Class 12

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Miscellaneous Exercise Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

MP Board Class 12th Chemistry Important Questions Chapter 2 Solutions

Solutions Important Questions

Very Short Answer Type Questions

Question 1.
Write formula of van’t Hoff factor ‘i’
Answer:
van’t Hoff factor, i = \(\frac {Observed value of colligative property}{Calculated value of colligative property}\)

Question 2.
Write down van’t Hoff equation. Give formula used for calculating molecular mass with its help.
Answer:
van’t Hoff equation, πV = nRT
or π = \(\frac { nRT}{V}\)
or π = \(\frac { WRT}{MV}\)
or π = \(\frac { WRT}{πV}\)
Where, W is mass of solute, R is Solution constant, T is temperature, n is osmotic pressure and V is volume of Solution.

The normality formula of a solution is the gram equivalent weight of a solute per liter of solution.

Question 3.
Define Normality.
Answer:
Number of gram equivalent of solute present in one litre of Solution: is called normality. It is represented by N.
Normality (N) = \(\frac {Number of gram equivalent of solute}{Volume of Solution: in litres}\)
or N = \(\frac {Mass of solute in grams }{Equivalent mass of solute}\) x \(\frac { 1000}{ Volume of Solution: in ml }\)
∵ [No. of gram equivalent =\(\frac {Mass of solute in grams}{Equivalent mass of solute}\)]
Normality of a Solution changes with temperature as it is based on mass – volume relationship and volume changes with change in temperature.

Question 4.
Differentiate between Molarity and Molality.
Answer:
Differences between Molarity and Molality :
Molarity (M):

• Molarity involves the total volume of Solution.
• In molarity, gram moles of solute are dissolved in 1 litre of Solution:.
• Molarity changes with temperature because volume changes with temperature.

Molality (m):

• Molality involves the mass of solvent.
• In molality, gram moles of solute are dissolved in 1 kg of solvent. Here volume of Solution is not considered.
• Molality is independent of temperature as it takes mass into consideration.

Question 5.
Explain the following term : Parts per million.
Answer:
Parts per million:
When a solute is present in very minute amounts (in traces), the concentration is expressed in parts per million abbreviated as ppm. The parts may be of mass or volume. It is the parts of a component per million parts of the Solution.
i.e., ppm = \(\frac {Mass of component A}{Total mass of Solution:}\) x 10⁶
or ppm = \(\frac {Volume of component A}{Total volume of Solution:}\) x 10⁶
Where, ppm. is the concentration of component A in parts per million.

Question 6.
What role does the molecular interaction play in a Solution: of alcohol and water? (NCERT)
Answer:
Alcohols dissolve in water due to formation of inter – molecular H – bonding with water.

Question 7.
Why do gases always tend to be less soluble in liquids as the temperature is raised? (NCERT)
Answer:
Gas + Liquid ⇌ Dissolved gas   ∆ H = – ve
The dissolution of a gas in a liquid is exothermic process. Therefore in accordance with Le – Chatelier’s principle with increase in temperature, the equilibrium shifts in back-ward direction. Therefore, the solubility of gas in Solution decreases with the rise in temperature.

Question 8.
State Henry’s. (NCERT)
Answer:
Henry’s law:
According to this law, ‘The mass of a gas dissolved per unit volume of a solvent at constant temperature, is proportional to the pressure of the gas with which the solvent is in equilibrium’. Let in unit volume of solvent, mass of the gas dissolved is m and equilibrium pressure be P, then m ∝P or m = KP, where K is a constant.

Question 9.
What are colligative properties?
Answer:
The physical properties of Solution which depend on the number of total particles present in Solution and the ratio of number of particles of solute not on the nature of solute particles are known as colligative properties.

Question 10.
Give an example of Solution: of solid in solid.
Answer:
Mixture of copper and gold is an example of solid in solid.

Question 11.
On dissolving ethanoic acid in benzene, experimental molecular mass of ethanoic acid is generally found to be double. Why?
Answer:
On dissolving ethanoic acid in benzene, it dimerizes due to the formation of hydrogen bond. Thus, its experimental value is generally found to be double.

Question 12.
What is vapour pressure? What is the effect of temperature on it?
Answer:
Pressure exerted by the vapours on the surface of a liquid at the state of equilibrium is known as vapour pressure. On increasing the temperature vapour pressure increases.

Question 13.
What type of deviation is represented by acetone and CS₂ Solution?
Answer:
Positive deviation.

Osmotic pressure calculator … The factor i is also called the dissociation factor or the van’t Hoff factor.

Question 14.
Why is CaCl₂ used for clearing off snow from the roads?
Answer:
On adding CaCl₂, freezing point of water decreases. Therefore, CaCl₂ is used for clearing off snow from the roads.

Question 15.
Based on solute – solvent interactions, arrange the following in order of increasing solubility in n – octane and explain :
Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
For solubility we know Tike dissolves like’, n – octane is a non – polar solvent, hence non-polar compounds will be more soluble.
KCl < CH₃OH < CH₃CN < Cyclohexane.

Question 16.
What is transition temperature?
Answer:
The temperature at which the nature of solubility changes (i.e., first it increases, then decreases) is known as transition temperature. Solubility of sodium sulphate in water increases upto 32.4, then it starts decreasing. Thus, 32.4°C is the transition temperature of sodium sulphate.

Question 17.
Define molarity and molality. (NCERT)
Answer:
Molarity “is defined as number pf gram mole§ of solute dissolved in a litre of Solution. It is denoted by M.
Molality (M) = \(\frac {Mass of solute in gram per litre}{ Molecular mass of solute}\)

Molality is defined as number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m.
Molality (m) = \(\frac {Mass of solute in kg of solvent}{Gram molecular mass of solute}\)

Solutions Short Answer Type Questions

Question 1.
Write down Raoult’s law.
Answer:
The vapour pressure of a solution containing non – volatile solute is directly pro – portional to the mole fraction of the solute.
Mathematically,
\(\frac { { X }_{ A } }{ { X }_{ A }+{ X }_{ B } } \) = \(\frac { P{ A }^{ 0 } }{ { P }_{ A }+{ P }_{ B } } \)
Where, P⁰_A = Vapour pressure of pure solvent, P_A = Vapour pressure of solvent in solution, X_B = Mole fraction of solute.

Question 2.
What is Azeotropic mixture? They are of how many types? (MP 2018)
Answer:
Azeotropic mixture is the mixture of liquids which boil at one temperature with-out any change in composition. For example, at the composition of 95 6% alcohol and 4.4% water. It form an azeotropic mixture which boils at 78.13°C. Components of this mixture cannot be separated fully by fractional distillation.

They are of two types :
1. Low boiling azeotropic mixture:
Such Solutions which represent positive deviation towards Raoulfs law i.e. their vapour pressure is high thus their boiling point is low are known as low boiling azeotropic mixture.

Example:

• CS₂ + Acetone
• C₂ H₅ OH + n – hexane.

2. High boiling azeotropic mixture:
Such Solutions which represent negative deviation towards Raoult’s law i.e. their vapour pressure is low thus their boiling point is high are known as high boiling azeotropic mixture.

Example:

• Acetone + Chloroform
• Ether + Chloroform.

Question 3.
What are ideal and non – ideal Solutions? Explain with example.
Answer:
Ideal Solutions:
Ideal Solutions are those Solutions in which Raoult’s law can be applied completely for all concentrations of the Solutions and at all temperatures.

Condition for ideal Solutions are following :

1. P_A = P_A⁰ X _A and P_B= P⁰_B X B
2. ∆V_mixing = 0 and
3. ∆H_mixing = 0 and

Example:
C₂H₅Br + C₂H₅Cl, C₆H₆ + C₆H₅CH₃, CCl₄ + SiCl₄etc.

Non – ideal Solutions:
Solutions in which Raoult’s law cannot be applied completely for all concentrations and temperatures are called non – ideal Solutions.
For these Solutions:

1. P_A ≠ P⁰_A X_A and p_B ≠ P⁰_BX_B 
2. ∆V_mixing ≠ 0 and
3. ∆H_mixing ≠ 0 and

Example:
Benzene + Acetone, CHCl₃ + HNO₃ etc.

