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In this article, we share MP Board Class 12th Maths Book Solutions Chapter 4 Determinants Ex 4.5 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 4 Determinants Ex 4.5

In this article, we will share MP Board Class 10th Hindi Solutions पद्य Chapter 7 सामाजिक समरसता Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Hindi Navneet Solutions पद्य Chapter 7 सामाजिक समरसता

सामाजिक समरसता अभ्यास

बोध प्रश्न

सामाजिक समरसता अति लघु उत्तरीय प्रश्न 

प्रश्न 1.
लोहा किसकी श्वांस से भस्म हो जाता है?
उत्तर:
लोहा मरे हुए जानवर की खाल से बनी हुई मसक की श्वांस (हवा) से भस्म हो जाता है।

प्रश्न 2.
भक्ति किस प्रकार के प्राणियों से नहीं हो सकती?
उत्तर:
कामी, क्रोधी और लालची प्रकृति के प्राणियों से भक्ति नहीं हो सकती।

प्रश्न 3.
गुण लाख रुपये में कब बिकता है?
उत्तर:
गुण लाख रुपये में तब बिकता है जब उसे गुण का ग्राहक मिल जाता है।

प्रश्न 4.
संसार में विद्यमान सचर-अचर प्राणी क्या कर रहे हैं?
उत्तर:
संसार में विद्यमान जितने भी सचर या अचर प्राणी हैं, वे सभी अपने-अपने कर्मों में लगे हुए हैं।

प्रश्न 5.
दयामयी माता के तुल्य किसे माना गया है?
उत्तर:
दयामयी माता के तुल्य पृथ्वी को माना गया है।

प्रश्न 6.
जीवन को सफल बनाने के लिए कवि क्या निर्देश देते हैं?
उत्तर:
जीवन को सफल बनाने के लिए कवि ने मनुष्यों को अपने-अपने उद्देश्य में लगे रहने का निर्देश दिया है।

सामाजिक समरसता लघु उत्तरीय प्रश्न

प्रश्न 1.
कबीर ईश्वर से क्या माँगते हैं?
उत्तर:
कबीर ईश्वर से यह माँगते हैं कि हे भगवान तुम कृपा करके मुझे इतना दे देना जिससे मैं अपने परिवार को पाल लूँ, मैं खुद भी भूखा न रहूँ और कोई साधु भी मेरे घर से खाली हाथ न जाये।

प्रश्न 2.
कवि के अनुसार किस प्रकार के वृक्ष के नीचे विश्राम करना चाहिए?
उत्तर:
कवि के अनुसार उस वृक्ष के नीचे विश्राम करना चाहिए जिसमें बारह महीने फल लगते हों, जिसमें शीतल छाया हो और घने फल हों तथा जिस पर पक्षी गण क्रीड़ा करते हों।

प्रश्न 3.
साधु से किस प्रकार के प्रश्न नहीं करने चाहिए?
उत्तर:
साधु से उसकी जाति नहीं पूछनी चाहिए, उससे तो केवल ज्ञान की बातें पूछनी चाहिए।

प्रश्न 4.
‘लोक कल्याण कामना’ से कवि का क्या आशय है?
उत्तर:
लोक कल्याण कामना से कवि का आशय है कि हमें संसार में रहकर ऐसे कार्य करने चाहिए जिनसे अधिक-से-अधिक लोगों का कल्याण हो। स्वार्थ के लिए हमें कार्य नहीं करने चाहिए।

प्रश्न 5.
कवि जग की विषम आँधियों के सम्मुख किस प्रकार के स्वभाव की अपेक्षा कर रहा है?
उत्तर:
कवि जग की विषम आँधियों के सम्मुख हिम्मत से डटे रहने के स्वभाव की अपेक्षा कर रहा है।

प्रश्न 6.
कर्मच्युत होने से क्या परिणाम होगा?
उत्तर:
कर्मच्युत होने से यह परिणाम निकलेगा कि तुम धोखे में पड़कर इस अलभ्य (अनमोल) अवसर से हाथ धो बैठोगे।

सामाजिक समरसता दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
दुर्बल को सताने का क्या दुष्परिणाम होता है?
उत्तर:
दुर्बल व्यक्ति को सताने का यह परिणाम होता है कि उसकी मोटी आहों से कोई बच नहीं पायेगा।

प्रश्न 2.
अवगुण का निवास कहाँ है?
उत्तर:
अवगुण का निवास मनुष्यों के हृदय में होता है।

प्रश्न 3.
किन विशेषताओं को खोकर भक्ति की जा सकती है?
उत्तर:
जाति एवं वर्ण जैसी विशेषताओं को खोकर ही भक्ति की जा सकती है।

प्रश्न 4.
मनुष्य किन-किन शक्तियों से सम्पन्न है? संसार में जीने का उद्देश्य लिखिए।
उत्तर:
मनुष्य अमित बुद्धि एवं बल से युक्त है, अतः उसका संसार में जीने का भी निश्चित उद्देश्य है; और वह है जीवन में प्रतिक्षण अपने उद्देश्य को पाने के लिए कर्म में जुटे रहना।

प्रश्न 5.
‘जीवन सन्देश’ कविता का केन्द्रीय भाव लिखिए।
उत्तर:
‘जीवन सन्देश’ कविता में कवि ने स्पष्ट किया है कि मनुष्य एक कर्त्तव्यपरायण व्यक्ति है, किन्तु कभी-कभी वह कर्तव्य से विमुख होकर भी काम करने लगता है और जीवन के उच्च उद्देश्यों से भटक जाता है। इस प्रकार के कार्यों से वह समाज की कोई भलाई नहीं कर सकता। अपने समाज और अपनी जन्म-भूमि के प्रति भी मनुष्य में कर्त्तव्य भावना होनी चाहिए, तभी उसकी जीवन यात्रा सार्थक है।

प्रश्न 6.
संसार मनुष्य के लिए एक परीक्षा स्थल है, ऐसा कवि ने क्यों कहा है ?
उत्तर:
संसार में अनेकानेक विषम परिस्थितियाँ और विपदाएँ आती रहती हैं, इसलिए कवि ने संसार को एक परीक्षा स्थल कहा है। इन विषम परिस्थितियों और आपदाओं से मनुष्य को घबड़ाना नहीं चाहिए बल्कि उनका डटकर सामना करना। चाहिए तथा अपने उद्देश्य को पाने के लिए निरन्तर संघर्ष करते रहना चाहिए।

