisation.

Question 11.
What type of oxides are formed when non-metals combine with oxygen?
Answer:
When non-metals are combined with oxygen then neutral or acidic oxides are formed. Examples of acidic oxides are NO2, SO2 and examples of neutral oxides are NO, CO etc.

Question 12.
Give reasons:

  1. Platinum, gold and silver are used to make jewellery.
  2. Sodium, potassium and lithium are stored under oil.
  3. Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
  4. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Answer:
1. Platinum, gold and silver are used to make jewellery because they are very lustrous. Also, they are very less reactive, ductile and do not corrode easily.

2. Sodium, potassium, and lithium are very reactive metals and react very vigorously with air and water. Therefore, they are kept immersed in oil.

3. Though aluminium is a highly reactive metal, it is resistant to corrosion. This is because aluminium reacts with oxygen present in the air to form a thin layer of aluminium oxide. This oxide layer is very stable and prevents further reaction of aluminium with oxygen. Also, it is light in weight and a good conductor of heat.
Hence, it is used to make cooking utensils.

4. Carbonate and sulphide ores are usually converted into oxides during the process of extraction because metals can be easily extracted from their oxides rather than from their carbonates and sulphides.

Lithium oxide formula graphical representation.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
Copper reacts with moist carbon dioxide in the air and slowly tosses its shiny brown surface and gains a green coat. This green substance is basic copper carbonate. Citric acid and tartaric acid neutralise copper carbonate. Hence citric or tartaric acid are effective in cleaning the vessel.

Question 14.
Differentiate between metal and non-metal on the basis of their chemical properties.
Answer:
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 5

Question 15.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument, the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
Answer:
The solution he had used was Aqua regia. Aqua regia is a Latin word which means ‘Royal Water’. It is the mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3:1. It is capable of dissolving metals like Gold and Platinum. Since the outer layer of the gold bangles is dissolved in aqua regia, so their weight was reduced drastically.

Question 16.
Give reasons why copper is used to making hot water tanks and not steel (an alloy of iron).
Answer:
Iron do not react with hot water, but reacts with steam and forms metallic oxide and Hydrogen. But copper do not reacts with water. Hence copper is used to make hot water tanks and not steel (an alloy of iron).

MP Board Class 10th Science Chapter 3 Additional Questions

MP Board Class 10th Science Chapter 3 Multiple Choice Questions

Question 1.
What kind of element is carbon?
(a) Metal
(b) Metalloid
(c) Non-metal
(d) An alloy
Answer:
(c) Non-metal

Question 2.
A metal of daily use which does not get rusted is:
(a) Steel
(b) Iron
(c) Gold
(d) Silver
Answer:
(a) Steel

MP Board Solutions

Question 3.
Liquid non-metal at room temperature:
(a) Oxygen
(b) Nitrogen
(c) Mercury
(d) Bromine
Answer:
(c) Mercury

Question 4.
Steel is primarily made up of:
(a) Fe and C
(b) Cu and C
(c) Fe and S
(d) Zn and C
Answer:
(a) Fe and C

Question 5.
Silver is coated over iron in electroplating, it represents:
(a) Silver is more reactive than Iron.
(b) Iron is more reactive than silver.
(c) Both are equally reactive
(d) None of above.
Answer:
(b) Iron is more reactive than silver.

Question 6.
Brass is an alloy of:
(a) Cu and Mn
(b) Cu and Zn
(c) Cu and Fe
(d) Cu, Fe and Zn.
Answer:
(b) Cu and Zn

Question 7.
During electrolytic refining of a metal, metal gets deposited at:
(a) Anode
(b) Cathode
(c) Solution
(d) None
Answer:
(b) Cathode

Question 8.
What basic physical property can differentiate metal and non-metal?
(a) Hardness
(b) Lustre
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 9.
We can cut ………. metal with an ordinary knife:
(a) Sodium
(b) Carbon
(c) Gold
(d) Silver
Answer:
(a) Sodium

Question 10.
Hardest metal present in our nature is:
(a) Gold
(b) Diamond
(c) Tungsten
(d) Copper
Answer:
(c) Tungsten

Question 11.
Lustrous non-metal is:
(a) Oxygen
(b) Nitrogen
(c) Iodine
(d) Gold
Answer:
(c) Iodine

Question 12.
Iron pyrites contain which constituent other than Fe:
(a) Co
(b) Cl
(c) S
(d) Pt
Answer:
(c) S

Question 13.
When a metal reacts with water, it forms:
(i) Metal
(ii) H2
(iii) Metal hydroxide
Choose the best combination:
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) All
Answer:
(c) (ii) and (iii)

Question 14.
Metal forms salts when it reacts with:
(a) Acid
(b) Water
(c) Another Metal
(d) Oxygen
Answer:
(a) Acid

Question 15.
What kinds of metals are coin metals (Cu, Ag, Au)?
(a) Most reactive
(b) Non-reactive
(c) Least reactive
(d) None
Answer:
(c) Least reactive

Question 16.
What kinds of metals are noble metals?
(a) Most reactive
(b) Non-reactive
(c) Least reactive
(d) None
Answer:
(b) Non-reactive

Question 17.
Arrange the following metals in descending order of reactivity – Na, Al, Au, H:
(a) Na > Al > Au>H.
(b) H < Au < Al < Na.
(c) Au > Al > Na > H.
(d) Na > Al > H > Au.
Answer:
(b) H < Au < Al < Na.

Question 18.
When a metal reacts with oxygen, it forms:
(a) Hydrated metals
(b) Metal oxides
(c) Non-metals
(d) Oxygen
Answer:
(b) Metal oxides

Question 19.
Which metal violently reacts with cold water?
(a) Cu and Ag
(b) Au and Ag
(c) K and Na
(d) Hg
Answer:
(c) K and Na

Question 20.
Metals that do not react with water at all are:
(a) Alkali metal
(b) Alkaline eater
(c) Lanthanides
(d) Coin metals
Answer:
(d) Coin metals

Question 21.
Aluminium develops a thin layer of oxide when exposed to air, this process is called:
(a) Anodization
(b) Amalgamation
(c) Corrosion
(d) Rancidity
Answer:
(c) Corrosion

Question 22.
When water reacts with metal, which gas is evolved in the reaction?
(a) Oxygen
(b) Hydrogen
(c) Nitrogen
(d) Sulphur dioxide
Answer:
(b) Hydrogen

Question 23.
What will be the missing product of the following reaction?
Ca(s) + 2H2O → Ca(OH)2(aq) + ….
(a) O2
(b) CaO
(c) H2
(d) O3
Answer:
(c) H2

Question 24.
Which solution or reagent can dissolve gold and platinum?
(a) Conc. H2SO4
(b) Conc. HCl
(c) Conc. HNO3
(d) Aqua regia
Answer:
(d) Agua regia

Question 25.
Benchmark element of the reactivity series is –
(a) Au
(b) H
(c) Na
(d) Fe
Answer:
(b) H

Question 26.
Elements more electropositive in nature are:
(a) Metals
(b) Non-metals
(c) Metalloids
(d) None
Answer:
(a) Metals

Question 27.
Ionic solids are:
(a) Solid and hard
(b) Having high boiling and melting point
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 28.
Which one of the following represents electron dot structures of sodium?
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 6
Answer:
(d) M

Question 29.
Carefully observe the given diagram below:
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 7
Choose the best compound to show above electron-dot structure bonding:
(a) NaCl
(b) MgCl2
(c) H2O
(d) None
Answer:
(c) H2O

Question 30.
Ore of metals with high reactivity can be separated by:
(a) Roasting
(b) Calcination
(c) Electrolysis
(d) Reduction process
Answer:
(c) Electrolysis

Question 31.
Thermionic reactions are:
(a) Displacement reactions of highly exothermic nature
(b) Endothermic reaction
(c) Both (a) and (b)
(d) Electrolytic reduction reactions.
Answer:
(c) Both (a) and (b)

Question 32.
Rusting is an example of:
(a) Electrolysis
(b) Calcination
(c) Corrosion
(d) None
Answer:
(c) Corrosion

Question 33.
Which compounds of metals are basic in nature?
(a) Hydrides
(b) Chlorides
(c) Cyanides
(d) Oxides
Answer:
(d) Oxides

Question 34.
When metal reacts with water which gas is liberated?
(a) Oxygen
(b) Hydrogen
(c) Metal oxide
(d) All of these
Answer:
(b) Hydrogen

Question 35.
Which of the following gas is produced when dilute H2SO4 reacts with iron?
(a) Oxygen gas
(b) Hydrogen gas
(c) Both
(d) None of these
Answer:
(b) Hydrogen gas

Question 36.
What happen when Zinc is added to iron (II) sulphate Zn(s) + Fe(s):
(a) ZnSO2 + FeO2
(b) ZnSO2(aq) + Fe(s)
(c) ZnFe + 2Ov
(d) None
Answer:
(b) ZnSO2(aq) + Fe(s)

Question 37.
Choose the best reason from the following for why ionic compounds have high melting points:
(a) Electrostatic forces
(b) Low magnetic forces
(c) (a) and (b) both
(d) None of these
Answer:
(a) Electrostatic forces

Question 38.
Which of the following is gangue?
(a) Water
(b) Metals
(c) Rocks
(d) CO2
Answer:
(c) Rocks

Question 39.
Sodium, potassium and lithium are:
(a) Very less reactive
(b) More reactive and react with air
(c) Easily converted to oxides
(d) None
Answer:
(b) More reactive and react with air

Question 40.
Choose from following ways to prevent the rusting of iron:
(a) Oiling
(b) Greasing
(c) Galvanisation
(d) All (a), (b) and (c)
Answer:
(d) All (a), (b) and (c)

Fill in the blanks:

  1. Metal generally give ……………Oxides when dissolved in water.
  2. Li, Na and K are …………… metals.
  3. K and Na catch fire when exposed to ……….
  4. Ag and Au ………….. react with water at all.
  5. Aluminium is positioned ……………. hydrogen in reactivity series.
  6. …………………….. is called royal water.
  7. Metal is displaced from their ……………… or …………….. form.
  8. Metal gives salt when it reacts with…….
  9. Hydrogen is not liberated when metal reacts with ……….. acid.
  10. Ionic compounds are ……………………….. in water.
  11. Ionic compounds are hard but …………….. in nature hence break into pieces when pressure is applied.
  12. Earth’s …………….. is main source of metal.
  13. Minerals containing metals are called ……………….
  14. Sulphide ores contain metals with ………………. reactivity.
  15. Carbonate ore is converted to their metal oxides by the process of …………
  16. Electrolysis is done to the ores of ……………….. reactivity to obtain pure metal.
  17. Impurities such as soil, sand etc. are called …………… in the ore.
  18. Bronze is a homogeneous mixture of ……………… and …………….