Question 4.
Establish van’t Hoff Solution: equation. (MP 2018)
Answer:
Osmotic pressure of dilute Solution of a non – volatile solute is proportional to absolute temperature of the Solution at constant concentration. This is known as van’t Hoff law.
π ∝ T .. (1)
Derivation : Osmotic pressure n of a Solution: is directly proportional to a molar concentration.
π ∝ C
By eqn.(l) and eqn.(2) π ∝ C (T is constant) .. (2)
or π = RCT .. (3)
Where, R = Gas constant
or
C = \(\frac {1}{V}\)
π = \(\frac {RT}{V}\) .. (4)
πV = RT .. (5)
This is known as van’t Hoff Ideal Solution equation.

Question 5.
Define the following:

1. Molal elevation boiling point constant.
2. Molal freezing point depression constant. (MP 2018)

Answer:
1. Molal elevation boiling point constant : Molal elevation constant can be defined as “The elevation in boiling point of the Solution in which 1 gm of solute is dissolved in 1000 gm of solvent.”
∴ Elevation in boiling point
∆ T_b ∝ m
∆ T_b = K_bm
m= 1, ∆ T_b = K_b
Where K_b = Molal boiling point elevation constant.

2. Molal freezing point depression constant : Molal depression constant may be defined as “The depression in freezing point for 1 molal Solution: i.e., Solution: in which 1gm mole of solute is dissolved in 1000 gm of solvent.”
∴ Depression in freezing point
∆ T_f ∝ m
∆ T_f = k_f
If m = 1, ∆ T_f = k_f
Where K_f = Molal freezing point depression constant.

Question 6.
What is Raoult’s law? Establish its mathematical expression.
Or,
What is Raoult’s law? How can molar mass of a non – volatile solute be determined with its help?
Answer:
Raoult’s law:
For a Solution in which solute is non – volatile, the Raoulf s law may be stated as following :
“At any constant temperature, vapour pressure of solvent collected above the Solution of non – volatile solute, is directly proportional to the mole fraction of solute.” If a non – volatile solute is added to a volatile solvent, the vapour pressure of the solvent decreases. The vapour pressure of the solvent is directly proportional to its mole fraction. As the solute is non – volatile, the vapour pressure of the Solution (P) will be equal to the vapour pressure of the solvent (PA).
P = P_A ∝ X_A
or P_A=KX_A .. (1)
Where, K = Proportionality constant.
Apply eqn. (1) for pure solvent if X_A= 1 and P_A = P⁰_A
Then, P⁰_A = K x 1 .. (2)
Where, p⁰_A= Vapour pressure of pure solvent.
Putting the value of K from eqn. (2) in eqn. (1),
P_A=P⁰_AX_A .. (3)
Where, X_A is the mole fraction of the solvent in the Solution.
The mole fraction of solute is represented as XB.
So, X_A+X_B=1
or X_A= 1 – X_B .. (4)
From eqns. (3) and (4),
P_A= P⁰_A(1-X_B) = P⁰ – P⁰_AX_B
X_B = \(\frac { { P }_{ A\quad }^{ 0 }-\quad { P }_{ A } }{ { P }_{ A }^{ 0 } } \)
P⁰_A is lowering in vapour pressure and \(\frac { { P }_{ A\quad }^{ 0 }-\quad { P }_{ A } }{ { P }_{ A }^{ 0 } } \) is relative lowering in vapour pressure.

On the basis of equation (5) Raoult’s law can be defined as “The relative lowering in vapour pressure of a Solution containing non – volatile solute is equal to mole fraction of solute”.

Question 7.
Give relation between elevation in boiling point and molecular mass of solute.
Answer:
Relation between elevation in boiling point and molecular mass of solute:
Suppose, W_B gram of non – volatile solute dissolve in WA gram of solvent and the molecular mass of non – volatile solute is M_B gram. Then, molality, m will be

Question 8.
What is van’t Hoff factor? How does it equation?
Answer:
van’t Hoff Equation : van’t Hoff equation

Question 9.
What are constant boiling mixture? Write three differences in Ideal solution and Non – ideal Solutions. (MP 2011,15,16)
Answer:
Constant boiling mixtures or azeotropic mixture. A Solution which distils with out change in composition is called azeotropic mixture.
Differences between Ideal and Non – ideal Solution:

Question 10.
Define the following :

1. Reverse osmosis
2. Isotonic Solution
3. Semi – permeable membrane.

Answer:
1. Reverse osmosis:
It may be noted that if a pressure higher than the osmotic pressure is applied on the Solution the solvent will flow from the Solution into the pure solvent through the SPM. The process is called reverse osmosis.

2. Isotonic Solutions:
The Solution:s which have same osmotic pressure are called as isotonic Solution:s. For example, 0.91% Solution of NaCl called as saline water is isotonic with human blood corpuscles. That is why medicines are mixed with saline water before intravenous injections.

Osmotic pressure is a colligative property and depends on the number of solute particles in a Solution So, the isotonic Solution:s must have same number of solute particles in a given volume of Solution Consequently, the isotonic Solution:s are those which have same molality, if that the solute neither associate nor dissociate in the Solution.

3. Semipermeable membrane:
These are the membrane which allow the movement of the solvent molecules through them. The membrane appear to be continuous sheets or films. But they have very tiny holes or pores which are semimicro in nature. Through the holes, only the molecules of solvent can pass while those of bigger solute molecules cannot pass through.

Solutions Long Answer Type Questions

Question 1.
Write five differences in Solution: having Positive deviation and Negative deviation.
Answer:
Differences between Positive deviation and Negative deviation :

Question 2.
Explain in brief Berkeley and Hartley’s method of osmotic pressure measurement and state its uses.
Answer:
Berkeley and Hartley’s method:
In this method, pressure is applied over the Solution to stop the flow of solvent. This pressure is equivalent to osmotic pressure.
In this method, the apparatus consists of a strong vessel made up of steel in which porous pot is fitted. In the porous pot, copper ferrocyanide semipermeable membrane is deposited. The porous pot is fitted with a capillary tube on one side and a water reservoir on the other side. A piston and pressure gauge are fitted to the steel vessel.

The porous pot and steel vessel are filled with water and Solution respectively. Osmosis takes place and water moves into the steel vessel from the porous pot through the semipermeable membrane. This is shown by fall in water level in the capillary tube. This flow of water is stopped by applying external pressure on the Solution with the help of piston.

This method has the following advantages :

1. It takes comparatively lesser time to determine osmotic pressure.
2. Concentration of Solution: does not change, hence better results are obtained.
3. As high pressure is not exerted over semipermeable membrane, it does not break.
4. High osmotic pressure can be measured.

Question 3.
What is molal freezing point depression constant? Derive the formula to establish relation between molal freezing point depression constant and molecular mass of solute.
Or,
What is molal freezing point depression constant? Show that depression in freezing point is a colligative property. How can molecular mass of solute be determined from depression in freezing point?
Answer:
Molal freezing point depression constant is equal to depression in freezing point of the Solution when 1 gm mole is dissolved in 1000 gm of solvent. It is represented by i.e.,
K_f = \(\frac {depression in freezing point}{Number of moles}\)
or ∆ T_f = K_f x m
As K_f is molal freezing point depression constant, then
∆ T_f ∝m
Where, m = Number of moles dissolved in 1000 gm solvent.
Thus, depression in freezing point is proportional to molality of Solution. Molality is directly proportional to number or molecules of solute substance. Therefore, depression in freezing point is a colligative property.
Calculation of molecular mass of solute:
By determination of depression in freezing point, the Molecular mass of non – volatile solute can be determined.
For a Solution: of non – volatile solute,
∆T_f = K_f x m .. (1)
Let WB gram non – volatile solute is dissolved in w_A gram solvent and molecular mass of solute is M_B.

Question 4.
What is elevation in boiling points? How addition of a non – volatile solute elevates the boiling point of a solvent? Explain it with the help of graph diagram.
Answer:
The vapour pressure of the Solution:
containing a non – volatile solute is always less than that of pure solvent. Therefore, the Solution has to be heated to higher temperature so that its vapour pressure become equal to the atmospheric pressure. Thus, the boiling point of Solution (T_b)is always higher than the boiling point of solvent (T_b⁰). The difference T_b – T_b⁰ is called elevation in boiling point.