प्रश्न 7.
निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए
(क) ऊँचै कुल का ………. निंद्या सोई॥
उत्तर:
कबीरदास जी कहते हैं कि ऊँचे कुल में जन्म लेने भर से कोई व्यक्ति ऊँचा नहीं हो जाता। यदि उसके कार्य ऊँचे नहीं हैं, तो वह कभी ऊँचा नहीं हो सकता है। ऊँचे अर्थात् श्रेष्ठ कार्य करने वाला व्यक्ति ही.ऊँचा होता है। वे उदाहरण देते हुए कहते हैं कि यदि सोने के कलश में मदिरा भरी हुई है, तो साधु लोग उस कलश की प्रशंसा न करके निन्दा ही करेंगे।

(ख) जब गुण ………… बदलै जाइ॥
उत्तर:
कबीरदास जी कहते हैं कि जब गुण के ग्राहक मिल जाते हैं तो गुण लाख रुपये में बिकता है और यदि गुण के ग्राहक न मिलें तो वह गुण कौड़ियों में बिक जाता है।

(ग) तुम मनुष्य हो ………….. निज जीवन में?
उत्तर:
कविवर त्रिपाठी जी कहते हैं कि तुम मनुष्य हो और तुम्हारा जन्म इस संसार में अत्यधिक बुद्धि एवं बल से युक्त है। ऐसी दशा में क्या तुमने अपने मन में कभी इस बात पर विचार किया है कि तुम्हारा जीवन क्या उद्देश्य रहित है? अर्थात् नहीं। कवि पुनः कहता है कि हे मनुष्यो! तुम बुरा मत मानना। तुम अपने मन में एक बार सोचो तो कि क्या तुमने अपने जीवन के सभी कर्त्तव्य पूरे कर लिये हैं अर्थात् अभी नहीं किये हैं।

सामाजिक समरसता काव्य सौन्दर्य

प्रश्न 1.
करुण रस को उदाहरण सहित समझाइए।
उत्तर:
करुण रस-किसी प्रिय वस्तु अथवा व्यक्ति की अनिष्ट की आशंका या विनाश से हृदय में उत्पन्न क्षोभ या दुःख को करुण रस कहते हैं। इसका स्थायी भाव शोक’ है।

उदाहरण :
अभी तो मुकुट बँधा था माथ हुए कल ही हल्दी के हाथ।
खुले भी न थे लाज के बोल, खिले भी चुंबन शून्य कपोल। हाय रुक गया यहीं संसार,बना सिन्दूर अंगार।
बातहत लतिका वह सुकुमार पड़ी है छिन्न धार।
स्पष्टीकरण स्थायी भाव-शोक।
विभाव :
(क) आलम्बन विनष्ट पति।
(ख) आश्रय-पत्नी।
(ग) उद्दीपन-मुकुट का बाँधना, हल्दी के हाथ होना,लाज के बोल न खलना।
अनुभाव :
वातहत लतिका के समान नायिका का बेहाल पड़े रहना। संचारी भाव-जड़ता,स्मृति, दैन्य और विषाद आदि।

प्रश्न 2.
गीतिका छन्द को अन्य उदाहरण सहित समझाइए।
उत्तर:
गीतिका छन्द में 26 मात्रा, 14-12 पर यति तथा प्रत्येक चरण के अन्त में ‘लघु-गुरु’ (15) होता है।

उदाहरण :
हे प्रभो! आनन्ददाता, ज्ञान हमको दीजिए।
शीघ्र सारे दुर्गुणों कों दूर हमसे कीजिए।
लीजिए हमको शरण में, हम सदाचारी बनें।
ब्रह्मचारी धर्मरक्षक, वीर व्रत धारी बनें।।

प्रश्न 3.
हरिगीतिका छन्द के लक्षण देते हुए एक अन्य उदाहरण लिखिए।
उत्तर:
हरिगीतिका सममात्रिक छन्द है। इसमें चार चरण होते हैं। प्रत्येक चरण में 16 एवं 12 पर यति के साथ कुल 28 मात्राएँ होती हैं। चरण के अन्त में लघु-गुरु होता है।

उदाहरण :
‘गाते प्रियाओं के सहित रस राग यक्ष जहाँ तहाँ। प्रत्यक्ष दो उत्तर दिशा की दीखती लक्ष्मी यहाँ। कहते हुए यों पार्श्व में सहसा उदासी छा गई। उत्तर दिशा से याद सहसा उत्तरा की आ गई।’

सामाजिक समरसता महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

सामाजिक समरसता बहु-विकल्पीय प्रश्न

प्रश्न 1.
लोहा किसकी श्वांस से भस्म हो जाता है?
(क) कोयले की
(ख) सोने की
(ग) लकड़ी की
(घ) मरी हुई खाल की।
उत्तर:
(ग) लकड़ी की

प्रश्न 2.
दयामयी माता के तुल्य किसे माना गया है?
(क) माँ
(ख) भारत की धरती को
(ग) भारत
(घ) प्रान्त को।
उत्तर:
(ख) भारत की धरती को

प्रश्न 3.
कवि के अनुसार किस प्रकार के वृक्ष के नीचे विश्राम करना चाहिए?
(क) छायादार
(ख) घना
(ग) फलयुक्त
(घ) उपर्युक्त सभी।
उत्तर:
(घ) उपर्युक्त सभी।

प्रश्न 4.
सज्जन व्यक्ति से क्या पूछना चाहिए? (2009)
(क) जाति
(ख) नाम
(ग) ज्ञान
(घ) काम।
उत्तर:
(ग) ज्ञान

रिक्त स्थानों की पूर्ति

1. जाति न पूछो साधु की पूछि लीजिए ………..
2. कबीर ने बाह्य आडम्बरों तथा ………… का व मूर्तिपूजा का विरोध किया।
3. ऊँचे कुल में जन्म लिया हो पर करनी ऊँची न हो, ऐसा व्यक्ति ……….. स्वर्ण कलश के समान है।
4. लोक कल्याण की कामना ही सच्ची ……….. है।

उत्तर:

1. ज्ञान
2. अन्ध विश्वास
3. मदिरा से भरे
4. लोक सेवा।

सत्य/असत्य

1. कबीर की भाषा को खिचड़ी या सधुक्कड़ी भाषा कहा जाता है।
2. कबीर ने बाहरी आडम्बरों का समर्थन किया। (2009)
3. ‘सबसे अधिक अवगुण औरों में ही होते हैं’ कबीर का यह कथन है।
4. रामनरेश त्रिपाठी स्वच्छन्दतावादी काव्यधारा के प्रतिष्ठित कवि हैं। इन्होंने स्वच्छन्दतावादी परम्परा को बढ़ाया।