Answers:

  1. basic
  2. soft
  3. air
  4. do not
  5. above
  6. Aqua regia
  7. solution, molten
  8. alkali
  9. nitric acid
  10. soluble
  11. brittle
  12. crust
  13. Ores
  14. low
  15. Calcination
  16. high
  17. gangue
  18. copper, tin

MP Board Class 10th Science Chapter 3 Very Short Answer Type Questions

Question 1.
Which one is the most abundant element in our earth crust?
Answer:
Oxygen.

Question 2.
Write the names of two diatomic gaseous elements.
Answer:

  1. Oxygen, O2.
  2. Nitrogen, N2.

Question 3.
Is any metal known in gaseous form, in its natural conditions?
Answer:
No.

Question 4.
Name two naturally occurring soft metals.
Answer:
Sodium, magnesium.

Question 5.
Which metal shows poor conductivity?
Answer:
Tungsten [W] or Bismuth (Bi].

Question 6.
Which non-metal conducts electricity?
Answer:
Graphite.

Question 7.
Which metal forms amphoteric oxides when reacted with oxygen?
Answer:
Aluminium, Zinc, etc.

Question 8.
Draw an electron dot structure of SO2.
Answer:
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 8

Question 9.
Give an example of metal oxide’s reaction with water.
Answer:
All metal oxides do not react with water but some metal oxides react with water to give alkali. For example, Na and K.
Na2O(s) + H2O(l) → 2NaOH(aq)

Question 10.
Give an example of amphoteric oxides.
Answer:
Al2O3.

Question 11.
Give a term to the following:
A process in which a carbonate ore is heated at very high temperature to get metal oxide.
Answer:
Roasting.

MP Board Solutions

Question 12.
How less reactive metal oxides are reduced?
Answer:
Less reactive metal oxides are reduced by reducing agents like aluminium.

Question 13.
Give a reaction in which metal hydroxide is formed.
Answer:
‘Na’ directly reacts with hydrogen gas and forms sodium hydride.
2Na + H2 → 2NaH.

Question 14.
Give an example for each an alloy and amalgam.
Answer:

  1. Alloy – Steel.
  2. Amalgam – Sodium Amalgam. (Alloy of mercury and sodium).

Question 15.
What do we call the removal of gangue from the ore?
Answer:
Enrichment of ores.

Question 16.
What is the place of less reactive metals in metal reactivity series?
Answer:
Less reactive metals are arranged at the bottom of the series.

Question 17.
Name two metals which can replace iron in electroplating.
Answer:
Gold and silver.

Question 18.
What kind of compounds has the highest melting and boiling points?
Answer:
Ionic compounds.

Question 19.
What corrodes copper?
Answer:
Moist carbon-dioxide corrodes copper.

Question 20.
Iron pillar of Qutub Minar, Delhi is prevented against rusting. Why?
Answer:
The corrosion-resistant nature is due to protective film at the non-rust interface because of high phosphorus content.

Question 21.
Why every metal has a different rate of reactivity with the same chemical or reactant? (HOTS)
Answer:
Every metal has different electronic configuration and different electro-positivity or ion forming capability, so metal reactivity differs from each other.

MP Board Class 10th Science Chapter 3 Short Answer Type Questions

Question 1.
Why can we draw gold to thin wire form?
Answer:
Gold is a metal and metal has the ability to be drawn in very thin wire or sheets without being broken, this property of a metal is called ductility. Gold is the most ductile metal.

Question 2.
What is a semiconductor?
Answer:
There are some elements known to us which are neither metal nor nonmetals, they alter their properties with the different physical and chemical environment provided. Normally, they behave as non-metals and show insulation property but when provided with extra energy, they start behaving as metal and show conduction. Example: Silicon.

Question 3.
What do you know about amphoteric oxides?
Answer:
Metal oxides which produce salts in both cases, either react with acid or base are termed amphoteric oxides.

Question 4.
What is gangue? Why is it important to remove them before the extraction process?
Answer:
Impurities associated with ores such as sand, soil etc. are called gangue. The original compound becomes bulkier with their presence and extraction of pure metal consumes lots of energy and time. Hence, gangue is removed before the extraction process.

Question 5.
Give an example of displacement reaction used for extraction of metal from its oxide.
Answer:
Highly reactive metals are used to reduce metal oxide to the metal in an electrolytic displacement reaction.

Example:
3MnO2(s) + 4Al(s) → 3Mn(l) + 2Al2O3
Here, aluminium helps in extraction of pure Mn.

Question 6.
Explain the reactivity series of metals in brief.
Answer:
All metals are arranged in an order on the basis of reactivity. This series represents displacing ability of one metal to displace other from its compound form when it undergoes an electrolytic displacement reaction.

Question 7.
How metal and non-metal interact to form salts?
Answer:
When metal and non-metal interact, they form ionic compounds. Metal tends to lose an electron and form positive ion while non-metal forms a negative ion and a bond is formed due to strong electrostatic forces of attraction among two ions.

Question 8.
How can we protect metals corrosion?
Answer;
Prevention from corrosion can be done by painting, oiling, greasing, galvanizing and electroplating etc. on the metal. All these processes stop the interaction of outer layer of air with metal.

Question 9.
What is galvanisation?
Answer:
Galvanisation is a protection technique against metal corrosion. In this process, metal is coated with a thin layer of zinc.

Question 10.
Why pure gold is mixed with silver or copper while making jewellery?
Answer:
Pure gold metal is very soft. So, jewellery will be brittle if used in the same form. Impurity of other metal in nominal percentage makes the metal harder and ready to use and it does not appear very different than the pure one.

Question 11.
What is an amalgam?
Answer:
When one component of any alloy is mercury it is called amalgam.

Question 12.
Why alloys are not used as conductance medium of electricity?
Answer:
Alloys are not used as conductance medium of electricity because of electrical conductivity and melting point of an alloy is lesser than the pure metal.

Question 13.
Why cooking utensils are made up of metals and their handle or knobs with non-metal materials? (HOTS)
Answer:
Metals are good conductors of heat. So, when utensil made of metals are pulled over a flame, they spread heat energy evenly and food gets cooked properly. But handles or knobs are made of non-metals because they are bad conductors of heat and give us ease to work with highly heated utensils so that we can hold them easily while cooking.

Question 14.
Why ageing is a natural phenomenon? (HOTS)
Answer:
The oxygen of the atmosphere react with the outer skin of body fat and carbohydrate, hence continuously degrade it. Similarly inside the body for the energy need, carbohydrate and body fat undergo oxidation process and get continuously decomposed which causes ageing, so ageing is a natural phenomenon.

Question 15.
Neha went to market with her grandmother to purchase some cooking utensils for their family’s wedded couple. Grandmother started buying cooking utensils made of copper and iron but Neha suggested her to take utensils made of steel. (VBQ)

  1. How is steel better than copper?
  2. What kind of corrosion occurs in Cu?
  3. What values does Neha show in this act?

Answer:

  1. Steel is an alloy. It does not get corroded or spoiled with time and it is cheap too.
  2. Cu forms its oxide when it comes in contact with oxygen and turns green in colour.
  3. Neha shows awareness about metal of more usability and durability.

MP Board Class 10th Science Chapter 3 Long Answer Type Questions

Question 1.
Draw line diagrams for the steps involved in the extraction of metals.
Answer:
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 9

Question 2.
Write short notes on the following:

  1. Roasting
  2. Calcination
  3. Mineral
  4. Anodising

Answer:
1. Roasting:
To extract metal of medium reactivity from its sulphide ore, the ore is heated strongly in the presence of air and this extraction process is called roasting.

2. Calcination:
When medium reactivity metal is extracted from its carbonate ore by heating moderately, the process is known as calcination.

3. Mineral:
Naturally occurring compounds of metals and non-metals in various combinations are called minerals.

4. Anodising:
When aluminium is exposed to air it develops a thick layer of oxides and turns green in colour.

Question 3.
Rupam was painting the garden chair and other pots made with iron at his home, his younger brother asked him why he is doing so and wasting his time:

  1. What is your view about his work?
  2. What are other methods to protect the metal from corrosion?
  3. What values does Rupam express in this way?

Answer:

  1. Rupam was painting the garden’s metal objects to protect them from corrosion.
  2. Oiling, euchre plating, anodizing etc. are other methods.
  3. Rupam was doing a great job protecting the metal from degrading because it can give a long life to objects and save money.

MP Board Class 10th Science Chapter 3 Textbook Activities

Class 10 Science Activity 3.1 Page No. 37

  1. Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample.
  2. Clean the surface of each sample by rubbing them with sandpaper and note their appearance again.

Observations:
Iron, copper, aluminium and magnesium have lustre which clearly appears on rubbing them with sandpaper.

Class 10 Science Activity 3.2 Page No. 37

  1. Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations.
  2. Hold a piece of sodium metal with a pair of tongs.

Caution:

  1. Always handle sodium metal with care. Dry it by pressing between the folds of a filter paper.
  2. Put it on a watch-glass and try to cut it with a knife.
  3. What do you observe?

Observations:
Only Na and Mg are soft metals and we can cut them with a knife but other metals are hard. de

Class 10 Science Activity 3.3 Page No. 38

  1. Take pieces of iron, zinc, lead and copper.
  2. Place any one metal on a block of iron and strike it four or five times with a hammer.
  3. What do you observe?
  4. Repeat with other metals.
  5. Record the change in the shape of these metals.

Observations:
Fe, Zn, Pb and Cu are metals and show proper malleability when hammered.

Class 10 Science Activity 3.4 Page No. 38

List the metals whose wires you have seen in daily life.

Observations:
Lead cannot be drawn to a thin wire. But Fe, Cu and Al are ductile.

Class 10 Science Activity 3.5 Page No. 38

  1. Take an aluminium or copper wire. Clamp this wire on a stand, as shown in the figure.
  2. Fix a pin to the free end of the wire using wax.
  3. Heat the wire with a spirit lamp, candle or a burner near the place where it is clamped.
  4. What do you observe after some time?
  5. Note your observations. Does the metal wire melt?

Observations:

  1. The pin fell down because heat is being conducted.
  2. The metal wire does not melt.

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 10

Class 10 Science Activity 3.6 Page No. 39

  1. Set up an electric circuit as shown in the figure.
    Place the metal to be tested in the circuit between terminals A and B as shown.
  2. Does the bulb glow? What does this indicate?

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 11

Observations:
Yes, the bulb glows as metals are a good conductor of heat and electricity.

Class 10 Science Activity 3.7 Page No. 39

  1. Collect samples of carbon (coal or graphite), sulphur and iodine.
  2. Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample.
  3. Clean the surface of each sample by rubbing them with sandpaper and note their appearance again.
  4. Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations.
  5. Hold a piece of sodium metal with a pair of tongs.