If we plot graph between temperature and vapour pressure of a pure solvent and its Solution, then following curve is obtained. Curve AB gives the vapour pressure for the pure solvent and the curve CD gives the vapour pressure of the Solution: at different temperature. At temperature T_b⁰ the vapour pressure of the solvent becomes equal to the atmospheric pressure hence it boils at T_b⁰.

Now, by the addition of non – volatile solute, lowering of vapour pressure of the Solution takes place. And to increase the vapour pressure of the Solution to become equal to atmospheric pressure, the temperature rises. Hence, at T_b the Solution boils. Thus, the boiling point is now elevated from T_b⁰ to T_b. The rise in temperature that results by the addition of a non – volatile solute in a solvent is termed as elevation in boiling point. It is represented by ∆T_b.
So, elevation in boiling point (∆T_b) = T_b – T_b⁰.

Question 5.
What are Non – Ideal Solution? How many types of are they? Explain with giving examples.
Answer:
Non – ideal Solutions:
Solutions in which Raoult’s law cannot be applied completely for all concentrations and temperatures are called non – ideal Solutions.
For these Solutions:

1. P_A ≠ P⁰_A X_A and p_B ≠ P⁰_BX_B,
2. ∆V_mixmg ≠ 0 and
3. ∆H_mixmg ≠ 0 and

Non – ideal Solution are of two types :

1. Solutions showing positive deviations:
For such Solutions the total vapour pressure will be greater than the corresponding vapour pressure according to the Raoult’s law. The boiling point of such Solutions are lowered. Because interactions between the mol-ecules of Solution: is less than the interactions between pure solvents and solutes. Thus the vapour pressure greater than expected. In these type of Solution formed enthalpy and volume increases.
Characteristics:

• P_A > P⁰_A X_A and P_B > P⁰_B X_B
• ∆V_mixlng > 0
• ∆H_mixlng > 0

Example:
Solution of Cyclohexane and Ethanol:
In Ethanol, its molecules are held together by hydrogen bond

On adding cyclohexane, the molecules tend to occupy the space between ethyl alcohol molecules. As a result the attractive forces between alcohol molecules becomes less. Thus, formation of such Solution: is slight increase in vapour pressure, endothermic and an increase in volume.

2. Solutions showing negative deviations:
For such Solutions the total vapour pressure becomes less than expected according to the Raoult’s law. Because interaction between molecules of Solution is greater than interaction between the pure solvent or solute molecules. Thus the vapour pressure of Solution is less. In these type of Solutions formed enthalpy and volume decreases.
Characteristics:

• P_A < P⁰_A X_A and P_B < P⁰_B X_B
• ∆V_mixlng < 0
• ∆H_mixlng < 0

Example:
Solution of Acetone and Chloroform:
When acetone and chloroform are mixed, there are new attractive forces due to intermolecular hydrogen bonding. These force become stronger. Thus the formation of such Solution:s is an exothermic and the vapour pressure of Solution is less and decrease in volume.

Solutions Numerical Questions

Question 1.
Calculate the mass of urea (NH₂ CONH₂ ) required in making 2.5 kg of 0.25 molal aqueous Solution. (NCERT)
Solution:
0.25 molal aqueous Solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14+ 2 +12+ 16 +14+ 2 = 60 g mol^-1
∴ 0 .25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the Solution = 1000 + 15 g
= 1015 g = 1.015 g
∵ 1.015 kg of Solution contain urea = 15 g
∴ 2.5 kg of Solution will require urea = \(\frac {1.5}{1.015}\) x 2.5 kg = 37 g.

Question 2.
Henry’s law constant for CO₂ in water is 1.67 x 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 ml of soda water when packed under 2.5 atm CO₂ pressure at 298 K. (NCERT)
Solution:
According to Henry’s law,
P = K_HX .. (1)
P = 2.5 atm = 2.5 x 101325 Pa, KH= 1.67 x 10⁸ Pa
Putting these values in equation (1), we get

Question 3.
Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_fK kg mol^-1. (NCERT)
Solution:
We know that ∆T_f = K_f x \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\) .. (1)
Given ∆T_f = 1.5, K_f = 3.9 K kg mol^-1, W_A = 75g, MB (ascorbic acid, C₆H₈O6)
= 6 x 12 + 8 x l + 16 x 6 = 176
Putting, these values in equation (1),
1.5 = 3.9 x \(\frac{\mathrm{W}_{\mathrm{B}} \times 1000}{176 \times 75}\)
W_B = 5.077g.

Question 4.
Calculate the osmotic pressure in pascals exerted by a Solution prepared by dissolving 1.0 g of polymer of molar mass 1,85,000 in 450 ml of water at 37°C. (NCERT)
Solution:
π = CRT
= \(\frac {n}{V}\) RT
Here, number of moles of solute dissolved (n)
= \(\frac {1.0}{185,000}\) mol^-1
= \(\frac {1}{185,000}\) g mol^-1
v = 450mL = 0.450 L
T = 37°C = 37 + 273 = 310 K
R = 8.314 K Pa LK^-1 mol^-1
= 8.314 x 103 Pa LK^-1 mol^-1
Substituting these values we get,

Question 5.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous Solution. What should be the molarity of such a sample of the acid if the density of the Solution is 1.504 g ml^-1? (NCERT)
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of Solution = 100 g
Molar mass of HNO₃ = 63 g mol^-1
∴ 68 g HNO₃ = \(\frac {68}{63}\) mole = 1.079 mole
Density of solution = 1.504 g mL^-1
∴ Volume of solution = \(\frac {100}{1.504}\)mL = 66.5 mL = 0 0665 L
Molarity of the solution = \(\frac {Moles of the solute}{Volume of Solution in L}\)
= \(\frac {1.079}{0.0665}\) M = 16.23 M.

Question 6.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293K when 25g of glucose is dissolved in 450g of water. (NCERT)
Solution:
\(\frac { { p }^{ 0 }-{ p }^{ 0 } }{ { p }^{ 0 } } \) = X₂ = \(\frac { { W }_{ B }{ M }_{ A } }{ { M }_{ B }{ W }_{ A } } \)
or \(\frac { 17.535-{ P }_{ s } }{ 17.535 }\) = \(\frac { 25×18 }{ 180×450 }\)
or \(\frac { 17.535-{ P }_{ s } }{ 17.535 }\) = 5.56×10^-3
17.535 – p_s = 0.0975
P_s = 17.438 mm Hg.

Question 7.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵mm Hg. Calculate the solubility of methane in benzene at 298K under 760 mm Hg. (NCERT)
Solution:
Here, KH = 4.27 x 10⁵ mm, P = 760 mm
Applying Hemy’s law, P = K_HX
\(\mathrm{X}=\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}=\frac{760}{4 \cdot 27 \times 10^{5}}\) =1.78 x 10^-3
∴ Mole fraction of methane in benzene = 1.78 x 10^-3

Question 8.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. (NCERT)
Solution:
A → Benzene (C₆H₆); B —> Toluene (C₇H₈)
Number of moles of benzene n_A = \(\frac {80}{78}\)
Number of moles of toluene = n_B = \(\frac {100}{92}\)
X_A = \(\frac { { n }_{ A } }{ { n }_{ A }+{ n }_{ B } } \quad\) = \(\frac {1.026}{1.026+1.087}\) = 0.486
X_B = 1 – X_A = 1 – 0.486 = 0.514
Given P°_A = 50.71 mm Hg and P°_B = 32.06 mm Hg
We know, P = P_A + P_B
= P°_AX_A + P°_AX_B
= (50.71 x 0.486) + (32.06 x 0.514)
= 24.65+ 16.48 = 41-13
Mole fraction of components in vapour phase may be calculated by using Dalton’s law,
P_A = y_A x P
y_A = \(\frac { { P }_{ A } }{ p } \quad \) = \(\frac {24.65}{41.13}\) = 0.60

Question 9.
Determine the osmotic pressure of 5% glucose Solution at 25°C. Molecular mass of glucose = 180, R = 0.0821 litre atmosphere. (MP 2013,16)
Solution:
∴5 gm glucose is dissolved in 100 ml.
∴ 180 gm glucose will be dissolved in \(\frac {100}{5}\) x 180
= 3600 ml = 3.6 litre.
We know that,
PV = RT
P x 3.6 = 0.0821 x (25 + 273) = 298
or P = \(\frac {0.52×12.5×1000}{0.63×170}\)
= 6.80 atmospheres.