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. सत्य।

सही जोड़ी मिलाइए

उत्तर:
1. → (घ)
2. → (ग)
3. → (क)
4. → (ख)

एक शब्द/वाक्य में उत्तर

1. कबीर सच्चे अर्थों में कवि होते हुए भी क्या थे?
2. साधु से क्या नहीं पूछना चाहिए? (2012)
3. कवि ने जीवन सन्देश में अपने कर्म में कैसी तत्परता बतायी है?
4. अटल निश्चय व भय रहित संसार में कौन है?
5. जीवन संदेश कविता का मूल स्वर क्या है? (2012)
6. गुण कब लाख रुपये में बिकता है? (2018)

उत्तर:

1. समाज सुधारक
2. जाति
3. तुच्छ पत्र जैसी
4. ध्रुव जैसा
5. कर्म में रत रहना
6. जब गुणों को समझने वाला ग्राहक मिलता है।

कबीर की साखियाँ भाव सारांश

कबीर ने अपनी साखियों में जीवन से सम्बन्धित अनेक उपयोगी सीखें दी हैं। मानव को भगवान से उतनी ही याचना करनी चाहिए जितनी आवश्यकता हो। दुर्बल को सताना अपने लिए काँटे बोना है। साधु की जाति नहीं पूछनी चाहिए अपितु उसके गुणों पर दृष्टिपात करना चाहिए। भक्ति के मार्ग में काम,क्रोध तथा लोभ बाधक हैं, अत: इनका त्याज्य अनिवार्य है। उच्च कुल में जन्म लेने वाले मानव का आचरण भी पवित्र होना आवश्यक है। फलदायी वृक्ष का आश्रय सुखद होता है। गुणी मानव हीं गुणों की परख कर सकता है। हीरा की परख जौहरी ही जानता है। भगवान गुणों के अक्षय कोष हैं, अवगुण तो हमारे मन-मानस में पल्लवित हैं। कबीर की साखियों में उच्चादर्शों से सम्बन्धित उत्कृष्ट काव्य सृष्टि है।

कबीर की साखियाँ संदर्भ-प्रसंगसहित व्याख्या

साईं इतना दीजिए, जामैं कुटुम समाय।
मैं भी भूखा न रहूँ, साधु न भूखा जाय॥ (1)

शब्दार्थ :
साईं = स्वामी, भगवान। कुटुम = कुटुम्ब। समाय = भरण-पोषण हो जाए। साधु = अतिथि।

सन्दर्भ :
प्रस्तुत साखी कबीरदास द्वारा रचित ‘कबीर की साखियाँ’ शीर्षक से ली गयी है।

प्रसंग :
इसमें कवि ने भगवान से प्रार्थना की है कि हे भगवान! हमें इतना दे दो कि हमारा भली-भाँति गुजारा हो जाए।

व्याख्या :
कबीरदास जी कहते हैं कि हे भगवान! आप हमें इतना भर दे दें जिसमें हमारे कुटुम्ब का भली-भाँति भरण-पोषण हो जाए। मैं भी भूखा न रहूँ और मेरे घर पर जो भी साधु-संन्यासी आएँ, वे भी खाली हाथ न लौटें।

विशेष :

1. कवि ने मात्र कुटुम्ब भरण तक की सुविधा ही भगवान से माँगी है।
2. दोहा छन्द।

दुर्बल को न सताइए, जाकी मोटी हाय।
मरी चाम की स्वांस से, लौह भसम है जाय॥ (2)

शब्दार्थ :
दुर्बल = कमजोर व्यक्ति। चाम = चमड़े।

सदर्भ :
पूर्ववत।

प्रसंग :
कबीर ने दुर्बल व्यक्ति को न सताने का उपदेश दिया है।

व्याख्या :
कबीरदास जी कहते हैं कि दुर्बल व्यक्तियों को मत सताओ। इनकी आहे बड़ी मोटी अर्थात् भारी होती हैं। एक उदाहरण देकर वे कहते हैं कि जब मरे हुए चमड़े से बनाई गई मसक की श्वांस से लोहा भी भस्म हो जाता है तो जीवित व्यक्ति की आहों से कितना अहित हो जायेगा, इसे अच्छी तरह सोच लो और दुर्बलों को मत सताओ।

विशेष :

1. कवि ने दुर्बलों को न सताने का उपदेश दिया।
2. दोहा छन्द।

जाति न पूछो साधु की, पूछि लीजिए ज्ञान।
मोल करो तलवार का, पड़ा रहन दो म्यान॥ (3)

शब्दार्थ :
मोल = मोल-भाव।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कबीरदास जी साधु संन्यासी की जाति न पूछने और ज्ञान जानने की बात कहते हैं।

व्याख्या :
कबीरदास जी संसारी मनुष्यों से कहते हैं कि हे मनुष्यो! तुम कभी भी किसी साधु-संन्यासी की जाति के बारे में मत पूछो। यदि तुम्हें पूछना ही है तो उससे उसके ज्ञान के बारे में जानकारी लो। एक उदाहरण देकर कबीर बताते हैं कि मोल-भाव तो तलवार का किया जाता है, म्यान को कौन पूछता है।

विशेष :

1. गुणों को जानना चाहिए, जाति को नहीं।
2. दोहा छन्द।

जाति हमारी आतमा, प्रान हमारा नाम।
अलख हमारा इष्ट है, गगन हमारा ग्राम॥ (4)

शब्दार्थ :
अलख = अदृश्य। इष्ट = भगवान। गगन = आकाश।

सन्दर्भ ;
पूर्ववत्।

प्रसंग :
इस साखी में कबीरदास जी अपना परिचय देते हुए कह रहे हैं।

व्याख्या :
कबीरदास जी कहते हैं कि हमारी आत्मा ही हमारी जाति है, हमारा नाम ही हमारे प्राण हैं। अदृश्य अर्थात् न दिखाई देने वाला निर्गुण निराकार ईश्वर ही हमारा इष्ट है और आकाश हमारा ग्राम है।
विशेष :

1. कवि ने अपनी व्यापकता का परिचय दिया है।
2. दोहा छन्द है।

कामी क्रोधी लालची, इन पै भक्ति न होय।
भक्ति करै कोई शूरमाँ, जाति वरण कुल खोय॥ (5)