Caution:

  1. Always handle sodium metal with care. Dry it by pressing between the folds of a filter paper.
  2. Put it on a watch-glass and try to cut it with a knife.
  3. What do you observe?
  4. Take pieces of iron, zinc, lead and copper.
  5. Place any one metal on a block of iron and strike it four or five times with a hammer.
  6. What do you observe?
  7. Repeat with other metals.
  8. Record the change in the shape of these metals.
  9. Set up an electric circuit as shown in the figure.
  10. Place the metal to be tested in the circuit between terminals A and B as shown.
  11. Does the bulb glow? What does this indicate?

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 12

Observations:

  1. Yes, the bulb glows as metals are a good conductor of heat and electricity.
  2. List the metals whose wires you have seen in daily life.
  3. Iron, copper, aluminium and magnesium have lustre which clearly appears on rubbing them with sandpaper.
  4. Only Na and Mg are soft metals and we can cut them with a knife but other metals are hard.
  5. Fe, Zn, Pb and Cu are metals and show proper malleability when hammered.

Class 10 Science Activity 3.8 Page No. 40

  1. Take a magnesium ribbon and some sulphur powder.
  2. Burn the magnesium ribbon. Collect the ashes formed and dissolve them in water.
  3. Test the resultant solution with both red and blue litmus paper.
  4. Is the product formed on burning magnesium acidic or basic?
  5. Now burn sulphur powder. Place a test tube over the burning sulphur to collect the fumes produced.
  6. Add some water to the above test tube and shake.
  7. Test this solution with blue and red litmus paper.
  8. Is the product formed on burning sulphur acidic or basic?
  9. Can you write equations for these reactions?

Observations:
(i)
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 13
Result: Oxide of metal is basic.
(ii)
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 14
Result: Oxide of non-metal is acidic.

Class 10 Science Activity 3.9 Page No. 41

Caution:

  1. The following activity needs the teacher’s assistance. It would be better if students wear eye protection.
  2. Hold any of the samples taken above with a pair of tongs and try burning over a flame. Repeat with the other metal samples.
  3. Collect the product if formed.
  4. Let the products and the metal surface cool down.
  5. Which metals burn easily?
  6. What flame colour did you observe when the metal burnt?
  7. How does the metal surface appear afterburning?
  8. Arrange the metals in the decreasing order of their reactivity towards oxygen.
  9. Are the products soluble in water?

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 15
Only Zn and Al form amphoteric oxide (in nature) other metals forms basic oxides.

Class 10 Science Activity 3.10 Page No. 42

Caution:

  1. This Activity needs the teacher’s assistance.
  2. Collect the samples of the same metals as in Activity 3.9.
  3. Put small pieces of the samples separately in beakers half-filled with cold water.
  4. Which metals reacted with cold water?
  5. Arrange them in the increasing order of their reactivity with cold water.
  6. Did any metal produce fire on water?
  7. Does any metal start floating after some time?
  8. Put the metals that did not react with cold water in beakers half-filled with hot water.
  9. For the metals that did not react with hot water, arrange the apparatus as shown in the figure and observe their reaction with steam.
  10. Which metals did not react even with steam?
  11. Arrange the metals in the decreasing order of reactivity with water.

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 16

Observations:

  1. Metals which reacted with cold water → Na, K and Ca.
  2. Metals which produced fire → Na and K.
  3. Metals which started floating after some time → Ca and Mg.
  4. Metals which reacted with hot water → Mg.
  5. Metals which did not react with steam also → Pb, Cu, Ag and Au.
  6. Arranging reactivity of metals in ascending order:
    K > Na > Ca > Mg > Al > Zn > Fe

Class 10 Science Activity 3.11 Page No. 44

Caution:

  1. Do not take sodium and potassium as they react vigorously even with cold water.
  2. Put the samples separately in test tubes containing dilute hydrochloric acid.
  3. Suspend thermometers in the test tubes, so that their bulbs are dipped
    in the acid.
  4. Observe the rate of formation of bubbles carefully.
  5. Which metals reacted vigorously with dilute hydrochloric acid?
  6. With which metal did you record the highest temperature?
  7. Arrange the metals in the decreasing order of reactivity with dilute acids.
    MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 17
  8. Collect all the metal samples except sodium and potassium again. If the samples are tarnished. rub them clean with sandpaper.

Class 10 Science Activity 3.12 Page No. 44-45

  1. Take a clean wire of copper and an iron nail.
  2. Put the copper wire in a solution of iron sulphate and the iron nail in a solution of copper sulphate taken in test tubes figure.
    MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 18
  3. Record your observations after 20 minutes.
  4. In which test tube did you find that a reaction has occurred? On what basis can you say that a reaction has actually taken place?
    MP Board Class 10th Science Solutions Chapter 3 21
  5. Only Zn and Al form amphoteric oxide (in nature) other metals forms basic oxides.
  6. Metals which reacted with cold water → Na, K and Ca.
  7. Metals which produced fire → Na and K.
  8. Metals which started floating after some time → Ca and Mg.
  9. Metals which reacted with hot water → Mg.
  10. Metals which did not react with steam also → Pb, Cu, Ag and Au.
  11. Arranging reactivity of metals in ascending order:
    K > Na > Ca > Mg > Al > Zn > Fe
    MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 19
  12. Write a balanced chemical equation for the reaction that has taken place.
  13. Name the type of reaction.

Observations:

  1. In these above observations show iron is more reactive than copper.
    Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq)
  2. In the given set-up (A) and (B) reactivity of metals is as follows:
    MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 20

Class 10 Science Activity 3.13 Page No. 48

  1. Taķe samples of sodium chloride, potassium iodide, barium chloride or any other salt from the science laboratory.
  2. What is the physical state of these salts?
  3. Take a small amount of a sample on a metal spatula and heat directly on the flame figure. Repeat with other samples.

Observations:
MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 21

  1. What did you observe? Did the samples impart any colour to the flame? Do these compounds melt?
  2. Try to dissolve the samples in water, petrol and kerosene. Are they soluble?
  3. Make a circuit as shown in figure (testing the conductivity of a salt solution) and insert the electrodes into a solution of one salt. What did you observe? Test the other salt samples too in this manner.
  4. What is your inference about the nature of these compounds? Sodium chloride, potassium iodide and barium chloride give the following observation:

Class 10 Science Activity 3.14 Page No. 53

  1. Take three test tubes and place clean iron nails in each of them.
  2. Label these test tubes A, B and C. Pour some water in test tube A and cork it.
  3. Pour boiled distilled water in test tube B, add about 1 ml of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
  4. Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture, if any, from the air. Leave these test tubes for a few days and then observe (Figure).

MP Board Class 10th Science Solutions Chapter 3 Metals and Non-metals 22

Observations:
Set of 3 test tube A, B and C are given below:

  1. Test tube A: Iron nail became rusty.
    Presence of water and air in test-tube A: Air and water are both exposed.
  2. Test tube B: Iron nail do not become rusty.
    Presence of water and air in test-tube B: Due to layer of oil, air does not expose.
  3. Test tube C: Iron nail do not become rusty.
    Presence of water and air in test-tube C: Air and water both are not present.

Conclusion:

  1. Test-tube A: Iron nail become rusty due to exposure to air and water.
  2. Test-tube B: Iron nail exposed with water but not with air, so the iron nail does not rust.
  3. Test-tube C: Iron nail not exposed to air and water, so iron nail do not rust.

So, air and water both are required for an iron nail to rust.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations

In this article, we will share MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations

MP Board Class 10th Science Chapter 1 Intext Questions

Intext Questions Page No. 6

Question 1.
Why should a magnesium ribbon be cleaned before burning in air?
Answer:
Magnesium metal is highly reactive. In stored conditions, it reacts with oxygen to form magnesium oxide over its outer layer. To remove this layer and to expose the underlying metal into air, the magnesium ribbon is cleaned by sandpaper.

MP Board Solutions

Take help of Chemical Equation Calculator to balance the equation calculator your chemical equation.

Question 2.
Write the balanced equation for the following chemical reactions:

  1. Hydrogen + Chlorine → Hydrogen Chloride.
  2. Barium Chloride + Aluminium Sulphate → Barium Sulphate + Aluminium Chloride.
  3. Sodium + Water Sodium Hydroxide + Hydrogen.

Answer:
Balanced equations for the chemical reactions are as follows:

  1. H2(g) + Cl2(g) → 2HCl(g)
  2. 3BaCl2(aq) + Al2(SO4)3(aq) → 3BaSO4(s) + 2AlCl3(aq)
  3. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Question 3.
Write a balanced chemical equation with state symbols for the following reactions:

  1. Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
  2. Sodium hydroxide solution (in water) reacts with a hydrochloric acid solution (in water) to produce sodium chloride solution and water.

Answer:
Balanced chemical equations with state symbols for the required reactions are as follows:

  1. BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
  2. NaOH(aq) + HCl(aq) + NaCl(aq) + H2O(l)

Intext Questions Page No. 10

Question 1.
A solution of a substance ‘X’s used for white-washing.
(i) Name the substance ‘X’ and write its formula.
(ii) Write the reaction of the substance ‘X’ named in (i) above with water.
Answer:
(i) ‘X’ is calcium oxide. It is used in white-washing and its chemical formula is CaO.
(ii) Calcium oxide reacts vigorously with water to form calcium hydroxide (slaked lime).
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 1

Question 2.
Why is the amount of gas collected in one of the test tubes in double of the amount collected in the other? Name this gas.
Answer:
Water (H2O) contains two parts of hydrogen and one part of oxygen. The ratio of water components i.e., hydrogen and oxygen is 2:1. Therefore, the amount of hydrogen and oxygen produced after water electrolysis is in a ratio of 2:1. This is why during electrolysis the amount of gas collected in hydrogen’s test tubes is double the amount collected in the oxygen’s test tube.

  1. Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in Figure.
  2. Connect these electrodes to a 6-volt battery.
  3. Fill the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water.
  4. Take two test tubes filled with water and invert them over the two carbon electrodes.
  5. Switch on the current and leave the apparatus undisturbed for some time.
  6. You will observe the formation of bubbles at both electrodes. These bubbles displace water in the test tubes.
  7. Is the volume of the gas collected the same in both the test tubes?
  8. Once the test tubes are filled with the respective gases. Remove them carefully.
  9. Test these gases one by one by bringing a burning candle close to the mouth of the test tubes.

Caution:

  1. This step must be performed carefully by the teacher.
  2. What happens in each case?
  3. Which gas is present in each test tube?

Observations:
On electrolysis of water, hydrogen and oxygen gas is produced. Hydrogen gas has twice the volume of oxygen gas. Oxygen gas let the burning continue while hydrogen gas burns with a pop sound. The gas collected at the cathode is oxygen while gas collected at the anode is hydrogen.

Intext Questions Page No. 13

Question 1.
Why does the colour of copper sulphate solution change when an iron nail is dipped in it?
Answer:
The colour of copper sulphate solution changes when an iron nail is dipped in it because iron is highly reactive than copper. Therefore Iron displaces copper from copper sulphate.