Question 10.
12.5 gm of urea dissolved in 170 gm of water. The elevation in boiling point was found to be 0.63 K. If KA for water = 0.52 Km^–1, calculate the molecular mass of urea. (MP 2010)
Solution:
Formula: M_B\(\frac { { K }_{ b }x{ W }_{ B }x1000 }{ ∆{ T }_{ b }x{ W }_{ A } }\)
Given: W_A = 170gm,W_B= 12.5gm, ∆T_B=0.63K, K_b=0.52Km^-1
M_B = \(\frac {0.52×12.5×1000}{0.63×170}\)
M_B = 60.7 gm mol^-1

Question 11.
If 6.84 gram sucrose is dissolved in 100 ml. Solution then what will be its osmotic pressure at 20°C? (R = 0.082 litre atmosphere K^-1mol^-1) (Molecular mass of sucrose M_B = 342)
Solution:
Given : Weight of solute (W_B) = 6.84 gm
Molecular mass of solute (M_B) = 342
Volume of Solution: (V) =\(\frac {100}{1000}\) =0.1 litre
Temperature (T) = 20° C = 273 + 20 = 293 K
Solution constant R = 0.082 litre atmosphere K^-1 mol^-1
Osmotic pressure π = \(\frac { { W }_{ B }RT }{ { M }_{ B }V } \)
= \(\frac {6.84×0.082×293}{342×0.1}\)
= 4.8 atmosphere.

MP Board Class 12th Chemistry Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1

MP Board Class 12th Physics Important Questions Chapter 13 Nuclei

Nuclei Important Questions 

Nuclei Objective Type Questions

Question 1.
Choose the correct answer of the following:

Half-life formula is the time required for the amount of something to fall to half its initial value.

Question 1.
The relation between half life time Tm and decay constant of a radioactive material is:

Answer:

Question 2.
The decayed part of a radioactive sample in two half lives will be :
(a) One fourth
(b) Half
(c) Three fourth
(d) Full part
Answer:
(c) Three fourth

Question 3.
The α – particle is a :
(a) Hydrogen nucleus
(b) Deuterium nucleus
(c) Helium nucleus
(d) Tritium.
Answer:
(c) Helium nucleus

Question 4.
The value of A and Z in the nuclear reaction \(_{ 238 }^{ 92 }{ U }\) → \(_{ A }^{ Z }{ Th }\) + \(_{ 4 }^{ 2}{ He }\) will be :
(a) 234,94
(b) 234,90
(c) 238,94
(d) 238,90.
Answer:
(b) 234,90

Question 5.
The source of energy in the sun is :
(a) Nuclear fission
(b) Nuclear fusion
(c) Chemical reaction
(d) Photo electric reaction.
Answer:
(b) Nuclear fusion

Question 6.
For the nuclear fusion which of the following is suitable :
(a) Heavy nucleus
(b) Lighter nucleus
(c) Atom bomb
(d) Radio active decay.
Answer:
(b) Lighter nucleus

Question 7.
In the symbol of nucleus \(_{ Z }^{ A }{ X }\) there are :
(a) Z – neutrons, (A – Z) protons
(b) Z – protons, (A – Z) neutrons
(c) Z – protons, A neutrons
(d) A protons. Z neutrons.
Answer:
(b) Z – protons, (A – Z) neutrons

Question 8.
In the gamma ray emission of a nucleus :
(a) Only proton number varies
(b) Number of neutrons and protons both varies
(c) There is no change in number of protons and neutrons
(d) Only number of neutrons varies.
Answer:
(c) There is no change in number of protons and neutrons

Question 2.
Fill in the blanks :

1. In the controlled chain reaction the fast moving neutrons are slowed down by using moderator. These neutrons are called …………..
2. The Einstein’s mass energy equivalence relation is …………..
3. The unstable nucleus obtained by artificial radioactivity is called …………..
4. The mass of neutrons in a nucleus is nearly equal to mass of …………..
5. In the process of ………….. the lighter nuclei together makes a heavy nucleus.
6. In nuclear reactor the heavy water is used as …………..
7. The SI unit of radio activity is …………..
8. In the range mass number A = 30 to A = 170 the value of ………….. is nearly constant.

Answer:

1. Thermal neutrons
2. E = me²
3. Radio active isotopes
4. Protons
5. Fusion
6. Moderator
7. Becquerel (Bq)
8. Per nucleon binding energy.

Question 3.
Match the columns :

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (b)

Question 4.
Write the answer in one word/sentence :

1. Write the equation showing the decay of free neutron.
2. The energy distribution of P – rays is continuous. Why?
3. In the α and β particles, which one has more ionization power?
4. The ionization power of α – particles is high. Why?
5. The penetrating power of γ(gamma) – rays is very high. Why?
6. Write an equation showing nuclear fusion.
7. Which part of the electromagnetic spectrum has highest penetrating power?
8. What is order of size of nucleus?
9. Does the density of all nuclei remain same? If yes then write its order.
10. How is nuclear radius related to atomic mass?

Answer:

1. ₀n¹ → ₁H¹ + _-1B⁰ + \(\overline { ν }\)
2. Because during the β – decay anti – neutrino is emitted
3. α – particles
4. Because α – particles have large mass and large size and their velocity is low
5. Because their velocity is very high and they do not have any charge,
6. ₁H² + ₁H² → ₂He⁴ + Q
7. γ – rays
8. 10^-15 m
9. Yes, 10¹⁷ per cubic metre
10. R = R₀ A^1/3

Nuclei Very Short Answer Type Questions

Question 1.
What will be the ratio of radii of two nucleus of mass number A₁ and A₂?
Solution:
From the formula

Question 2.
What will be the energy equivalent to 10 milligram?
Solution:
E = mc²
= 10 x 10^-6 x (3 x 10⁸)²
= 9 x 10⁹ joule.

Question 3.
What are thermal neutrons?
Answer:
In the process of controlled nuclear fission, the fast moving neutrons are slowed down with the help of moderation. These slow stray neutrons are called as thermal neutrons.

Question 4.
Does the ratio of neutron and proton after emission of α – particle in a nucleus increase, decrease or remain constant?
Answer:
The ratio can increase, decrease or can remain same. It depends upon the nature of nucleus.

Question 5.
Nuclear fusion is not possible in laboratory. Why?
Answer:
For nuclear fusion very high temperature (≈ 10⁷K) and very high pressure is required which is not possible in laboratory.

Question 6.
Is a free neutron a stable particle?
Answer:
No, free neutron is decayed in a proton, an electron and an antineutrino (\(\overline { ν }\)).
₀n¹ → ₁P¹ + _-1e⁰ + \(\overline { ν }\)

Question 7.
In heavy nucleus the number of neutrons is more than number of protons. Explain why?
Answer:
In a nucleus there is a coulomb’s repulsive force between protons in addition to a strong attraction nuclear force. On the other hand in case of neutrons, within the nucleus, there is only a short range attractive nuclear force. For heavy nuclei to be stable the repulsive force must be less. This is possible only if the number of neutrons is more than number of protons.

Question 8.
If a radioactive substance that is capable to emit a, p and y rays is kept on a piece of paper, then which rays has the maximum possibility to be stoped?
Answer:
As the mass of α – particles is more and its penetrating power is least, so it will be stopped.

Question 9.
Why is heavy water used as moderator in a nuclear reactor?
Answer:
Heavy water contains protons (of mass nearly equal to the mass of neutrons). Fast moving neutrons undergo elastic collision with these slow moving neutrons and thus get slowed down. Hence heavy water is used as moderator.

Question 10.
An atom expressed by _YX^A emits n α – particles. How many proton will remain in it? What will be the new atomic number of new atom?
Answer:
In an α – particles there are 2 protons. If n α – particles are emitted then 2n proton will be reduced.
∴ No. of remaining protons = Z – 2n
Atomic number = Y – 2n.