शब्दार्थ :
कामी = काम वासना में रत रहने वाला। जाति = जाति, कुल। वर्ण = वर्ण व्यवस्था। खोय = खोकर।

सन्दर्भ-:
पूर्ववत्।

प्रसंग :
इस छन्द में कबीरदास जी ने बताया है कि कामी-क्रोधी व्यक्तियों से ईश्वर की भक्ति नहीं हो सकती है।

व्याख्या :
कबीरदास जी कहते हैं कि कामी, क्रोधी और लालची स्वभाव के व्यक्तियों से भगवान की भक्ति नहीं हो सकती है जो व्यक्ति जाति एवं वर्ण भूल जाता है, वही शूरमाँ होता है और वही भगवान की भक्ति कर सकता है।

विशेष :

1. विषय-वासनाओं से मुक्त होकर ही भक्ति की जा सकती है।
2. दोहा छन्द।

ऊँचै कुल का जनमियाँ, जे करणी ऊँच न होइ।।
सोवन कलस सुरै भऱ्या, साधू निंद्या सोई॥ (6)

शब्दार्थ :
जनमियाँ = जन्म लेने वाला। करणी = कार्य। सोवन = सोने के। सुरै = मदिरा। भऱ्या = भरी हुई है। निंद्या = निंदा करता है। सोइ = उसकी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कबीरदास जी कहते हैं कि ऊँचे कुल में जन्म लेने से कोई व्यक्ति ऊँचा नहीं होता है।

व्याख्या :
कबीरदास जी कहते हैं कि ऊँचे कुल में जन्म लेने भर से कोई व्यक्ति ऊँचा नहीं हो जाता। यदि उसके कार्य ऊँचे नहीं हैं, तो वह कभी ऊँचा नहीं हो सकता है। ऊँचे अर्थात् श्रेष्ठ कार्य करने वाला व्यक्ति ही.ऊँचा होता है। वे उदाहरण देते हुए कहते हैं कि यदि सोने के कलश में मदिरा भरी हुई है, तो साधु लोग उस कलश की प्रशंसा न करके निन्दा ही करेंगे।

विशेष :

1. व्यक्ति अच्छे कामों से ऊँचा होता है, ऊँचे कुल में जन्म लेने से नहीं।
2. दोहा छन्द।

तरवर तास बिलंबिए, बारह मास फलंत।
सीतल छाया गहर फल, पंधी केलि करत॥ (7)

शब्दार्थ :
तरवर = श्रेष्ठ वृक्ष। तास = उसी का। बिलंबिए = सहारा लीजिए। फलंत = फलता है। गहर = घने। पंषी = पक्षी। केलि = क्रीड़ाएँ।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कबीरदास जी ने ऐसे व्यक्ति का आश्रय लेने का उपदेश दिया है, जिससे हर व्यक्ति सुखी हो।

व्याख्या :
कबीरदास जी तरुवर के माध्यम से उस श्रेष्ठ व्यक्ति का आश्रयं ग्रहण करने का उपदेश देते हैं कि उसी श्रेष्ठ वृक्ष का आश्रय लेना चाहिए जो बारह महीने फल देता हो, जिसकी छाया शीतल हो, जिसमें घने फल लगते हों और जिस पर बैठकर पक्षीगण अपनी क्रीड़ाएँ करते हों।

विशेष :

1. सज्जन लोगों का आश्रय लेने की बात कही है।
2. दोहा छन्द।

जब गुण कँगाहक मिलै, तब गुण लाख बिकाइ।
जब गुण कौं गाहक नहीं, कौड़ी बदले जाइ॥ (8)

शब्दार्थ :
गाहक = ग्रहण करने वाला, जानकार।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि कहता है कि गुणों का महत्त्व तभी तक है जब तक उसके ग्राहक मिलते हैं।

व्याख्या :
कबीरदास जी कहते हैं कि जब गुण के ग्राहक मिल जाते हैं तो गुण लाख रुपये में बिकता है और यदि गुण के ग्राहक न मिलें तो वह गुण कौड़ियों में बिक जाता है।

विशेष :

1. गुण के ग्राहक का महत्व है।
2. दोहा छन्द।

सरपहि दूध पिलाइये, दूधैं विष है जाइ।
ऐसा कोई नां मिलै, सौं सरपैं विष खाइ॥ (9)

शब्दार्थ :
सरपहि = साँप को। लै जाइ = हो जायेगा।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि कहता है कि सर्प को दूध पिलाने से कोई लाभ नहीं है, क्योंकि वह दूध भी विष बन जायेगा।

व्याख्या :
कबीरदास जी कहते हैं कि सर्प को दूध पिलाने से कोई लाभ नहीं है क्योंकि सर्प के गले में जाकर दूध भी विष बन जायेगा। कबीर कहते हैं कि मुझे आज तक ऐसा कोई व्यक्ति नहीं मिला, जो सर्प के विष को खा जाये।

विशेष :

1. साँप से आशय दुष्ट लोग हैं।
2. दोहा छन्द।

करता करे बहुत गुण, आगुण कोई नांहि।
जे दिल खोजौं आपणौं, तो सब औगुण मुझ माहि॥ (10)

शब्दार्थ :
करता = कर्ता, सृष्टि का निर्माणकर्ता। केरे = तेरे। आगुण = अवगुण। आपणौ = अपना। मांहि = अन्दर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कर्ता (भगवान) में गुण ही गुण होते हैं, अवगुण नहीं।

व्याख्या :
कबीरदास जी कहते हैं कि कर्ता अर्थात् भगवान में केवल गुण ही गुण हैं, उनमें कोई भी अवगुण नहीं है। कबीर दास जी कहते हैं कि जब मैंने अपना दिल खोजा तो मैंने पाया कि सभी अवगुण मेरे अन्दर हैं।

विशेष :

1. ईश्वर (कर्ता) की महत्ता का वर्णन तथा अपने अवगुणों का वर्णन कवि ने किया है।
2. दोहा छन्द।

जीवन संदेश भाव सारांश

प्रस्तुत कविता ‘जीवन संदेश’ में श्री रामनरेश त्रिपाठी ने मानव को कर्म में रत रहने का संदेश दिया है। संसार में जड़,चेतन सभी पदार्थ अपने-अपने कर्म में लगे रहते हैं। सूर्य संसार में नवीन शोभा विकीर्ण करता है। चन्द्रमा रात्रि में अमृत की वर्षा करता है।