Question 2.
Give an example of a double displacement reaction other than the one.
Answer:

Above two reactions represents double displacement reaction. As both electropositive ions of two reactants exchange their electronegative ions and form two new and more stable products, show two replacements at a time and thus called double displacement reactions. In these reactions, acid and base exchange ions to form two new compounds. Hence, it is a double displacement reaction.

MP Board Solutions

Question 3.
Identify the substances that are oxidized and the substances that are reduced in the following reactions.

  1. Na(s) + O2(g) → 2NaO(s)
  2. CuO(s) + H2(g) → Cu(s) + H2O(l)

Answer:

  1. Sodium (Na) is oxidized as it gains oxygen and oxygen gets reduced.
  2. Copper oxide (CuO) is reduced to copper (Cu) while hydrogen (H) gets oxidised to water (H2O).

MP Board Class 10th Science Chapter 1 NCERT Textbook Exercises

Question 1.
Which of the statements about the reaction below are incorrect?
2PbO(s) + C(s) → 2Pb(s) + CO2(g)
(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced.
(i) (a) and (b)
(ii) (a) and (c)
(iii) (a), (b) and (c)
(iv) all
Answer:
(i) (a) and (b)
Lead is getting reduced. & Carbon dioxide is getting oxidised.

Question 2.
Fe2O3 + 2Al → Al2O3 + 2Fe. The given reaction is an example of a:
(a) Combination reaction.
(b) Double displacement reaction.
(c) Decomposition reaction.
(d) Displacement reaction.
Answer:
(d) The reaction is an example of a displacement reaction.

Question 3.
What happens when dilute hydrochloric acid is added to iron filings? Tick the correct answer.
(a) Hydrogen gas and iron chloride are produced.
(b) Chlorine gas and iron hydroxide are produced.
(c) No reaction takes place.
(d) Iron salt and water are produced.
Answer:
(a) Hydrogen gas and iron chloride are formed.

Question 4.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer;
A balanced chemical equation is an equation having an equal number of atoms on both sides (reactant and product side) of the reaction. According to, “the law of conservation of mass”, mass can neither be created nor it can be destroyed. So, in a chemical reaction, the total mass of reactants should be equal to the total mass of the products formed.

MP Board Solutions

Question 5.
Translate the following statements into chemical equations and then balance them.
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer;
(a) 3H2(g) + N2(g) → 2NH3(g).
(b) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g).
(c) 3BaCl2(aq) + Al2(SO4)3(aq) → 2AlCl3(aq) + 3BaSO4(s).
(d) 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g).

Question 6.
Balance the following chemical equations:
(a) HNO3 + Ca(OH)2 → Ca(NO3)2 +H2O.
(b) NaOH + H2SO4 → Na2SO4 + H2O.
(c) NaCl + AgNO3 → AgCl + NaNO3
(d) BaCl2 + H2SO4 → BaSO4 + HCl.
Answer:
(a) 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O.
(b) 2NaOH + H2SO4 → Na2SO4 + 2H2O.
(c) NaCl + AgNO3 → AgCl + NaNO3.
(d) BaCl2 + H2SO4 → BaSO4 + 2HCl.

Question 7.
Write the balanced chemical equations for the following reactions.
(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water.
(b) Zinc + Silver nitrate → Zinc nitrate + Silver.
(c) Aluminium + Copper chloride → Aluminium chloride + Copper.
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride.
Answer:
(a) Ca(OH)2 + CO2 → CaCO3 + H2O.
(b) Zn + 2AgNO3 → Zn(NO3)2 + 2Ag.
(c) 2Al + 3CuCl2 → 2AlCl3 + 3Cu.
(d) BaCl2 + K2SO4 → BaSO4 + 2KCl.

Question 8.
Write the balanced chemical equation for the following and identify the type of reaction in each case.
(a) Potassium bromide (aq) + Barium iodide (aq) → Potassium iodide(aq) + Barium bromide (s).
(b) Zinc carbonate (s) → Zinc oxide (s) + Carbon dioxide (g).
(c) Hydrogen (g) + Chlorine (g) → Hydrogen chloride (g).
(d) Magnesium (s) + Hydrochloric acid (aq) → Magnesium chloride (aq) + Hydrogen (g).
Answer:
(a) 2KBr (aq) + BaI2 (aq) → 2KI (aq) + BaBr2 (s); Double displacement reaction.
(b) ZnCO3(s) → ZnO(s) + CO2(g); Decomposition reaction.
(c) H2(g) + Cl2(g) → 2HCl(g); Combination reaction.
(d) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g); Displacement reaction.

Question 9.
What does one mean by exothermic and endothermic reactions? Give examples.
Answer:
Exothermic reactions:
Reactions in which energy is evolved are called exothermic reactions.
Chemicalreactionsmayreleaseenergyintheformofheat, light or sound.

Examples: Decomposition of glucose, the formation of hydroxides:
Mg + 2H2O → Fe Mg(OH)2 + H2 + Heat
C + O2 → CO2 + Heat
N2 + 3H2 → 2NH3 + Heat

Reactions in which energy is absorbed are known as endothermic reactions.
These kinds of reactions require energy in order to proceed.
For example: In photosynthesis, plants take photo-energy from the sun to convert carbon dioxide and water into glucose and oxygen,
N2 + O2 → 2NO + Heat
H2SO4(g) → SO3(g) + H2O(g) + Heat
C + 2S → CS2 + Heat

Question 10.
Why is respiration considered an exothermic reaction? Explain.
Answer:
During respiration at the cellular level, food is decomposed to simpler substances releasing carbon dioxide (that we exhale) and water with a release of huge amount of energy. Hence, it is called exothermic reaction. Glucose combines with oxygen in the cells and provides energy.
C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O (l) + Energy.
Since a large amount of energy is released, it is an exothermic reaction.

MP Board Solutions

Question 11.
Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.
Answer:
Decomposition reactions involve breaking down of compounds to form two or more substances. These reactions require energy to proceed. Thus, they are the exact opposite of combination reactions in which two or more substances combine to give a new substance.
Examples:

  1. ZnCO3(s) → ZnO(s) + CO2(g); Decomposition reaction.
  2. H2(g) + Cl2(g) → 2HCl(g); Combination reaction.

In the first equation, since ZnCO3 is broken down into ZnO and CO2 it is a decomposition reaction. In the second equation, H2 and Cl2 combine to give a new substance HCl. Therefore, it is a combination reaction.

Question 12.
Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
Answer:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 3

Question 13.
What is the difference between displacement and double displacement reactions? Write equations for these reactions.
Answer:
In a displacement reaction, a more reactive element displaces a less reactive element from a compound.
A + BX → AX + B; where A is more reactive than B.
In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.
AB + CD → AD + CB
For example:
Displacement reaction:
CuSO4(aq) + Zn(s) → ZnSO4(aq) + Cu(s)

Double displacement reaction:
Na2SO4(aq) + BaCl2 (aq) → BaSO4(s) + 2NaCl(aq)

Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.
Answer;
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 4

Question 15.
What do you mean by a precipitation reaction? Explain by giving examples.
Answer:
When a compound of product settles down as precipitate at the end of the reaction, it is called a precipitation reaction. In this type of reaction, an ‘insoluble solid called precipitate is formed.
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 5
In this reaction, silver is obtained as a precipitate. Hence, it is a precipitation reaction.

Question 16.
Explain the following in terms of gain or loss of oxygen with two examples each?
(a) Oxidation
(b) Reduction
Answer:
(a) A reaction which involves gain of oxygen is called oxidation reaction.
For example:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 6

(b) A reaction which involves loss of oxygen is called a reduction reaction.
For example:
(i) CuO + H2 → Cu + H2O, CuO is reduced to Cu.
(ii) ZnO + C → Zn + CO, ZnO is reduced to Zn.

Question 17.
A shiny brown coloured element ‘X’ on heating in the air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
Answer:
‘X’ is copper (Cu) and the black coloured compound is a copper oxide (CuO). The equation of the reaction involved in heating copper is given below:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 7

Question 18.
Why do we apply paint on iron articles?
Answer:
Iron articles are painted to prevent them from rusting. When painted, the contact of iron articles with atmospheric moisture and the air is cut off. Hence, rusting is prevented.

Question 19.
Oil and fat containing food items are flushed with nitrogen. Why?
Answer:
Oil and fat containing food items are flushed with nitrogen to prevent the items from getting oxidised which may result in rancidity of such products. When fats and oils are oxidised, they become rancid and their smell and taste change. Nitrogen provides an inert atmosphere for them.

Question 20.
Explain the following terms with one example each.

  1. Corrosion
  2. Rancidity.

Answer:

  1. Corrosion:
    When a metal is attacked by substances around it such as moisture, acids etc. it gets corroded and the process is called corrosion. For example, rusting of iron products.
  2. Rancidity:
    The Process in which fats and oils or food products made from fats or oils get oxidised resulting in a change of smell and the taste is called rancidity. For example, food items made from oil like chips becomes rancid if kept open for some time.

MP Board Class 10th Science Chapter 1 Additional Questions

MP Board Class 10th Science Chapter 1 Multiple Choice Questions

Question 1.
The burning of magnesium in air results in weight gain of products, it shows:
(a) Physical change in magnesium
(b) A chemical reaction between magnesium and oxygen of air shift
(c) No change
(d) Both (a) and (b) are correct.
Answer:
(b) A chemical reaction between magnesium and oxygen of air shift

Question 2.
Which one of the following is not a chemical reaction?
(a) Rusting of iron materials
(b) Cooking food
(c) Melting of ice
(d) Burning paper
Answer:
(c) Melting of ice

Question 3.
Which one of the following is not a characteristic of a chemical change?
(a) Change in colour
(b) Evolution of gas
(c) Change in temperature
(d) Change in size
Answer:
(d) Change in size

Question 4.
The balanced chemical reaction is represented with:
(a) The exact amount of material used.
(b) Correct formulation and their ratio being used.
(c) Products only.
(d) Reactants only.
Answer:
(b) Correct formulation and their ratio being used.