Question 11.
Penetrating power of β – particles is more than that of α – particles but its penetrating power is less. Why?
Answer:
β – particles have small size and high velocity so its penetrating power is more than α – particles. Due to small size and high velocity the probability of collision with gas molecules decreases and hence ionization power becomes less.

Question 12.
The nuclear fusion could not be used as an experimental and controlled source of energy till now. Why?
Answer:
The nuclear fusion takes place at very high temperature. Due to such very high temperature it is not possible to control it.

Question 13.
Why is a neutron preferred as a bombarding particle in nuclear fission?
Answer:
Neutron is a neutral particle i.e., it has no charge. So it is neither repelled nor attracted by the nucleus and hence can penetrate deep into the nucleus to cause a nuclear reaction.

Question 14.
β – rays are emitted from nucleus. They are made of fast moving electron though there are no electrons in the nucleus. Why it is so?
Answer:
Nucleus has no electron in it. Actually due to decay of neutron into proton and electron, the β – rays (particle) are produced.

Question 15.
Justify that gamma rays have more penetrating power and α – particles have more ionizing power.
Answer:
Gamma rays are electromagnetic waves. Its speed is equal to the speed of light. So they can penetrate deeply into the matter i.e., their penetrating power is high, α – particles has high mass and posses greater kinetic energy. So when they collide with any atom, it transfer its energy to orbital electrons and they are ejected. Hence α – particles has high ionizing power.

Question 16.
The mass number of two nucleus are 1 : 2. What is the ratio of their nuclear density?
Answer:
1 : 1 because nuclear density does not depend on mass number.

Question 17.
What is effect of temperature and pressure or radio activity?
Answer:
No, effect.

Question 18.
Due to the emission of α – particles does the ratio of neutron to proton decrease, increase or remains the same?
Answer:
The ratio \(\frac {n}{p}\) increases, because number of protons decreases by 2.

Nuclei Short Answer Type Questions

Question 1.
Which particles remains inside the nucleus? What will be the number of neutrons in the nucleus _zX^A?
Answer:
The neutrons and protons remains inside the nucleus. In the nucleus _zX^A number of neutrons be A – Z.

Question 2.
Define decay constant and mean life of a radioactive substance write their units.
Answer:
Decay Constant:
The decay constant is defined as the reciprocal of that time interval during which the number of active nuclei in a given sample of a radioactive substance reduces to times of the initial value of number of nuclei.
If T is half life time and X is decay constant then
T = \(\frac {0.6931}{λ}\)
Its unit is “per second”.

Mean life:
The reciprocal of decay constant is called mean life, its unit is ‘second’.

Question 3.
What is radioactivity? Write the names of the rays emitted from radio – active substances.
Answer:
Radioactivity is that property by virtue of which the nucleus of a heavy element disintegrates itself with the emission of radiation without being forced by any external agent to do so.

Following rays are emitted from a radioactive substance :

1. α – rays
2. β – rays
3. γ – rays.

Question 4.
A radioactive nucleus decays as given below :

If the mass number of A₂ is 176 and atomic number is x then what is the atomic number and mass number of A₁ and A? Are they isotopes or isobars?
Answer:
The mass number of A₂ = 176,
atomic number of A₂ = 71.
Due to decay of α – particle mass number decreases by 4 and atomic number decreases by 2.
∴ Mass number of A₁ = 176 + 4 = 180 and
its atomic number = 71 + 2 = 73
Similarly
The mass number of A = 180 + 1 = 181 and
atomic number of A = 73 + 0 = 73
Since, the atomic number of A and A₁ are same so they are isotopes.

Question 5.
Define decay constant of a radioactive substance. Write its relation with half life?
Answer:
The decay constant is defined as the reciprocal of that time interval during which the number of active nuclei in a given sample of a radioactive substance reduces to times of the initial value of number of nuclei.
If T is half life time and X is decay constant then
T = \(\frac {0.6931}{λ}\).

Question 6.
What is nuclear fission? Give an example.
Answer:
When a heavier nucleus is bombarded with neutrons then it get splitted into two lighter nuclei. This process is called as nuclear fission.
A huge amount of energy is released in this process.
₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆kr⁹² + 3 ₀n¹ + 200 MeV.

Question 7.
What is nuclear fusion? Give an example.
Answer:
When two lighter nuclei fused together to form a heavier nucleus then a huge amount of energy is released in this process. This process is called nuclear fusion.
₁H² + ₁H² → ₂He⁴+ 24MeV.

Question 8.
A radioactive element with mass number 218 and atomic number 84 emits a β – particle. What would be the mass number and atomic number after decay?
Answer:
Since, after the emission of β – particle there is no change in the mass number of element but its atomic number is increased by 1.
Therefore the mass number will be 218 and atomic number will be 84 + 1 = 85.

Question 9.
The ratio of radii of two nuclei are 1 : 2. Find out the ratio of their mass number.
Solution:
Radius of nucleus

Question 10.
The nuclear fusion process is difficult as compared to nuclear fission. Why?
Answer:
For nuclear fusion a high temperature (≈ 10⁷ K) is required. It is very difficult to achieve this temperature, so nuclear fusion is difficult.

Question 11.
Write the important properties of nuclear forces.
Answer:

1. Nuclear forces are attraction force
2. These forces does not depend on charge
3. These forces are short range forces
4. They are strong forces
5. The nuclear forces are not central forces.

Question 12.
Write properties of nuclear forces. Prove that the density of nucleus is independent of mass number.
Answer:
Properties of nuclear forces :

1. Nuclear forces are attraction force
2. These forces does not depend on charge
3. These forces are short range forces
4. They are strong forces
5. The nuclear forces are not central forces.

Let the mass number of a nucleus is A and radius is r then volume of nucleus be

Thus, it is clear that material density of a nucleus is independent of its mass number.

Question 13.
Write difference between nuclear fusion and nuclear fission.
Answer:
Difference between nuclear fusion and nuclear fission :
Nuclear fission:

• In this prosses a heavy nucleus splitted into two lighter nuclei.
• This process is possible at normal temperature.
• Per fission released energy is very high (200 MeV)

Nuclear fusion:

• In this process two lighter nuclei fused together to form a heavy nucleus.
• This process is possible at very high temperature.
• Per fusion released energy is comparatively less (24 MeV).

Nuclei Long Answer Type Questions

Question 1.
Define ‘electron volt’ and ‘Atomic mass unit”. Find out the energy equivalent to mass of a proton in joule.
Answer:
Electron volt:
The energy acquired by an electron when it is being accelerated by a potential difference of 1 volt is called ‘electron volt’.
1 electron volt = 1.6 x 10^-19J.

Atomic mass unit:
An atomic mass unit is defined as (\(\frac {1}{12}\)) of the mass of 1 atom of carbon – 12.
1 atomic mass unit = 1.66 x 10^-27kg
(1 a.m.u. or only 1u)
The mass of proton = 1.673 x 10^-27 x (3 x 10⁸ )²
= 15.05 x 10^-11J
= 1.505 x 10^-10J
∴ The energy equivalent mass of 1 proton is 1.505 x 10^-10J

Question 2.
Draw the graph showing the variation of binding energy per nucleon with the mass number. Write important conclusions drawn from it. (CBSE 1994)
Or
Draw the graph showing the variation of binding energy per nucleon with mass number. Highlight the region where the nuclei are more stable. (CBSE 1996)
Or
Draw the graph showing the variation of binding energy per nucleon with the mass number. Highlight the region where the nuclear fusion takes place. (CBSE 1996)
Answer:
From the binding energy curve, it follows that:

1. The average binding energy/ nucleon for nearly all elements is 8 MeV

2. The maximum value of binding energy per nucleon is 8.79 MeV and it is for Fe. This explains the large abundance of Fe in nature

3. Since for nuclei of intermediate mass numbers, the value of the binding energy per nucleon is around the maximum value, these nuclei are the most stable

4. The binding energy/ nucleon decreases for nuclei of mass number more than 56 and its least value is 7.6 MeV.

5. Below mass number 28, there are peaks in the curve corresponding to those nuclei whose mass numbers are multiples of four. These nuclei contain an equal number of protons and neutrons. Though peaks occur corresponding to \(_{ 4 }^{ 2 }{ He}\), \(_{ 8 }^{ 4 }{ Be }\), \(_{ 12 }^{ 6 }{ C }\), \(_{ 16 }^{ 8 }{ O }\) and \(_{ 20}^{ 10 }{ Ne }\), we have shown only the first peak corresponding to \(_{ 4 }^{ 2 }{ He }\) Obviously, these nuclei have more binding energy per nucleon than their neighbours. This explains the stability of an α – particle (\(_{ 4 }^{ 2 }{ U }\))

6. The smallest value of the binding energy per nucleon is in the case of a deuteron (\(_{ 2 }^{ 1 }{ U }\)) and its value (2.2 MeV)/2 = 1.1 MeV. This is confirmed by the fact that a photon whose energy is 2.2 MeV or more, can split a deuteron into a free neutron and a free proton. This phenomenon is called the photodisintegration of a deuteron.