तुच्छ तिनका भी कर्म में रत रहकर अपना सर्वस्व न्योछावर कर देता है। कवि मनुष्य को सचेत करता हुआ कहता है कि तुम मनुष्य हो, तुम्हारे पास बुद्धि,बल है। तुम्हें भी सोद्देश्य जीवन बिताना चाहिए।

मातृभूमि के प्रति भी अपने कर्तव्य का दृढ़ता से पालन करना चाहिए। जीवन में लोक कल्याण व लोक सेवा आवश्यक है। संसार में मानव को विषम परिस्थितियों का दृढ़ता से सामना करना चाहए, चाहे कितनी ही विषम आँधियाँ आयें। व्यक्ति को चन्द्रमा सा शान्त व ध्रुव तारे की भाँति अचल भावना रखनी चाहिए।

यह संसार ईश्वर के द्वारा निर्मित है। उसी के द्वारा समस्त सृष्टि का संचालन होता है। अतः कवि का कथन है कि जिस भूमि पर तुमने जन्म लिया है उसके प्रति अपने कर्तव्य का पालन अन्तिम श्वांस तक करना चाहिए। यही मानव का परम धर्म है।

जीवन संदेश संदर्भ-प्रसंगसहित व्याख्या

(1) जग में सचर अचर जितने हैं सारे कर्म निरत हैं।
धुन है एक न एक सभी को सबके निश्चित व्रत हैं।
जीवनभर आतप सह वसुधा पर छाया करता है।
तुच्छ पत्र की भी स्वकर्म में कैसी तत्परता है।

शब्दार्थ :
सचर = चलने-फिरने वाले प्राणी। अचर = न चलने-फिरने वाली जड़ वस्तुएँ। कर्म निरत हैं अपने-अपने आप में लगे हुए हैं। व्रत = उद्देश्य, निश्चय। आतप = धूप। वसुधा = पृथ्वी। पत्र = पत्ता। स्वकर्म= अपने काम में। तत्परता = लगन।

सन्दर्भ :
प्रस्तुत छन्द ‘जीवन सन्देश’ शीर्षक कविता से लिया गया है। इसके रचयिता श्री रामनरेश त्रिपाठी हैं।

प्रसंग :
कवि ने यहाँ बताने का प्रयास किया है कि संसार में जड़ एवं चेतन सभी अपने-अपने काम में लगे हुए हैं।

व्याख्या :
श्री रामनरेश त्रिपाठी कहते हैं कि इस संसार। में जितने भी जड़ और चेतन हैं, वे सभी अपने-अपने कामों में लगे हुए हैं। सभी को कोई-न-कोई धुन होती है और उनका कोई-न-कोई व्रत (उद्देश्य) होता है। कवि पत्ते का उदाहरण देते हुए कहता है कि वह तुच्छ पत्ता जीवन भर धूप को सहता हुआ पृथ्वी पर छाया करता रहता है। देखिए उस तुच्छ पत्ते की भी अपने काम में कितनी तत्परता है।

विशेष :

1. कवि का मानना है कि जड़ और चेतन सभी अपने कामों में लगे हुए हैं।
2. अनुप्रास की छटा।

(2) रवि जग में शोभा सरसाता सोम सुधा बरसाता।
सब हैं लगे कर्म में कोई निष्क्रिय दृष्टि न आता॥
हैं उद्देश्य नितान्त तुच्छ तृण के भी लघु जीवन का।
उसी पूर्ति में वह करता है अन्त कर्ममय तन का॥

शब्दार्थ :
रवि = सूर्य। सोम = चन्द्रमा। सुधा = अमृत। निष्क्रिय = बिना काम के। नितान्त = पूरी तरह से। तुच्छ = छोटे। तृण = घास। कर्ममय = काम में लगते हुए।

सन्दर्भ-:
पूर्ववत्।

प्रसंग :
कवि का कथन है कि सूर्य, चन्द्रमा आदि सभी अपने-अपने कर्म में लगे हुए हैं।

व्याख्या :
कविवर त्रिपाठी जी कहते हैं कि सूर्य संसार में शोभा का विस्तार करता है तो चन्द्रमा पृथ्वी पर अमृत बरसाता है। इस संसार में सभी जड़, चेतन अपने-अपने कामों में लगे हुए हैं। कोई भी व्यक्ति हमें बिना काम के नजर नहीं आता है। तुच्छ तिनके के लघु जीवन का भी उद्देश्य है और उसी उद्देश्य की पूर्ति हेतु वह अपने शरीर का अन्त कर देता है।

विशेष :

1. सभी संसार में कार्यरत हैं, बिना काम के कोई नहीं है।
2. अनुप्रास की छटा।

(3) तुम मनुष्य हो, अमित बुद्धिबल-विलसित जन्म तुम्हारा।
क्या उद्देश्य रहित है जग में तुमने कभी विचारा?
बुरा न मानो, एक बार सोचो तुम अपने मन में।
क्या कर्त्तव्य समाप्त कर लिए तुमने निज जीवन में॥

शब्दार्थ :
अमित = अत्यधिक। बुद्धि-बल विलसित = बुद्धि और बल से सुशोभित।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि मनुष्यों को सचेत करते हुए कहता है कि तुम्हें भी अपने जीवन को काम करते हुए सार्थक बनाना है।

व्याख्या :
कविवर त्रिपाठी जी कहते हैं कि तुम मनुष्य हो और तुम्हारा जन्म इस संसार में अत्यधिक बुद्धि एवं बल से युक्त है। ऐसी दशा में क्या तुमने अपने मन में कभी इस बात पर विचार किया है कि तुम्हारा जीवन क्या उद्देश्य रहित है? अर्थात् नहीं।

कवि पुनः कहता है कि हे मनुष्यो ! तुम बुरा मत मानना। तुम अपने मन में एक बार सोचो तो कि क्या तुमने अपने जीवन के सभी कर्त्तव्य पूरे कर लिये हैं अर्थात् अभी नहीं किये हैं।

विशेष :

1. कवि मनुष्य का कर्म क्षेत्र में निरन्तर लगे रहने की प्रेरणा देता है।
2. अनुप्रास की छटा।

(4) वह सनेह की मूर्ति दयामयि माता-तुल्य मही है।
उसके प्रति कर्त्तव्य तुम्हारा क्या कुछ शेष नहीं है।
हाथ पकड़कर प्रथम जिन्होंने चलना तुम्हें सिखाया।
भाषा सिखा हृदय का अद्भुत रूप स्वरूप दिखाया।