Question 5.
Which one of the following is not mentioned in a balanced chemical reaction?
(a) Temperature and pressure
(b) State of reactants and products
(c) Mass of reactants and products
(d) Formulae of reactants and products
Answer:
(b) State of reactants and products

Question 6.
The breaking down of food to CO2 and H2O is a type of:
(a) Exothermic reaction
(b) Endothermic reaction
(c) Reactions without temperature change
(d) Reactions without energy change
Answer:
(a) Exothermic reaction

Question 7.
The burning of magnesium in the air is represented by the following equation:
2Mg + O2 → 2MgO
It is a type of:
(a) Displacement reaction
(b) Double displacement
(c) Combination reaction
(d) Corrosion
Answer:
(c) Combination reaction

Question 8.
Which one of the following is a kind of combination reaction?
(a) 4Na(s) + O2(g) → 2Na2O(s)
(b) CuO(s)+ H2(g) → Cu(s) + H2O
(c) ZnO(s) + C(s) → Zn(s) + CO(g)
(d) None of above.
Answer:
(a) 4Na(s) + O2(g) → 2Na2O(s)

Question 9.
Photosynthesis is a:
(i) Endothermic reaction
(ii) Exothermic reaction
(iii) Catalytic reaction
Choose correct combination:
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) only
Answer:
(c) (ii) and (iii)

MP Board Solutions

Question 10.
Dry fruits if not consumed for a long time get rotten because of:
(a) Temperature change
(b) Rusting
(c) Rancidity
(d) None of the above
Answer:
(c) Rancidity

Question 11.
A chemical reaction involves:
(a) breaking of bonds
(b) formation of bonds
(c) no change
(d) both breaking and formation of bonds
Answer:
(d) both breaking and formation of bonds

Question 12.
A balanced chemical equation has:
(a) Equality of mass at both side of reaction.
(b) Equality of atoms at both side of reaction.
(c) Equality of numbers of elements at both side of reaction.
(d) All of the above.
Answer:
(d) All of the above.

Question 13.
Combustion of food at the cellular level of our body is:
(a) Oxidation
(b) Reduction
(c) Redox
(d) Heating
Answer:
(a) Oxidation

Question 14.
Lemon contains:
(a) Formic acid
(b) Citric acid
(c) Succinic acid
(d) Ascorbic acid.
Answer:
(b) Citric acid

Question 15.
Food gives a bad taste and a bad smell sometimes, because of:
(a) Rancidity
(b) Displacement
(c) Heating
(d) None
Answer:
(a) Rancidity

Question 16.
The formation of water from its components gases is an example of:
(a) Combination reaction
(b) Oxidation reaction
(c) Decomposition reaction
(d) Reduction reaction
Answer:
(a) Rancidity

Question 17.
The sign in a reaction indicates:
(a) release of gas
(b) equilibrium
(c) formation of a precipitate
(d) lowering of temperature
Answer:
(b) equilibrium

Question 18.
Silver develops a black colour layering after some time in the open air, because of:
(a) Rancidity
(b) Displacement
(c) Heating
(d) Corrosion
Answer:
(d) Corrosion

MP Board Class 10th Science Chapter 1 Very Short Answer Type Questions

Question 1.
Is given reaction a balanced reaction?
2Mg + O2 → 2MgO.
Answer:
Yes, it is.

Question 2.
Why gold and silver do not corrode?
Answer:
It is because they are not very reactive.

Question 3.
Why does copper vessel acquire green coating in the open atmosphere?
Answer:
It reacts with CO2 in the atmosphere and forms a layer of basic copper carbonate.

Question 4.
What type of reaction is the reaction between an acid and a base?
Answer:
Neutralization reaction.

Question 5.
What type of reaction is rusting of iron?
Answer:
Oxidation reaction.

Question 6.
What is the name of the gas which burns with a pop sound?
Answer:
Hydrogen gas.

Question 7.
Why is hydrogen peroxide stored in coloured bottles?
Answer:
To prevent photolytic decomposition of hydrogen peroxide, it is stored in coloured bottles.

Question 8.
Give two examples from everyday life situations where redox reactions are taking place.
Answer:
Corrosion and rancidity.

Question 9.
In electrolysis of water, what is the volume of gas collected over electrode?
Answer:
In water (H2O), hydrogen and oxygen are present in the ratio of 2:1 by volume.

Question 10.
What is the nature of change during a chemical reaction?
Answer:
It is a permanent change.

Question 11.
What kind of change occurs during the dissolution of sugar in the water?
Answer:
Physical change.

Question 12.
Give two examples of exothermic reactions.
Answer:
Formation of ammonia and digestion of food.

Question 13.
Write three forms of energy evolution during the exothermic reaction.
Answer:
Heat, sound and light evolution.

Question 14.
Where do we write or indicate the value of temperature used during a chemical reaction in a chemical equation?
Answer:
Over the arrow or with the arrow.

Question 15.
What kind of chemical reaction occurs during the formation of nitrogen mono-oxide from nitrogen and oxygen gas?
Answer:
Combination reaction.

Question 16.
Which catalyst is used in a given reaction?
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 8
Answer:
Pt or platinum.

Question 17.
Give an example of a decomposition reaction.
Answer:
Formation of calcium oxide from the decomposition of calcium carbonate,
CaCO3 \(\underrightarrow { \Delta } \) CaO + CO2

Question 18.
What kind of chemical reaction occurs while rusting of silver?
Answer:
Oxidation reaction.

MP Board Class 10th Science Chapter 1 Short Answer Type Questions

Question 1.
Write two important uses of the balancing equation.
Answer:

  1. It gives an idea of the ratio of reactants and products formed.
  2. It verifies the law of conservation of mass.

Question 2.
Write two important checklists to analyse whether the equation is balanced or not.
Answer:

  1. Equal number of atoms on both sides.
  2. Correct formulae of atoms and molecules on both sides.

Question 3.
Write two examples of each for physical and chemical change.
Answer:
Physical change:

  1. Melting of ice.
  2. Evaporation.

Chemical change:

  1. Cooking food.
  2. Photosynthesis.

Question 4.
Balance the following equations:

  1. HgO \(\underrightarrow { \Delta } \) Hg + O2
  2. KClO3 \(\underrightarrow { Mn{ O }_{ 2 } } \) KCl + O2

Answer:
Balanced equations are as follows:

  1. 2HgO \(\underrightarrow { \Delta } \) 2Hg + O2
  2. 2 KClO3 \(\underrightarrow { Mn{ O }_{ 2 } } \) 2KCl3 + 3O2

Question 5.
Name at least three types of chemical reactions with examples.
Answer:

  1. Combination reaction – Formation of water from gases.
  2. Decomposition reaction – Digestion.
  3. Oxidation reaction – Rusting.

Question 6.
What do you know about reversible reactions? Give an example of it.
Answer:
Reactions in which products reform the reactant under suitable conditions are called reversible reaction.
Example:
Formation of ammonia:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 9

Question 7.
Why gold plated articles over silver are available?
Answer:
Gold is more reactive than silver, hence during electrodeposition, it replaces silver and gets deposited as a coating. During electroplating, displacement reaction occurs.

Question 8.
What do you mean by balanced chemical equation?
Answer:
An equation that has an equal number of atoms of each element on both the sides of the equation is called the balanced chemical equation, i.e., the mass of the reactants is equal to the mass of the products.
e.g., 2Mg + O2 → 2MgO

Question 9.
Define rancidity.
Answer:
When fats and oils are oxidized, they become rancid and their smell and taste change. This process is known as rancidity.

Question 10.
Give an example of a decomposition reaction where energy is supplied in the form of light.
Answer:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 10

Question 11.
Name the oxidizing agent and reducing agent in the following equation:
3MnO2(s) + 4Al(s) → 3Mn(s) + 2Al2O3(s)
Answer:

  1. Reducing agent: Al
  2. Oxidizing agent: MnO2.

MP Board Solutions

Question 12.
Explain how respiration is an exothermic reaction.
Answer:
During digestion, food is broken down into simpler substances. For example, rice, potatoes and bread contain carbohydrates. These carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and provides energy. Hence, respiration is an exothermic process,
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 11

Question 13.
Write factors influencing the rate of a chemical reaction.
Answer:
Following are the factors that can influence the rate of reaction:

  1. Nature of reactants (i.e., fast or slow).
  2. The concentration of reactants.
  3. Surface Area.
  4. Temperature.
  5. Catalyst.

Question 14.
Name one metal that when placed in ferrous sulphate solution will discharge its green colour.
Answer:
Potassium (K):
‘K’ is more reactive than iron and will replace it. The green colour is due to the formation of a new compound.
2K + FeSO4 → K2SO4 + Fe

Question 15.
What is being added to the first reactant in reaction given below? (HOTS)

  1. 2FeCl2 + Cl2 → 2FeCl3.
  2. 2KI + H2O2 → I2 + 2KOH.

Answer:

  1. Cl2 is added.
  2. OH is added.

Question 16.
Which reactant is being oxidized in the following reactions? (HOTS)

  1. 2Cu + O2 → 2CuO
  2. H2S + Cl2 → 2HCl + S

Answer:

  1. ‘Cu’ is being oxidized.
  2. ‘H’ is being oxidised.

MP Board Class 10th Science Chapter 1 Long Answer Type Questions

Question 1.
Write balanced chemical equations for the following reactions:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 12
Answer:
Balanced chemical equations are:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 13

Question 2.
Write balanced chemical equations for the following reactions and write the name of reactions and ratio of products formed for each of them.
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 12
Answer:
Balanced chemical equations are:
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 13

Reactions and ratio of products:
(i) Redox or oxidation reaction ratio 2:1.
(ii) Oxidation reaction ratio 2:1.
(iii) Decomposition ratio 2:1.
(iv) Reversible and redox ratio 1 : 4.
(v) Decomposition ratio 2:1.

Question 3.
Rama bought a new set of an iron chair for her open garden space. After some months she found a dirty, brown coating over it. She painted the iron surface with colour to make it better for use. (Value-Based)

  1. Why the chair gets brown, dirty?
  2. Why did she use the paint?
  3. What value does Rama show?

Answer:

  1. The outer surface which is made of iron reacts with weather’s moisture and gets rusted. Hence, gets the brown dirty coating.
  2. Paint protects the outer surface of iron from getting rusted.
  3. She protected the garden chair and showed her sense of awareness.

MP Board Class 10th Science Chapter 1 Textbook Activities

Class 10 Science Activity 1.1 Page No. 1

Caution:

  • This activity needs the teacher’s assistance. It would be better if students wear suitable eyeglasses.
  • Clean a magnesium ribbon about 3-4 cm long by rubbing it with sandpaper.
  • Hold it with a pair of tongs. Burn it using a spirit lamp or burner and collect the ash so formed in a watch-glass as shown in below Figure. Burn the magnesium ribbon keeping away as far as possible from your eyes.
  • What do you observe?

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 14
Observations:
When we burn magnesium ribbon, it burns with crackling sound, dazzling light and produces ash. This ash contains oxide of magnesium,
2Mg(s) + O2(g) → 2MgO(s)

Class 10 Science Activity 1.2 Page No. 2

  1. Take a lead nitrate solution in a test tube.
  2. Add potassium iodide solution to this.
  3. What do you observe?

Observations:
Lead iodide is formed when lead nitrate reacts with potassium iodide. Lead iodide is insoluble in water and yellow in colour.
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 15

Class 10 Science Activity 1.3 Page No. 2

  1. Take a few zinc granules in a conical flask or a test tube.
  2. Add dilute hydrochloric acid or sulphuric acid to this below figure.