Question 3.
Calculate binding energy per nucleon of ₂₆Fe⁵⁶.
Given : m(\(_{ 26 }^{ 56 }{ Fe }\)) = 55.934939 a.m.u., m (proton ) = 1.007825 a.m.u., m (neutron) = 1.008665 a.m.u. (CBSE 1993,95, 2000 Supp.)
Solution:
Number of protons in = \(_{ 26 }^{ 56 }{ He }\) is 26 and number of neutrons is 56 – 26 = 30.
∴ Mass of 26 protons = 1 0077825 x 26
= 26.202345 a.m.u.
Mass of 30 neutrons = 1.008665 x 30
= 30.25995 a.m.u.
Total mass of nucleons = 26.202345 + 30.25995
= 56.4623 a.m.u.
Mass of ₂₆Fe⁵⁶ = 55.934939.a.m.u.,
∴ Mass defect ∆m = 56.4623 – 55.934939
= 0.527361 a. m. u.
∴ Binding energy of ₂₆Fe⁵⁶ = 0.527361 x 931
= 490.973091 MeV
Binding energy per nucleon = \(\frac {490.973091}{56}\)
= 8.7674 MeV.

Question 4.
Define the term half-life period and decay constant of a radioactive sample. Derive a relation between these terms.
Answer:
The half – life of a radioactive substance is the time it takes half of a given number of radioactive nuclei to decay. (CBSE 1995, 2001)
The SI unit of Tm is second (s).
Example:
Half life of radium is 1600 years. It means that if 1 gm of radium is taken then after 1600 years \(\frac {1}{2}\) gm of radium will decay.
Let us assume that at time t = 0, the number of radioactive nuclei present is N₀ and after time t, the number of radioactive nucleus remaining is N, then according to law of radioactive decay,
N = N₀e^-λt
Where, λ is decay constant.
If the half – life of radioactive sample is T_1/2, then at t = T_1/2
N = \(\frac { { N }_{ 0 } }{ 2 }\)
Substituting in equation (1), we get ,

From equation (2), it is clear that half life of a radioactive sample is inversely proportional to decay constant.

Question 5.
State law of radioactive decay and derive the following formula :
N = N₀ e^-λt, where symbols have their usual meaning.
Answer:
According to the radioactive decay law :
The rate at which a particular decay process occurs in a radioactive sample is proportional to the number of radioactive nuclei present (that is, those nuclei that have not yet decayed).
Let N₀ = Total number of radioactive nuclei present originally at time t = 0,
(t = 0 refers to the time when the radioactive element is freshly separated from its by products), and N = Total number of radioactive nuclei present at any time t.

According to the radioactive decay law,
– \(\frac {dN}{dt}\) ∝ N
(- ve sign indicated that the number of radioactive nuclei decreases with time).
Thus, – \(\frac {dN}{dt}\) = λN
Where A is a constant of proportionality and is called the disintegration constant or decay constant.
Equation (1) can be written as
\(\frac {dN}{dt}\) = – λdt
When t = 0, N = N₀ and when t = t,N = N.
Integrating equation (2) within proper limits,

From equation (3) it is clear that (i) the radioactive decay is an exponential process i.e., the decay is fast initially and later on it becomes slow.

Nuclei Numerical Questions

Question 1.
An imaginary fission reaction is given below :
\(_{ 92 }^{ 236 }{ X }\) → \(_{ a}^{ 141 }{ Y }\) + \(_{ 36 }^{ b }{ Z }\) + 3\(_{ 0 }^{ 1 }{ n }\)
Find out value of a and b.
Solution:
a + 36 + 0 = 92
a = 56.
and 141 + b + 3 = 236
b = 92.

Question 2.
The half life of a radioactive substance is 30 days. What will be the number of radioactive atoms after 90 days?
Solution:
Given : T = 30 days, t = 90 days
n = \(\frac {t}{T}\) = \(\frac {90}{30}\) = 3
Formula : N = N₀
= N₀(\(\frac {1}{2}\))ⁿ
= N₀(\(\frac {1}{2}\))³
= \(\frac { { N }_{ 0 } }{ 8 }\)
Thus, the number of radioactive atoms be \(\frac {1}{8}\) times of their initial value.

Question 3.
How many fission per second takes place if its half life for α – decay is 1.42 x 10¹⁷sec?
Solution:

MP Board Class 12th Physics Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Ex 1.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.2

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 संबंध एवं फलन Ex 1.4 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Book Solutions Chapter 1 संबंध एवं फलन Ex 1.4

प्रश्न 1.
निर्धारित कीजिए कि क्या निम्नलिखित प्रकार से परिभाषित प्रत्येक संक्रिया से एक द्विआधारी संक्रिया प्राप्त होती है या नहीं। उस दशा में जब एक द्विआधारी संक्रिया नहीं है, औचित्य भी बतलाइए।
(i) Z⁺ में, a * b = a – b द्वारा परिभाषित संक्रिया
(ii) Z⁺ में, a* b = ab द्वारा परिभाषित संक्रिया
(iii) R में, संक्रिया *, a* b = ab² द्वारा परिभाषित
(iv) Z⁺ में, संक्रिया *, a* b = |a – b| द्वारा परिभाषित
(v) Z⁺ में, संक्रिया *, a* b = a द्वारा परिभाषित
हल:
(i) Z⁺ में, a* b = a – b द्वारा परिभाषित संक्रिया है
यदि a > b, a * b = a – b ϵ Z⁺
परन्तु यदि a < b, a * b = a – b < 0, Z⁺ में नहीं है।
अत:* संक्रिया द्विआधारी संक्रिया नहीं है।

(ii) Z⁺ पर * संक्रिया, a * b = ab द्वारा परिभाषित है।
यदि a, b ϵ Z⁺ ⇒ a और b दोनों धनात्मक हैं।
a * b = ab भी धनात्मक है।
ab ϵ Z⁺
अतः यह संक्रिया द्विआधारी है।

(iii) R पर * संक्रिया a* b = ab⁺ द्वारा परिभाषित है।
यदि a, b ϵ R, ab² भी R* में है।
अतः यह संक्रिया द्विआधारी है।

(iv) Z⁺ पर * संक्रिया a * b = |a – b| द्वारा परिभाषित है।
यदि a, b ϵ Z⁺ , |a – b | ϵ Z⁺
अंत: यह संक्रिया द्विआधारी है।

(v) Z⁺ पर * संक्रिया a* b = a द्वारा परिभाषित है।
यदि a, b ϵ Z⁺, ∴ a * b = a ϵ Z⁺
अत: यह संक्रिया द्विआधारी है।

प्रश्न 2.
निम्नलिखित परिभाषित प्रत्येक द्विआधारी संक्रिया के लिए निर्धारित कीजिए कि क्या द्वि आधारी क्रमविनिमेय है तथा क्या साहचर्य है।
(i) Z में, a * b = a – b द्वारा परिभाषित
(ii) Q में, a * b = ab + 1 द्वारा परिभाषित
(iii) Q में, a * b = \(\frac{a b}{2}\) द्वारा परिभाषित
(iv) Z⁺ में, a * b = 2^ab द्वारा परिभाषित
(v) Z⁺ में, a* b = a^b द्वारा परिभाषित
(vi) R – { – 1} में, a* b = \(\frac{a}{b+1}\) द्वारा परिभाषित
हल:
(i) Z पर संक्रिया a* b = a – b द्वारा परिभाषित है।
(a) यदि a * b = a – b और b * a = b – a
परन्तु a – b ≠ b – a ⇒ a* b + b * a
∴ यह संक्रिया क्रमविनिमेय नहीं है।