शब्दार्थ :
सनेह = स्नेह, प्रेम। दयामयि = दया करने वाली। मही = पृथ्वी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि कहता है कि तुमने जन्म तो ले लिया पर जन्म देने वालों के प्रति तुम्हारे जो कर्त्तव्य हैं, उन्हें पूरा करो।

व्याख्या :
कविवर त्रिपाठी कहते हैं कि वह प्रेम की मूर्ति। और दयामयी पृथ्वी माता के समान है। क्या उसके प्रति तुम्हारा। कोई कर्त्तव्य नहीं है? अर्थात् निश्चय ही उसके प्रति हमारा कर्तव्य है? हाथ पकड़कर जिन्होंने सर्वप्रथम तुम्हें पृथ्वी पर चलना सिखाया है, बाद में भाषा का ज्ञान देकर हृदय के अद्भुत रूप। एवं स्वरूप को दिखाया है; क्या उनके प्रति तुम्हारा कर्त्तव्य नहीं? हाँ अवश्य है।

विशेष :

1. कवि का सन्देश है कि जिन्होंने हमारे साथ कुछ भी उपकार किया है, उनके प्रति हमें कर्त्तव्य पूरा करना चाहिए।
2. अनुप्रास की छटा।

(5) जिनकी कठिन कमाई का फल खाकर बड़े हुए हो।
दीर्घ देह की बाधाओं में निर्भय खड़े हुए हो॥
जिनके पैदा किए, बुने वस्त्रों से देह ढके हो।
आतप-वर्षा-शीत-काल में पीडित हो न सके हो॥

शब्दार्थ :
बाधाओं = रुकावटों। निर्भय = निडर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि कहता है कि पृथ्वी ने तुम्हें जो विभिन्न उपहार। प्रदान किये हैं उनके प्रति तुम्हें पूर्ण रूप से समर्पित होना चाहिए।

व्याख्या :
कविवर त्रिपाठी कहते हैं कि जिन माता-पिता की कमाई खाकर तुम बड़े हुए हो और अनेकानेक शारीरिक कष्टों से तुम्हें उबार कर जिन्होंने निर्भय बनाकर तुम्हें खड़ा कर दिया है। जिनके पैदा किए हुए तथा बुने हुए वस्त्रों से तुम अपना शरीर ढके हुए हो तथा जिनकी कृपा से धूप, वर्षा, शीत आदि की मुसीबतों से बचे रहे हो, क्या उनके प्रति तुम्हारा कोई कर्त्तव्य नहीं है? भाव यह है कि उनके प्रति तुम्हारा पूर्ण समर्पण होना चाहिए।

विशेष :

1. कवि ने मनुष्यों को अपने कर्तव्य के प्रति सचेत करना चाहा है।
2. अनुप्रास की छटा।

(6) क्या उनका उपकार-भार तुम पर लवलेश नहीं है।
उनके प्रति कर्तव्य तुम्हारा क्या कुछ शेष नहीं है।
सतत् ज्वलित दुःख दावानल में जग केदारुण रन में।
छोड़ उन्हें कायर बनकर तुम भाग बसे निर्जन में।

शब्दार्थ :
उपकार = कृपा। लवलेश = जरा भी। सतत् = निरन्तर। ज्वलित = जलते हुए। दावानल = जंगल की आग। दारुण = भयंकर। निर्जन = एकान्त।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि उन लोगों को फटकार लगाता है, जो संसारिक मुसीबतों से घबड़ाकर जंगल में चले जाते हैं।

व्याख्या :
कविवर त्रिपाठी कहते हैं कि क्या जिन लोगों ने तुम्हें जन्म दिया और पाल-पोस कर बड़ा किया, उनके प्रति तुम्हारा कोई कर्त्तव्य नहीं है। निरन्तर जलते हुए दुःख के दावानल में तथा संसार के भयानक रण-क्षेत्र में उन सबको छोड़कर और कायर बनकर तुम निर्जन स्थान में भाग कर बस गए हो, क्या यह तुम्हें उचित लगता है? अर्थात् बिल्कुल नहीं।

विशेष :

1. कवि ने कृतघ्न लोगों को फटकारा है।
2. अनुप्रास की छटा।

(7) यही लोक-कल्याण-कामना यही लोक-सेवा है।
यही अमर करने वाले यश-सुरतरु की मेवा है।
जाओ पुत्र! जगत् में जाओ, व्यर्थ न समय गँवाओ।
सदालोक-कल्याण-निरतहोजीवनसफलबनाओ।

शब्दार्थ :
यश-सुरतरु = शयरूपी कल्प वृक्ष। निरत = लगे रहो।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि मनुष्यों को व्यर्थ में ही समय न गँवाकर लोक कल्याण में लग जाने का उपदेश देता है।

व्याख्या :
श्री त्रिपाठी जी कहते हैं कि लोक कल्याण की भावना से जो व्यक्ति लोक सेवा करता है, इससे भी अमर होता है तथा इसी से उसे यशरूपी कल्पतरु से मिलने वाली मेवा प्राप्त होती है। इसलिए हे मानवी पुत्रो! तुम व्यर्थ में अपना समय मत गँवाओं और सदैव लोक कल्याण में रत रहकर अपने जीवन को सफल बनाओ।

विशेष :

1. कवि ने मनुष्यों को लोक कल्याणकारी कार्यों में लगे रहने का महत्व बताया है।
2. अनुप्रास की छटा।

(8) जनता के विश्वास कर्म मन ध्यान श्रवण भाषण में।
वास करो, आदर्श बनो, विजयी हो जीवन-रण में।
अति अशांत दुखपूर्ण विश्रृंखल क्रान्ति उपासक जग में।
रखना अपनी आत्म शक्ति पर दृढ़निश्चय प्रति पग में।

शब्दार्थ :
वास करो = निवास करो। जीवन-रण = जीवन रूपी रण में। विशृंखल = बिखरे हुए।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि का संदेश है कि अपने कर्मों, मन में, ध्यान में, श्रवण में और भाषण में सदैव ऐसी बातें रखना जिससे तुम। जनता के हृदय में वास कर सको, उनके आदर्श बन सको। इस प्रकार के क्रिया-कलाप करते हुए तुम जीवन रूपी रण में सदैव विजयी होगे। आज संसार का वातावरण अति अशान्त, दुःखपूर्ण बिखरा हुआ, क्रान्ति की उपासना करने वाला बना हुआ है। इन परिस्थितियों में भी तुम अपनी आत्म शक्ति पर प्रत्येक पग में दृढ़ निश्चय रखना अर्थात् कभी विचलित मत। हो जाना।