Caution:

  1. Handle the acid with care.
  2. Do you observe anything happening around the zinc granules?
  3. Touch the conical flask or test tube. Is there any change in its temperature?

Observations:
Zinc chloride is formed when zinc metal reacts with hydrochloric acid, bubbles of hydrogen gas are also observed in this process. The feat is evolved because it is an exothermic process.
Zn(s) + 2HCl(dil.) → ZnCl2(aq) + H2(g) + Δ
MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 16

Class 10 Science Activity 1.4 Page No. 6

  1. Take a small amount of calcium oxide or quick lime in a busker.
  2. Slowly add water to this.
  3. Touch the beaker as shown in Figure.
  4. Do you feel any change in temperature?

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 17
Observations:
Slaked lime is produced when calcium oxide reacts with water, it is an exothermic reaction and produces a huge amount of energy as heat. It is a combination reaction,
CaO(s) + H2O(l) → Ca(OH)2(aq) + Δ

Class 10 Science Activity 1.5 Page No. 6

  1. Take about 2 g ferrous sulphate crystals in a dry boiling tube.
  2. Note the colour of the ferrous sulphate crystals.
  3. Heat the boiling tube over the flame of a burner or spirit lamp as shown in Figure.

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 18
Observations:
When we heat ferrous sulphate crystals in a boiling tube, a brownish-black ferric oxide is formed, SO2 and SO3 gas are escaped. Their presence is observed by the smell in the air. It is a decomposition reaction,
2FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g)

Class 10 Science Activity 1.6 Page No. 8

  1. Take about 2 g lead nitrate powder in a boiling tube.
  2. Hold the boiling tube with a pair of tongs and heat it over a flame. as shown in Figure.
  3. What do you observe? Note down the change. If any.

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 19
Observations:
We will observe brown fumes emission from the test tube. Brown fumes signify nitrogen dioxide (NO2). It is a decomposition reaction.
2Pb(NO3)2(s) \(\underrightarrow { Heat } \) 2PbO(s) + 4NO2(g) + O2(g)

Class 10 Science Activity 1.7 Page No. 8

  1. Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in Figure.
  2. Connect these electrodes to a 6-volt battery.
  3. Fill the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water.
  4. Take two test tubes filled with water and invert them over the two carbon electrodes.
  5. Switch on the current and leave the apparatus undisturbed for some time.
  6. You will observe the formation of bubbles at both electrodes. These bubbles displace water in the test tubes.
  7. Is the volume of the gas collected the same in both the test tubes?
  8. Once the test tubes are filled with the respective gases. Remove them carefully.
  9. Test these gases one by one by bringing a burning candle close to the mouth of the test tubes.

Caution:

  1. This step must be performed carefully by the teacher.
  2. What happens in each case?
  3. Which gas is present in each test tube?

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 20
Observations:
On electrolysis of water, hydrogen and oxygen gas is produced. Hydrogen gas has twice the volume of oxygen gas. Oxygen gas let the burning continue while hydrogen gas burns with a pop sound. The gas collected at the cathode is oxygen while gas collected at the anode is hydrogen.

Class 10 Science Activity 1.8 Page No. 9

  1. Take about 2 g silver chloride in a china dish.
  2. What is its colour?
  3. Place this china dish in sunlight for some time Figure.
  4. Observe the colour of the silver chloride after some time.

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 21
Observations:
White silver chloride turns grey in sunlight due to the decomposition of silver chloride into silver and chlorine gas by light
2AgCl(s) \(\underrightarrow { Sunlight } \) 2Ag(s) + Cl2(g)

Class 10 Science Activity 1.9 Page No. 10

  1. Take three iron nails and clean them by rubbing with sandpaper.
  2. Take two test tubes marked as (A) and (B). In each test tube, take about 10mL copper sulphate solution.
  3. Tie two iron nails with a thread and immerse them carefully in the copper sulphate solution in test tube B for about 20 minutes [Fig. 1.8(a)]. Keep one iron nail aside for comparison.
  4. After 20 minutes, take out the iron nails from the copper sulphate solution.
  5. Compare the intensity of the blue colour of copper sulphate solutions in test tubes (A) and (B), [Fig. 1.8 (b)].
  6. Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept aside [Fig. 1.8 (b)].

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 22
Observations:

  1. When we dip the iron nail in copper sulphate solution its blue colour changes due to the formation of iron sulphate and represents displacement reaction.
    Fe(s) + CuSO4 (aq) → FeSO4(aq) + Cu(s)
  2. Copper sulphate in test tube A is dark blue and its colour in test tube B is light blue.
  3. An iron nail becomes brownish when we put them in test tube B because of Cu deposition over it and solution turns green due to formation of FeSO4.

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 23

Class 10 Science Activity 1.10 Page No. 10

  1. Take about 3 ml of sodium sulphate solution in a test tube.
  2. In another test tube, take about 3 ml of barium chloride solution.
  3. Mix the two solutions
  4. What do you observe?

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 24
Observations:
We get a white precipitate of BaSO4 when sodium sulphate reacts with barium chloride.
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl

Class 10 Science Activity 1.11 Page No. 12

  1. Heat a china dish containing about 1 g copper powder Figure.
  2. What do you observe?

MP Board Class 10th Science Solutions Chapter 1 Chemical Reactions and Equations 25
Observations:
Black colour copper oxide is formed.
2Cu + O2 \(\underrightarrow { \Delta } \) 2CuO

MP Board Class 10th Science Solutions

MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4

MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4

Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5 m) and (2.5l +0.5 m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 1
Solution:
(i) (2x + 5) × (4x – 3) = 2x (4x – 3) + 5(4x – 3) = 8x2 – 6x + 20x – 15 = 8x2 + 14x – 15
(ii) (y – 8) × (3y – 4) = y(3y – 4) – 8(3y – 4)
= 3 y2 – 4y – 24y + 32 = 3y2 – 28y + 32
(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) – 0.5 m (2.5l + 0.5m)
= 6.25l2 + 1.25lm – 1.25lm – 0.25m2 = 6.25l2 – 0.25m2
(iv) (a + 3b) × (x + 5) = a(x + 5) + 3b(x + 5)
= ax + 5a + 3 bx + 15 b
(v) (2pq + 3q2) × (3pq – 2q2)
= 2pq (3pq – 2q2) + 3q2 (3pq – 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6 p2q2 + 5pq3 – 6q4
MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 2
MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 3

MP Board Solutions

Question 2.
Find the product
(i) (5 – 2x)(3 + x)
(ii) (x + 7y)(7x – y)
(iii) (a2 + b)(a + b2)
(iv) (p2 – q2)(2p + q)
Solution:
(i) (5 – 2x) × (3 + x) = 5(3 + x) – 2x(3 + x)
= 15 + 5x – 6x – 2x2 – 15 – x – 2x2
(ii) (x + 7y)(7x – y) = x(7x – y) + 7y(7x – y)
=7x2 – xy + 49xy – 7y2 = 7x2 + 48xy – 7y2
(iii) (a2 + b) (a + b2) = a2 (a + b2) + b (a + b2)
= a3 + a2b2) + ab + b3
(iv) (p2 – q2) (2p + q) = p2 (2p + q) – q2 (2p + q)
= 2p3 + p2q – 2q2p – q3

MP Board Solutions

Get R Programming MCQ‘s with Answers & Solutions that checks your basic knowledge of R Programming.

Question 3.
Simplify.
(i) (x2 – 5)(x + 5) + 25
(ii) (a2 + 5)(b3 + 3) + 5
(iii) (t + s2)(t2 – s)
(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x – y)
(vi) (x + y)(x2 – xy + y2)
(vii) (1 .5x – 4y)(1 .5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
Solution:
(i) (x2 – 5) (x + 5) + 25
= x2(x +5) – 5(x + 5) +25
=x3+ 5x2 – 5x – 25 + 25 = r3+ 5r2 – 5x

(ii) (a2 + 5)(b3 + 3) + 5
= a2 (b3 + 3) + 5(b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 5b3 + 20

(iii) (t + s2) (t2 – s) = t(t2 – s) + s2(t2 – s)
= t3 – ts +s2t2 – s3

(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
=a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2a + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
=2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2
= 3x2 + 4xy – y2

(vi) (x + y) (x2 – xy + y2)
= x (x2 – xy + y2) + y (x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3 = x3 + y3

(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x (1.5x + 4y + 3) – 4y (1.5x + 4y + 3) – 4.5x + 12y
= 2.25x2 + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y
= 2.25x2 – 16y2

(viii) (a + b + c) (a + b – c)
= a (a+ b – c) + b (a+ b – c) + c (a + b – c)
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + 2 ab + b2 – c2

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

An online step by step percentage difference calculator.

Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.
(b) ₹ 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
(c) ₹ 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:
(a) We have,
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 13

(b) We have,
P = ₹ 18000
R = 10 % per annum
n = 2\(\frac{1}{2}\) years or 2.5 years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 14

Now, we calculate S.I. on this amount for \(\frac{1}{2}\) year at 10 % per annum.
∴ Amount after 2.5 years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 15
∴ Interest = A – P = 22869 -18000 = ₹ 4869

(c) We have,
P = ₹ 62500
R = 8 % per annum = 4 % per half year
n = 1\(\frac{1}{2}\) years = 3 half years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 16
∴ Interest = A – P = 70304 – 62500 = ₹ 7804

(d) We have,
P = ₹ 8000
R = 9 % per annum
= 4 % per half year
n = 1 year = 2 half years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 17
∴ Interest = A – P = 8736.20 – 8000 = ₹ 736.20

(e) We have,
P = ₹ 10000
R = 8% per annum = 4 % per half year
n = 1 year = 2 half years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 18
∴ Interest = A – P = 10816 -10000 = ₹ 816

MP Board Solutions

Fibonacci Calculator is a free online tool that displays the Fibonacci sequence for the given limit.

Question 2.
Kamla borrowed ₹ 26,400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint : Find 4 for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\) years).
Solution:
We have,
P = ₹ 26400
R = 15 % per annum
n = 2 years 4 months
At the end of 2 years,
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 19
Now, P = ₹ 34914
R = 15 % per annum
n = 4 months = \(\frac{1}{3}\) years
At the end of 2 years and 4 months,
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 20

Question 3.
Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
For Fabina, P = ₹ 12500
R = 12 % per annum
n = 3 years
For simple interest,
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 21
Interest = 17000 – 12500 = ₹ 4500
For Radha, P = ₹ 12500
R = 10 % per annum
n = 3 years
As this is compound interest
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 22
Interest = 16637.50 -12500 = ₹ 4137.50
Hence, Fabina pays more interest by 4500 – 4137.50 = ₹ 362.50

MP Board Solutions

RSD Calculator is Step by step explanation.