(b) यदि a * (b * c) = a * (b – c) = a * (b – c)
a – (b – c) = a – b + c
(a * b) * c = (a – b) * c = a – b – c
स्पष्ट है कि a * (b * c) (a * b) * c
∴ संक्रिया साहचर्य नहीं है। अतः संक्रिया न तो क्रमविनिमेय है और न ही साहचर्य है।

(ii) Q पर * संक्रिया, a * b = ab + 1 से परिभाषित है।
(a) a * b = ab + 1, b * a = ba + 1 = ab + 1
∴ a * b = b * a
∴ यह संक्रिया क्रमविनिमेय द्विआधारी है।

(b) यदि a * (b * c) = a * (bc + 1) = a (bc + 1) + 1
= abc + a + 1
(a * b) * c= (ab + 1) * c =(ab + 1)c + 1
= abc + c + 1
∴ (a * b) * c ≠ a + (b + c)
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है। अतः यह संक्रिया क्रमविनिमेय है परन्तु साहचर्य नहीं है।

(iii) Q पर * संक्रिया, a * b = \(\frac{a b}{2}\) द्वारा परिभाषित है।

∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया है।
अतः यह संक्रिया क्रमविनिमेय और साहचर्य दोनों हैं।

(iv) Z⁺ पर * संक्रिया a* b = 2^ab से परिभाषित है।
(a) ∴ a * b = 2^ab, b * a = 2^ba = 2^ab
⇒ a * b = b * a
अतः संक्रिया क्रमविनिमेय संक्रिया है।

(b) a * (b * c) = a * 2^bc = a^a.2^bc
(a * b) * c = 2^ab * c = 2^2ab.c
∴ a * (b * c) ≠ * (a * b) *c
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है। अतः यह संक्रिया क्रमविनिमेय है परन्तु साहचर्य नहीं है।

(v) Z⁺ पर * संक्रिया, a * b = a^b से परिभाषित है।
(a) a * b = a^b, b * a = b^a
∴ a * b + b* a
अत: यह संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c)=a * b^c = a^(bc)
(a * b) * c = ad * c = a^(b)c = a^bc
∴ (a * b) * c * a * (b * c)
∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है।
अत: यह संक्रिया न तो क्रमविनिमेय है और न ही साहचर्य

(vi) R – {-1} पर * संक्रिया, a * b = \(\frac{a}{b+1}\) द्वारा परिभाषित है।

∴ यह संक्रिया साहचर्य द्विआधारी संक्रिया नहीं है।
अत: यह संक्रिया क्रमविनिमेय है और न ही साहचर्य है।

प्रश्न 3.
समुच्चय {1, 2, 3, 4,5} में a ^ b= निम्नतम {a, b} द्वारा परिभाषित द्विआधारी संक्रिया पर विचार कीजिए। संक्रिया के लिए संक्रिया सारणी लिखिए।
हल:
समुच्चय {1, 2, 3, 4, 5} पर संक्रिया ^ सारणी निम्न है-

प्रश्न 4.
समुच्चय {1, 2, 3, 4, 5} में, निम्नलिखित संक्रिया सारणी (सारणी 1.2) द्वारा परिभाषित द्विआधारी संक्रिया पर विचार कीजिए तथा
(i) (2 * 3) * 4 तथा 2 * (3 * 4) का परिकलन कीजिए।
(ii) क्या * क्रम विनिमेय है?
(iii) (2 * 3) * (4 * 5) का परिकलन कीजिए। (संकेत : निम्न सारणी का प्रयोग कीजिए।)

हल:
(i) दी गई सारणी से
(2 * 3) * 4 = 1 * 4 = 1
तथा 2 * (3 * 4) = 2 * 1 = 1

(ii) माना a, b ϵ {1, 2, 3, 4, 5}
∴ सारणी से, a * a = a (a ≠ b) तथा a, b विषम संख्या है
a * b = b * a = 1
2 * 4 = 4 * 2 = 2 जहाँ a तथा b सम संख्या तथा a ≠ b
अतः a * b = b * a
अतः द्विआधारी संक्रिया क्रम विनिमेय है।

(iii) सारणी से,
(2 * 3) * (4 * 5) = 1 *1
= 1

प्रश्न 5.
मान लीजिए कि समुच्चय {1, 2, 3, 4, 5} में एक द्विआधारी संक्रिया *’, a *’ b = a तथा b का HCF द्वारा परिभाषित है। क्या संक्रिया *’ उपर्युक्त प्रश्न 4 में परिभाषित संक्रिया * के समान है? अपने उत्तर का औचित्य भी बतलाइए।
हल:
यहाँ समुच्चय {1, 2, 3, 4, 5} संक्रिया a*’ b H.C.F. a, b द्वारा परिभाषित है।
इस संक्रिया की निम्न सारणी दी गयी है-

प्रश्न 4 में दी गई सारणी और यह सारणी समान है।
अतः संक्रिया *’ तथा * समान है।

प्रश्न 6.
मान लीजिए कि N में एक द्विआधारी संक्रिया, a * b = a तथा b का LCM द्वारा परिभाषित है। निम्नलिखित ज्ञात कीजिए:
(i) 5 * 7, 20 * 16
(ii) क्या संक्रिय * क्रम विनिमेय है?
(iii) क्या * साहचार्य है?
(iv) N में * का तत्समक अवयव ज्ञात कीजिए।
(v) N के कौन-से अवयव * संक्रिया के लिए व्युत्क्रमणीय है?
हल:
द्विआधारी संक्रिया (Binary Operations) * इस प्रकार परिभाषित है कि
a * b = a तथा b का L.C.M.
(i) 5 * 7 =5 तथा 7 का L.C.M.
= 35
तथा 20 * 16 = 20 तथा 16 का L.C.M.
= 80

(ii) a * b = a तथा b का L.C.M.
= b तथा a का L.C.M
a * b = b * a
अतः द्विआधारी संक्रिया क्रम विनिमेय है।

(iii) a * (b * c) = a * (b तथा c का L.C.M.)
= a तथा (b तथा c का L.C.M.) का L.C.M.
= a, b तथा c का L.C.M.
इसी प्रकार
(a * b) * c =(a तथा b का L.C.M.) * c
=a, b, c of L.C.M.
⇒ a *(b * c) =(a * b) * c
अतः द्विआधारी संक्रिया * साहचर्य है।

(iv) N में * संक्रिया की तत्समक अवयव 1 है।
∵ 1 * a = a * 1 = a
= 1 तथा a का L.C.M.

(v) माना * : N × N → N इस प्रकार परिभाषित है कि a * b = a तथा b का L.C.M.
∴ a = 1, b = 1 के लिए,
a * b = 1 = b * a
अत: 1a* संक्रिया के लिए व्युत्क्रमणीय है।

The step by step instructions on how to find the least common multiple of 12 and 16.