विशेष :

1. कवि ने जनता के कल्याण में रत रहने का उपदेश दिया है।
2. रूपक एवं अनुप्रास की छटा।

(9) जग की विषम आँधियों के झोंके सम्मुख हो सहना।
स्थिर उद्देश्य-समान और विश्वास सदृश दृढ़ रहना।
जाग्रत नित रहना उदारता-तुल्य असीम हृदय में।
अन्धकार में शान्त, चन्द्रसा, ध्रुव-सा निश्चय भय में।

शब्दार्थ :
विषय आँधियों = विपरीत परिस्थितियों। – सम्मुख = सामने से। जाग्रत = जागना। तुल्य = समान। असीम = सीमा रहित।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि उपदेश देता है कि चाहे संसार में कैसी भी विषम परिस्थितियाँ क्यों न आ जाएँ, तुम ध्रुव तारे के समान डटे रहना।

व्याख्या :
कविवर त्रिपाठी कहते हैं कि संसार में आने वाली विषम परिस्थितियों के झोंकों से तुम विचलित मत होना, अपितु उनका डटकर सामना करना। समान उद्देश्य में स्थिर तथा विश्वास के समान दृढ़ बने रहना। उदारता करते समय नित्य जागते रहना और अपने असीम हृदय में यह गुण धरते रहना। अन्धकार अर्थात् निराशा के समय चन्द्रमा के समान शान्त रहना और भय तथा विपत्ति में ध्रुव के समान निश्चल (अडिग) बने रहना।

विशेष :

1. कवि ने किसी भी स्थिति में विचलित न होने। का सन्देश दिया है।
2. उपमा और अनुप्रास की छटा।

(10) जगन्नियंता की इच्छा से यह संसार बना है।
उसकी ही क्रीड़ा का रूपक यह समस्त रचना है।
है यह कर्म-भूमि जीवों की यहाँ कर्मच्युत होना।
धोखे में पड़ना अलभ्य अवसर से है कर धोना॥
पैदा कर जिस देश जाति ने तुमको पाला पोसा।
किए हुए वह निजहित का तुमसे बड़ा भरोसा।
उससे होना उऋण प्रथम है सत्कर्त्तव्य तुम्हारा।
फिर दे सकते हो वसुधा को शेष स्वजीवन सारा॥

शब्दार्थ :
जगन्नियंता = ईश्वर। कर्मच्युत = काम से भागना। अलभ्य = जो सरलता से प्राप्त न हो सके। उऋण = ऋण मुक्त।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
कवि ने सन्देश दिया है कि जिस देश जाति ने तुमको पैदा कर तुम्हारा भरण-पोषण किया है, उसके प्रति नि:स्वार्थ भाव से सेवा करके तुम उऋण हो सकते हो।

व्याख्या :
कविवर त्रिपाठी जी कहते हैं कि परम सत्ता की इच्छा से ही इस संसार का निर्माण हुआ है तथा उसकी ही क्रीड़ा का रूपक यह संसार है। यह देश जीवों की कर्मभूमि है अतः यहाँ कभी भी कर्म से पीछे मत हटना। यदि किसी धोखे में पड़कर तुमने कर्त्तव्य को त्याग दिया, तो तुम्हारे हाथ से अलभ्य (अनमोल) अवसर निकल जायेगा। जिस देश और जाति ने तुम्हें पैदा कर तुम्हें पाला-पोसा है, वह अपने हित का तुम परं बहुत बड़ा भरोसा किए हुए है। तुम्हारा यह सत्कर्म तथा प्रथम कर्त्तव्य है कि तुम देश और जाति के ऋण से ऋण-मुक्त हो जाओ। इस प्रकार तुम अपना शेष (बचा हुआ) जीवन अपनी पृथ्वी को दे सकते हो।

विशेष :

1. कवि ने कर्म क्षेत्र से भागने वालों की निन्दा सकते हो। की है।
2. कवि का उपदेश है कि जिस देश और जाति में तुमने जन्म पाया है, उनके प्रति भी तुम्हारे कुछ कर्त्तव्य हैं।
3. अनुप्रास की छटा।

MP Board Class 10th Hindi Solutions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 10 Vector Algebra Ex 10.3Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 10 Vector Algebra Ex 10.3

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MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.1

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MP Board Class 12th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

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MP Board Class 12th Maths Solutions Chapter 8 Application of Integrals Ex 8.2

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MP Board Class 12th Chemistry Solutions Chapter 2 Solutions

Solutions NCERT Intext Exercises

Question 1.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Solution:
Mass of solution = Mass of benzene + Mass of CCl₄
= 22g + 122g = 144g

Alternatively : Mass % of CC14 = 100 – 15.28 = 84.72%.

This Combined Gas Law Calculator can help you estimate either the pressure, temperature or the volume of gas.

Question 2.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Solution:
30% of benzene in carbon tetrachloride by mass means that
Mass of benzene in the solution = 30 g
Mass of solution = 100 g
∴ Mass of carbon tetrachloride = 100 – 30 g = 70 g
Molar mass of benzene (C₆H₆) = 78 g mol^-1
Molar mass of CCl₄ = 12 + 4 × 35.5 = 154 g mol^-1

The standard unit of concentration in molarity calculator chemistry.

Question 3.
Calculate the molarity of each of the following solutions:
(a) 30 g of Co(NO₃)₂. 6H₂O in 4.3 L of solution
(b) 30 ml of 0.5 M H₂SO₄ diluted to 500 ml.
Solution:
(a) Molar mass of Co(NO₃)₂.6H₂O = M_B = 291, W_B of Co(NO₃)₂.6H₂O = 30g, V_sol = 4.3L = 4300ml

(b) Molanty of solution after dilution may be calculated as.
M₁V₁ (Concentrated) M₂,V₂ (Diluted)
0.5 x 30 = M₂ x 500
M₂ = \(\frac{0 \cdot 5 \times 30}{500}\) = 0.03

Question 4.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
W_B = ?, M_B = 60, W_A = 2.5 kg, m = 0.25
m = \(\frac{W_{B}}{M_{B} \times W_{A} \text { in } k g}\)
W_B = m × M_B × W_A in kg
= 0.25 × 60
= 37.5 kg

How to find molar concentration and molarity.