Question 4.
I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
We have, P = ₹ 12000
R = 6 % per annum
n = 2 years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 23
So, the extra amount he would have to pay = 1483.20 -1440 = ₹ 43.20

MP Board Solutions

Question 5.
Vasudevan invested X 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
We have,
P = ₹ 60000
R = 12 % per annum = 6 % per half year
(i) n = 6 months = 1 half year
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 24

Question 6.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\(\frac{1}{2}\) years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
Solution:
We have,
P = ₹ 80000
R = 10 % per annum = 5 % per half year
(i) If the interest is compounded annually
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 25

(ii) If interest is compounded half yearly
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 26
∴ Difference in amounts = 92610 – 92400
= ₹ 210

MP Board Solutions

Question 7.
Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:
We have,
P = ₹ 8000
R = 5 % per annum
(i) n = 2 years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 27

(ii) n = 3 years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 28
Hence, interest for 3rd year = 9261 – 8820 = ₹ 441

MP Board Solutions

Question 8.
Find the amount and the compound interest on ₹ 10,000 for 1\(\frac{1}{2}\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Solution:
We have,
P = ₹ 10000
R = 10 % per annum = 5 % per half year
n = 1\(\frac{1}{2}\) years = 3 half years
(i) If interest is compounded half yearly
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 29
Interest = 11576.25 – 10000 = ₹ 1576.25

(ii) If interest is compounded annually Amount after 1 year,
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 30
Thus, more interest would be generated if interest is calculated half yearly.

MP Board Solutions

Question 9.
Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\(\frac{1}{2}\) % per annum, interest being compounded half yearly.
Solution:
We have,
P = ₹ 4096
R = 12.5 % per annum = 6.25 % per half year
n = 18 months = 3 half years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 31

MP Board Solutions

Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
Solution:
We have, population in 2003 = 54000
Rate = 5% per annum
(i) Let population in 2001 be x.
The population in 2003
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 32
∴ Population = 48980 in 2001

(ii) For population in 2005
n = 2 years
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 33

MP Board Solutions

Question 11.
In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Initial count of bacteria = 506000
Rate = 2.5 % per hour
n = 2 hours
∴ Count after 2 hours
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 34

MP Board Solutions

Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Initial price = ₹ 42000
Rate of depreciation = 8% per annum
n = 1 year
MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 35

MP Board Class 8th Maths Solutions

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 14 Statistics Ex 14.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Percentage Decrease Calculator is a free online tool that displays the percentage decrease for the given amount.

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 1
Solution:
Median:
Let us prepare a cumulative frequency table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 2
Now, we have N = 68 ⇒ \(\frac{N}{2}=\frac{68}{2}\) = 34
The cumulative frequency just greater than 34 is 42 and it corresponds to the class 125 – 145.
∴ 125 – 145 is the median class.
∴ l = 125, cf = 22, f= 20 and h = 20
Using the formula,
Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 125 + \(\left[\frac{34-22}{20}\right]\) × 20
= 125 + \(\frac{12}{20}\) × 20 = 125 + 12 = 137 units.
Mean: Let assumed mean, a = 135
∵ Class size, h = 20
∴ ui = \(\frac{x_{i}-a}{h}=\frac{x_{i}-135}{20}\)
Now, we have the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 3
∴ \(\overline{x}\) = a + h × [\(\frac{1}{N}\) Σfiui] = 135 + 20 × \(\frac{7}{68}\)
= 135 + 2.05 = 137.05 units.
Mode:
∵ Class 125 – 145 has the highest frequency i.e., 20.
∴ 125 – 145 is the modal class.
We have: h = 20, l = 125 , f1 = 20, f0 = 13, f2 = 14
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 4
We observe that the three measures are approximately equal.

Use this median calculator to quickly find the median of a set of numbers. Just enter your numbers in the box and click on the button that says calculate.

Learn how to find Recursive Sequences Formula Calculator for arithmetic sequences.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 5
Solution:
Here, we have N = 60
Now, cumulative frequency table is:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 6
Since, median = 28.5 (Given)
∴ Median class is 20 – 30 and l = 20, f = 20, cf = 5 + x, N = 60
∴ l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20}\right]\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ 57 = 40 + 25 – x
⇒ x = 40 + 25 – 57 = 8
Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 45 – 8 = 7.
Thus x = 8, y = 7

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

The percentage difference calculator is here to help you compare two numbers.

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 7
Solution:
The given table is cumulative frequency distribution. We write the frequency distribution as given below :
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 8
∵ The cumulative frequency just greater than 50 is 78.
∴ The median class is 35 – 40.
Now, \(\frac{N}{2}\) = 50, l = 35, cf = 45, f = 33 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 9
Thus, the median age = 35.76 years.

Also you can check the age difference between your loved ones, friends using the Age difference Calculator.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

When the positive number a is rounded to the nearest tenth, the result is the number b.

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 10
Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5,
126.5 – 135.5 ………… 171.5 – 180.5.]
Solution:
After changing the given table as continuous classes we prepare the cumulative frequency table as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 11
The cumulative frequency just above 20 is 29 and it corresponds to the class 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
We have: \(\frac{N}{2}\) = 20, l = 144.5, f= 12, cf = 17 and h = 9
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 144.5 + \(\left[\frac{20-17}{12}\right]\) × 9
= 144.5 + \(\frac{3}{12}\) × 9 = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25 = 146.75
Median length of leaves = 146.75 mm.

The percentage off calculator generates lightning-fast results without asking the users to follow intricate procedures.

Question 5.
The following table gives the distribution of the life time of 400 neon lamps:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 12
Find the median life time of a lamp.
Solution:
To compute the median, let us write the cumulative frequency distribution as given below:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 13
Since, the cumulative frequency just greater than 200 is 216.
∴ The median class is 3000-3500 and so l = 3000, cf= 130, f = 86, h = 500
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\frac{70}{86}\) × 500 = 3000 + \(\frac{35000}{86}\)
= 3000 + 406.98 = 3406.98
Thus, median life time of a lamp = 3406.98 hours.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3

Online frequency distribution calculator tool makes the calculation faster and it displays the frequency distribution in a fraction of seconds.

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 14
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution:
Median: The cumulative frequency distribution table is as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 15
Since, the cumulative frequency just greater than 50 is 76.
∴ The class 7-10 is the median class.
We have, \(\frac{N}{2}\) = 50 , f = 7, cf = 36, f = 40 and h = 3
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 16
Mode:
Since the class 7 – 10 has the maximum frequency i.e., 40.
∴ The modal class is 7 – 10.
So, we have l = 7,h = 3, f1 = 40, f0 = 30, f2 = 16
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 17
Thus, the required median = 8.05, mean = 8.32 and mode = 7.88.

Guidelines for conversion of CGPA into percentage.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 18
Solution:
We have cumulative frequency table as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 19
The cumulative frequency just greater than 15 is 19, which corresponds to the class 55 – 60.
So, median class is 55-60 and we have \(\frac{N}{2}\) = 15,
l = 55, f = 6, cf = 13 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 20
Thus, the required median weight of the students = 56.67 kg.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Find the value of x in each of the given triangles.

Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14; x – y = 4
(ii) s – f = 3; \(\frac{s}{3}+\frac{t}{2}\) = 6
(iii) 3x – y = 3; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
(v) \(\sqrt{2} x+\sqrt{3} y\) = 0; \(\sqrt{3} x-\sqrt{8} y\) = 0
(vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2, \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
(i) x + y = 14 … (1),
x – y = 4 …. (2)
From (1) , we get x = (14 – y) …. (3)
Substituting value of x in (2) , we get
(14 – y) – y = 4 ⇒ 14 – 2y = 4 ⇒ -2y = -10 ⇒ y = 5
Substituting y = 5 in (3), we have
x = 14 – 5 ⇒ x = 9
Hence, x = 9, y = 5

(ii) s – t = 3 … (1)
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 1
From (1), we have s = (3 + t) … (2)
Substituting this value of s in (2), we get
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 2
Substituting, t = 6 in (3) we get,
S = 3 + 6 = 9
Thus, S = 9, f = 6

(iii) 3x – y = 3 … (1),
9x – 3y = 9 … (2)
From (1) , y = (3x – 3)
Substituting this value of y in (2),
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9 ⇒ 9 = 9 which is true,
Eq. (1) and eq. (2) have infinitely many solutions.

(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 …. (2)
From the equation (1),
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 3
Substituting the value of y in (2), we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 4
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 5
Substituting the value of x in (1), we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 6

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Solve using Substitution Calculator is a free online tool that displays the solution of the pair of linear equations using the substitution method.

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 13

Enter the equation A and B in the substitution calculator for solving the linear equations.

Question 3.
Form the pair of linear equations for the following problems and find their solution by
substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for 13800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let the two numbers be X and y such that x > y
It is given that
Difference between two numbers = 26
∴ x – y = 26 … (1)
Also one number = 3 [the other number]
⇒ x = 3y … (2)
Substituting x = 3y in (1) , we get 3y – y = 26 ⇒ 2y = 26
Now, substituting y = 13 in (2) , we have
x = 3(13) ⇒ x = 39
Thus, two numbers are 39 and 13.

(ii) Let the two angles be x and y such that x > y
∵ The larger angle exceeds the smaller by 18° (Given)
∴ x = y + 18°…. (1)
Also, sum of two supplementary angles = 180°
∴ x + y = 180° … (2)
Substituting the value of x from (1) in (2) , we get,
(18° + y) + y = 180°
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 7
Substituting, y = 81° in (1) , we get
x = 18° + 81° = 99°
Thus, x = 99° and y = 81°

(iii) Let the cost of a bat = ₹ x
And the cost of a ball = ₹ y
∵ [cost of 7 bats] + [cost of 6 balls] = ₹ 3800
⇒ 7x + 6y = 3800 … (1)
Also, [cost of 3 bats] + [cost of 5 balls] = ₹ 1750
3x + 5y = 1750 …. (2)
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 8
Substituting this value of y in (1) , we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 9

(iv) Let fixed charges = ₹ x
and charges per km = ₹ y
∵ Charges for the journey of 10 km = ₹ 105 (Given)
∴ x + 10y = 105 … (1)
and charges for the journey of 15 km = ₹ 155
∴ x + 15y = 155 … (2)
From (1) , we have, x = 105 – 10y …. (3)
Putting the value of x in (2) , we get
(105 – 10y) + 15y = 155
⇒ 5y = 155 – 105 = 50 ⇒ y = 10
Substituting y = 10 in (3) , we get
x = 105 – 10(10) ⇒ x = 105 – 100 = 5
Thus, x = 5 and y = 10
⇒ Fixed charges = ₹ 5
and charges per km = ₹ 10
Now, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255
∴ The charges for 25 km journey = ₹ 255

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

(v) Let the numerator = x
and the denominator = y
∴ Fraction = \(\frac{x}{y}\)
According to the given condition,
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 10
⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y + 4 = 0 … (1)
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 11
Substituting this value of x in (1) , we have
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 12

(vi) Let the present age of Jacob = x years
and the present age of his son = y years
∴ 5 years hence: Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
According to given condition,
[Age of Jacob] = 3[Age of his son]
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x – 3y -10 = 0 … (1)
5years ago : Age of Jacob = (x – 5) years,
Age of his son = (y – 5) years
According to given condition,
[Age of Jacob] = 7[Age of his son]
∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35
⇒ x – 7y + 30 = 0 … (2)
From (1) , x = [10 + 3y] … (3)
Substituting this value of x in (2) , we get
(10 + 3y) – 7y + 30 = 0
⇒ -4y = -40 ⇒ y = 10
Now, substituting y = 10 in (3) ,
we get x = 10 + 3(10)
⇒ x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and present age of his son = 10 years

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2

In this article, we will share MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2

Question 1.
If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get \(\frac{1}{8}\).What is the number?
Solution:
Let the number be x.
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 1

Solving systems of linear equations calculator using Elimination method calculator.

Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What is the length and the breadth of the pool?
Solution:
Assume that the breadth of the rectangular pool be x m.
∴ Length = 2 + 2x
Perimeter of the rectangular pool = 2(Length + Breadth)
⇒ 154 = 2(2 + 2x + x)
⇒ 154 = 2(2 + 3x)
⇒ 154 = 4 + 6x
Transposing 4 to L.H.S., we get
154 – 4 = 6x ⇒ 150 = 6x
Now, dividing both sides by 6, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 2
∴ The breadth of the pool is 25 m and length of the pool = (2 + 2 × 25) m = (2 + 50) m = 52 m.

Question 3.
The base of an isosceles triangle is \(\frac{4}{3}\) cm. The perimeter of the triangle is 4\frac{2}{15} cm. What is the length of either of the remaining equal sides?
Solution:
The base of an isosceles triangle = \(\frac{4}{3}\) cm
Let x cm be the length of both of the remaining equal sides of the isosceles triangle.
Perimeter of the isosceles ∆ABC = AB + BC + CA
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 3
Transposing \(\frac{4}{3}\) to L.H.S., we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 4
∴ The length of either of the remaining equal sides is 1\(\frac{2}{5}\) cm.

MP Board Solutions

Least Common Multiple (LCM) of 15 and 20.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let one number be x and other number is 15 + x.
Now, x + 15 + x = 95 ⇒ 2x + 15 = 95
Transposing 15 to R.H.S., we get
2x = 95 – 15 ⇒ 2x = 80
Now, dividing both sides by 2, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 5
∴ One number is 40 and other number is 15 + 40 = 55.

Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Since, two numbers are in the ratio 5 : 3.
Let the two numbers be 5x and 3x.
Now, 5x – 3x = 18 ⇒ 2x = 18
Dividing both sides by 2, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 6
⇒ x = 9
∴ The required numbers are 5x = 5 × 9 = 45 and 3x = 3 × 9 = 27.

MP Board Solutions

Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let three consecutive integers be x, (x + 1) and (x + 2).
Now, x+x + 1+ x + 2 = 51 ⇒ 3x + 3 = 51
Transposing 3 to R.H.S., we get
3x = 51 – 3
⇒ 3x = 48
Dividing both sides by 3, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 7
Thus, the required three consecutive integers are 16,17 and 18.

MP Board Solutions

Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be 8x, 8(x + 1) and 8(x + 2).
Now, 8x + 8x + 8 + 8x + 16 = 888
⇒ 24x + 24 = 888
Transposing 24 to R.H.S., we get
24x = 888 – 24 ⇒ 24x = 864
Dividing both sides by 24, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 8
Thus, the required three consecutive multiples of 8 are 8 × 36 = 288, 8 × 37 = 296 and 8 × 38 = 304.

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let three consecutive integers be x, (x + 1) and (x + 2).
When they are multiplied by 2, 3 and 4, we get 2x, 3(x +1) and 4(x + 2) i.e., 2x, (3x + 3) and (4x + 8) respectively.
Now, 2x + 3x + 3 + 4x + 8 = 74 × 9x + 11 = 74
Transposing 11 to R.H.S., we get
9x = 74 – 11 ⇒ 9x = 63
Now, dividing both sides by 9, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 9
Thus, the required three consecutive integers are 7, 8 and 9.

MP Board Solutions

Question 9.
The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Since, the ages of Rahul and Haroon are in the ratio 5 : 7.
Let Rahul’s present age be 5x years and Haroon’s present age be 7x years.
After 4 years, we have
Rahul’s age = 5x + 4 and Haroon’s age = 7x + 4
Now, 5x + 4 + 7x + 4 = 56 ⇒ 12x + 8 = 56
Transposing 8 to R.H.S., we get
12x = 56 – 8 ⇒ 12x = 48
Now, dividing both sides by 12, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 10
Thus, Rahul’s present age = 5 × 4 = 20 years and Haroon’s present age = 7 × 4 = 28 years

Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Solution:
We know that the number of boys and girls are in ratio 7 : 5.
Let the number of boys be 7x and the number of girls be 5x.
Now, 7x = 5x + 8
Transposing 5x to L.H.S., we get
7x – 5x = 8 ⇒ 2x = 8
Dividing both sides by 2, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 11
Thus, the required number of boys = 7 × 4 = 28 and the required number of girls = 5 × 4 = 20
Hence, total number of students = 28 + 20 = 48

MP Board Solutions

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let Baichung’s grandfather’s age be x years.
∴ Baichung’s father’s age = (x – 26) years
Baichung’s age = [(x – 26) – 29] years = (x – 55) years
Since, sum of the ages of all the three is 135 years. So, x + x – 26 + x – 55 = 135
⇒ 3x – 81 = 135
Transposing – 81 to R.H.S., we get
3x = 135 + 81 ⇒ 3x = 216
Dividing both sides by 3, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 12
Thus, Baichung’s grandfather’s age = 72 years
Baichung’s father’s age = (72 – 26) years = 46 years
Baichung’s age = (72 – 55) years = 17 years

MP Board Solutions

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let the present age of Ravi be x years.
After 15 years, Ravi’s age = (15 + x) years
Now, 15 + x = 4x
Transposing x to R.H.S., we get
15 = 4x – x
⇒ 15 = 3x
Dividing both sides by 3, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 13
Thus, the present age of Ravi is 5 years.

Question 13.
A rational number is such that when you multiply it by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, you get \(-\frac{7}{12}\).What is the number?
Solution:
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 14

MP Board Solutions

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of notes of ₹ 100, ₹ 50 and ₹ 10 be 2x, 3x and 5x, respectively.
∴ The amount Lakshmi has from ₹ 100 notes : ₹ (2x × 100) = ₹ 200x from ₹ 50 notes : ₹ (3x × 50) = ₹ 150x from ₹ 10 notes : ₹ (5x × 10) = ₹ 50x Now, 200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000 ⇒ x = 1000
∴ Number of notes of ₹ 100 = 2x
= 2 × 1000 = 2000
Number of notes of ₹ 50 = 3x = 3 × 1000 = 3000
and number of notes of ₹ 10 = 5x = 5 × 1000
= 5000

MP Board Solutions

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins be x.
Number of ₹ 2 coins = 3x and number of ₹ 1 coins = 160 – x – 3x
The total amount
from ₹ 5 coins : ₹ 5 × x = × 5x
from ₹ 2 coins : ₹ 2 × 3x = ₹ 6x
and from ₹ 1 coins : ₹ 1[160 – x – 3x]
= ₹ [160 – 4x]
Now, 5x + 6x + 160 – 4x = 300
⇒ 7x + 160 = 300
Transposing 160 to R.H.S., we get
7x = 300 – 160
⇒ 7x = 140
Dividing both sides by 7, we get
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 15
⇒ x = 20
The number of ₹ 1 coins = 160 – 20 – 3 × 20 = 160 – 20 – 60 = 80
Number of ₹ 2 coins = 3 × 20 = 60
Number of ₹ 5 coins = 20

MP Board Solutions

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be x.
Then, the number of losers will be 63 – x.
Since, the winner gets a prize of ₹ 100 and a loser gets a prize of ₹ 25.
Amount got by winners = ₹ 100x
And amount got by losers = ₹ (63 – x) × 25 But total prize money is ₹ 3000.
Therefore, 100x + (63 – x) × 25 = 3000
⇒ 100x + 63 × 25 – 25x = 3000
⇒ 75x + 1575 = 3000
MP Board Class 8th Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 16
So, the number of winners is 19.

MP Board Class 8th Maths Solutions

MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2

MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2

Question 1.
Determine if the following are in proportion.
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(c) 24, 28, 36, 48
(d) 32, 48, 70, 210
(e) 4, 6, 8, 12
(f) 33, 44, 75, 100
Solution:
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 1
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 2

Geometric Sequence Calculator is a free online tool that displays the geometric sequence for the given first term and the common ratio calculator.

Question 2.
Write True (T) or False (F) against each of the following statements:
(a) 16 : 24 :: 20 : 30
(b) 21 : 6 :: 35 : 10
(c) 12 : 18 :: 28 : 12
(d) 8 : 9 :: 24 : 27
(e) 5.2 : 3.9 :: 3 : 4
(f) 0.9 : 0.36 :: 10 : 4
Solution:
(a) True
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 3

MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2

Question 3.
Are the following statements true?
(a) 40 persons : 200 persons = Rs. 15 : Rs. 75
(b) 7.5 litres: 15 litres = 5 kg : 10 kg
(c) 99 kg : 45 kg = Rs. 44 : Rs. 20
(d) 32 m: 64 m = 6 sec : 12 sec
(e) 45 km : 60 km = 12 hours : 15 hours
Solution:
(a) 40 persons : 200 persons
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 4
∴ 40 persons : 200 persons = Rs. 15 : Rs. 75
Hence, the statement is true.
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 5

Question 4.
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and Rs. 40: Rs. 160
(b) 39 litres: 65 litres and 6 bottles: 10 bottles
(c) 2 kg : 80 kg and 25 g : 625 g
(d) 200 ml: 2.5 litre and Rs. 4: Rs. 50
Solution:
(a) 25 cm : 1 m = 25 cm : (1 × 100) cm
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 6
Since the ratios are equal, therefore these are in proportion.
Middle terms are 1 m and Rs. 40 and extreme terms are 25 cm and Rs. 160.
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 7
Since the ratios are equal, therefore these are in proportion.
Middle terms are 65 litres and 6 bottles and extreme terms are 39 litres and 10 bottles.

MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 8
Since the ratios are not equal, therefore these are not in proportion.
(d) 200 ml : 2.5 litres = 200 ml: (2.5 × 1000) ml
MP Board Class 6th Maths Solutions Chapter 12 Ratio and Proportion Ex 12.2 9
Since the ratios are equal, therefore these are in proportion.
Middle terms are 2.5 litres and Rs. 4 and extreme terms are 200 ml and Rs. 50.

MP Board Class 6th Maths Solutions