प्रश्न 7.
क्या समुच्चय {1, 2, 3, 4, 5} में a * b = a तथा b का LCM द्वारा परिभाषित * एक द्विआधारी संक्रिया है? अपने उत्तर का औचित्य भी बतलाइए।
हल:
दिया गया समुच्चय = {1, 2, 3, 4, 5} द्विआधारी संक्रिया द्वारा परिभाषित है कि a * b = a और b का LCM 2 * 6 = 6 जो कि समुच्चय {1, 2, 3, 4, 5} में नहीं है इसलिए * एक द्विआधारी संक्रिया है।

प्रश्न 8.
मान लीजिए कि N में a * b = a तथा b का HCF द्वारा परिभाषित एक द्विआधारी संक्रिया है। क्या * क्रमविनिमेय है? क्या * साहचर्य है? क्या N में इस द्विआधारी संक्रिया के तत्समक का अस्तित्व है?
हल:
यहाँ N, प्राकृत संख्याओं का समुच्चय है।
द्विआधारी संक्रिया a * b = a, b का H.C.F. द्वारा परिभाषित
(i) a, b का H.C.F. = b, a के H.C.F.
a * b = b * a
अतः संक्रिया क्रमविनिमेय है।

(ii) a * (b * c)= a * (b, c का H.C.F.)
=a व b, c का H.C.F.
= a, b, c का H.C.F.
(a * b) * c = (a, b का H.C.F.) * c
= a, b व c का H.C.F.
= a, b, c का H.C.F.
a * (b * c)= (a * b) * c (∵ संक्रिया साहचर्य है)

(iii) 1 * a = a * 1 = 1 ≠ a
अतः तत्समक अवयव का अस्तित्व नहीं है।

प्रश्न 9.
मान लीजिए कि परिमेय संख्याओं के समुच्चय में निम्नलिखित प्रकार से परिभाषित * एक द्विआधारी संक्रिया है:
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = \(\frac{a b}{4}\)
(vi) a * b = ab²
ज्ञात कीजिए कि इनमें से कौन-सी संक्रियाएँ क्रमविनिमेय हैं और कौन-सी साहचर्य हैं।
हल:
यहाँ परिमेय संख्याओं का समुच्चय Q दिया है।
(i) a * b = ab – b, द्विआधारी संक्रिया है।
(a) b * a = b – a
∴ a – b ≠ b – a ⇒ a * b ≠ b * a
अत: यह संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c
(a * b) * c = (a-b)* c = a – b -c
∴ a – b + c ≠ a – b – c = a * (b * c) * (a * b) * c
अतः यह संक्रिया साहचर्य नहीं है।

(ii) (a) a * b = a² + b²
∴ b * a = b² + a² = a² + b²
⇒ a * b = b * a
अत: यह संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b² + c²) = a² + (b² + c²)²
(a + b) * c = (a² + b²) * c = (a² + b²)² + c²
⇒ a * (b * c) ≠ (a * b) * c.
अतः यह * संक्रिया साहचर्य नहीं है।

(iii) संक्रिया a * b = a + ab द्वारा परिभाषित है।
(a) a * b = a (1 + b), b * a = b + ba = b (1 + a)
∴ a * b + b * a
अतः यह * संक्रिया क्रमविनिमेय नहीं है।
(a) a * (b * c) = a + (b + bc)= a + a (b + bc)
= a + ab + abc
(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c
= a + ab + ac + abc
∴ a * (b * c) (ab) * c
अतः यह * संक्रिया साहचर्य नहीं है।

(iv) दिया है : a * b = (a – b)²
(a) a * b = (a – b)², b * a = (b – a)² = (a – b)²
∴ a * b = b * a
अतः यह * संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b – c) = [a – (b – c)²]²
(a * b) * c = (a – b)² * c = [(a – b)² – c]²
∴ a * (b * c) ≠ (a * b) * c
अतः यह * संक्रिया साहचर्य नहीं है।

(v) a * b = \(\frac{a b}{4}\)

अतः यह * संक्रिया साहचर्य है।

(vi) a * b = ab²
(a) a * b = ab², b * a = ba
∴ a * b ≠ b * a
अत: यह * संक्रिया क्रमविनिमेय नहीं है।
(b) a * (b * c) = a + bc² = a (bc²)² = ab²c⁴
(a * b) * c = ab² * c = ab²c² = ab²c²
∴ a + (b * c) ≠ (a * b) * c.
अतः यह * संक्रिया साहचर्य नहीं है।

प्रश्न 10.
सिद्ध कीजिए कि प्रश्न 9 में दी गई संक्रियाओं में किसी का तत्समक है, वह बतलाइए।
हल:
यहाँ (i) a * b = a – b
यदि e तत्समक अवयव हो तो
a * e = a – e, e * a = e – a
∴ a – e ≠ e – a ⇒ a * e ≠ e * a
अतः e का अस्तित्व नहीं है।

(ii) a * b = a² + b²
∴ a * e = a² + e², e * a = e² + a²
a * e = e * a ≠ a
अतः e का अस्तित्व नहीं है।

(iii) a * b = a + ab
a * e = a + ae, e * a = e + ea
∴ a * e # e * a # a
अत: e का अस्तित्व नहीं है।

(iv) a * b = (a – b)
a * e = (a – e)² # a, e * a = (e – a)² # a
a * e = e * a # a
अतः e का अस्तित्व नहीं है।

(v) a * b = \(\frac{a b}{4}\)
a * e = \(\frac{ae}{4}\) # a, e * a = \(\frac{ea}{4}\) # a
∴ a * e = e * a # a
अतः e का अस्तित्व नहीं है।

(vi) a * b = ab²
a * e = ae² # a, e * a = ea² # a
∴ a * e # e * a # a
अतः e का अस्तित्व नहीं है।

प्रश्न 11.
मान लीजिए कि A = N × N है तथा A में (a, b) * (c, d) = (a + c, b + d)द्वारा परिभाषित एक द्विआधारी संक्रिया है। सिद्ध कीजिए कि * क्रम विनिमेय तथा साहचर्य है। A में * का तत्समक अवयव, यदि कोई है, तो ज्ञात कीजिए।
हल:
माना A = N × N
द्विआधारी संक्रिया (Binary operation) * इस प्रकार परिभाषित है कि
(a, b) * (c, d) = (a + c, b + d)
इसलिए (c, d) * (a, b) = (c + a, d + b)
=(a + c, b + d)
=(a, b) * (c, d)
अतः द्विआधारी संक्रिया * क्रम विनिमेय है
पुनः (a, b)* [(c, d) * (e, f)]
= (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
तथा [(a, b) * (c, d)] * (e, f)
= (a + c, b + d) * (e, f)
= (a + c + e, b + d + f)
= (a, b) * [(c, d) * (e, f)]
= [(a, b) * (c, d)]* (e, f)
अतः दी गई संक्रिया * साहचर्य है।
A में तत्समक अवयव का अस्तित्व नहीं है।

प्रश्न 12.
बतलाइए कि क्या निम्नलिखित कथन सत्य हैं या असत्य हैं। औचित्य भी बतलाइए।
(i) समुच्चय N में किसी भी स्वेच्छ द्विआधारी संक्रिया* के लिए a * a = a, ∀a ϵ N
(ii) यदि N में * एक क्रमविनिमेय द्विआधारी संक्रिया है तो a * (b * c)=(c * b) * a
हल:
यहाँ द्विआधारी संक्रिया समुच्चय N पर इस प्रकार परिभाषित की गयी है कि
a * a = a ∀ a ϵ N
(i) यहाँ पर * संक्रिया में केवल एक ही अवयव का प्रयोग किया गया है।
अतः यह कथन असत्य है।
(ii) वास्तविक संख्याओं में समुच्चय पर संक्रिया क्रमविनिमेय है।
b * c = c * b
= (c * b) * a = (b * c) * a = a * (b * c)
∴ a * (b * c) = (c * b) * a
अतः यह कथन सत्य है।

प्रश्न 13.
a * b = a³ + b³ प्रकार से परिभाषित N में एक द्विआधारी संक्रिया * पर विचार कीजिए। अब निम्नलिखित में से सही उत्तर का चयन कीजिए।
(A) * साहचर्य तथा क्रमविनिमेय दोनों है
(B) * क्रमविनिमेय है किन्तु साहचर्य नहीं है
(C) * साहचर्य है किन्तु क्रमविनिमेय नहीं है
(D) * न तो क्रमविनिमेय है और न साहचर्य है
हल:
यहाँ द्विआधारी संक्रिया को समुच्चय पर इस प्रकार परिभाषित किया गया है कि
a * b = a³ + b³
(a) a * b = a³ + b³, b * a = b³ + a³ = a³ * b³
∴ a * b = b * a
अत: यह संक्रिया क्रमविनिमेय है।
(b) a * (b * c) = a * (b³ + c³) = a³ + (b³ + c³)³
(a * b) * c= (a³ + b³) * c = (a³ + b³) + c³
∴ a * (b * c) ≠ (a * b) * c
अतः यह * संक्रिया साहचर्य नहीं है।
∴ संक्रिया क्रमविनिमेय परन्तु साहचर्य नहीं है।
अतः विकल्प (B) सही है।

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