Question 5.
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g ml^-1.
Solution:
Here, M_B = 166, W_B = 20, W_A = 80

Calculating Equilibrium Concentrations from Initial Concentrations.

Question 6.
H₂S, toxic gas with the rotten egg-like smell, is used for the qualitative analysis. If the solubility of H₂S in the water at STP is 0.195 m, calculate Henry’s law constant
Solution:
0.195m solution means that 0.195 moles of H₂S is dissolved in 1 kg of water.
Moles of H₂S = 0.195

Get Mass Transfer MCQ‘s with Answers & Solutions that checks your basic knowledge of Mass Transfer.

Question 7.
Henry’s law constant for CO₂ in water is 1.67 × 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.
Solution:
According to Henry’s law
P = K_HX …(1)
P = 2.5 atm = 2.5 × 101325 Pa, K_H = 1.67 × 10⁸ Pa
Putting these values in equation (1), we get
2.5 × 101325 = 1.67 × 10⁸ × X_CO2

Simply, you can calculate the mole fraction equation by just giving the moles of each reactant.

Question 8.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also And the composition of the vapour phase.
Solution:
Here, P°A = 450 mm, P°_B = 700 mm, P_Total = 600 mm
Applying Raoult’s law
P_A = X_AP°_A, P_B = X_BP°_B = (1 – X_A)P°_B
P_Total = P_A + P_B = X_AP°_A + (1 – X_A)P°_B =
P°_B + (P°_A – P°_B) X_A.
Substituting the value we get
600 = 700+ (450 – 700) X_A
or 250 X_A = 100
or X_A = \(\frac { 100 }{ 250 } \) = 0.40
Thus, the composition of the liquid mixture will be
X_A (mole fraction of A) = 0.40
X_B (mole fraction of B) = 1 – 0.40 = 0.60
∴ P_A = X_AP°_A = 0.40 × 450 = 180 mm
P_B = X_BP°_B = 0.60 × 700 = 420 mm

Question 9.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
We know that

Here, P° = 23-8mm, W₂ = 50g, M₂(urea) = 60 g mol^-1, W₁ = 850g, M₁ (water) = 18 g mol^-1

Thus, vapour pressure of water in the solution = 23.4 mm.

Question 10.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol^-1.
Solution:

Question 11.
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_F = 3.9 K kg mol^-1.
Solution:

Question 12.
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of poly-mer of molar mass 1,85,000 in 450 ml of water at 37°C.
Solution:
π = CRT
\(=\frac{n}{\mathrm{V}} \mathrm{RT}\)
Here, number of moles of solute dissolved (n) =

Solutions NCERT TextBook Exercises

Tools in Online Chemistry formula calculator ·

Question 1.
Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Answer:
A solution is a homogeneous mixture of two or more substances which are chemically non-reacting. On the basis of physical component solutions are of the following types:
Solid Solution

Liquid Solution

Gaseous Solution

Question 2.
Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What kind of solid solution is this likely to be?
Answer:
Solid in the solid type. eg. Copper in gold. This type of solution is called alloys.

Converting grams to moles Calculator is as simple as multiplying the total mass by the moles per unit of mass.

Question 3.
Define the following terms :
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Answer:
(i) Mole fraction: Ratio of moles of a component (solute or solvent) to the total number of moles of all the components of solution is called mole fraction. If moles of solute is n and that of solvent is N, then
Mole fraction of solute = \(\frac { n }{ n+N } \)
and mole fraction of solvent = \(\frac { N }{ n+N } \)
(ii) Molality: Molality is defined as a number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m.

(iii) Molarity: Molarity is defined as a number of gram moles of solute dissolved in a litre of solution. It is denoted by M.

(iv) Mass percentage: The mass percentage of a component in a given solution is the mass of the component per 100 gm of the solution.

This can be expressed as WAV. For example, 10% Na₂CO₃ WAV means 10g of Na₂CO₃ is dissolved in 100 g of the solution (It means 10 g Na₂CO₃ is dissolved in 90 g of H₂O).

Question 4.
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 ?
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

Question 5.
A solution of glucose in water is labelled as 10% WAV, what would be the molality and mole fraction of each component in the solution ? If the density of solution is 1.2 g ml^-1, then what shall be the molarity of the solution ?
Solution:
10% (W/W) glucose means 10g of glucose in 100g of solution i.e., 90 g of water = 0.090 kg of water

Question 6.
How many ml of 0.1M HCl are required to react completely with lg mixture of Na₂CO₃ and NaHCO₃ containing the equimolar amount of both?
Solution:
Let there, is x g Na₂CO3 and (1 – x)g NaHCO₃ in the mixture.
Molar mass of Na₂CO₃ = 106 g/mol
Molar mass of NaHCO₃ = 84 g/mol
Number of moles of Na₂CO₃ = Number of moles of NaHCO₃
\(\frac { x }{ 106 } \) = \(\frac { (1-x) }{ 84 } \)
On solving, x = 0.56.
Number of moles of Na₂CO₃ = Number of moles of NaHCO₃ = 5.283 × 10^-3
During the process of Neutralisation, following reactions takes place :
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ ↑
NaHCO₃ + HC1 → NaCl + H₂O + CO₂ ↑
Number of moles of HCl required = 2 × Number of moles of Na₂CO₃ + Number of moles of NaHCO₃
= 2 × 5.283 × 10^-3 + 5.283 × 10^-3 = 0.0158

Question 7.
A solution is obtained by mixing 300 g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Solution:
300 g of 25% solution contain solute = 75g
400g of 40% solution contain solute = 160g
Total solute = 160 + 75 = 235g
Total solution = 300 + 400 = 700g
∴ Mass % of solute = \(\frac { 235 }{ 700 } \) × 100 = 33.5%
and mass % of water = 100 – 33.5 = 66.5%.

Question 8.
An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1-072 g ml^-1, then what shall be the molarity of the solution?
Solution:
Mass of solute, C₂H₄(OH)₂ = 222.6g, Molar mass of solute = 62 g mol^-1

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Solution:
15 ppm (by mass) means 15 g of CHCl3 is present in 106 g of solution.

Question 10.
What role does the molecular interaction play in a solution of alcohol and water?
Answer:
There is strong hydrogen bonding in alcohol molecules as well as water molecules. On mixing the molecular interactions are the weekend. Hence, they show positive deviations from ideal behaviour. AS a result, the solution will have higher vapour pressure and lower boiling point than that of water and alcohol.

Question 11.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:

The dissolution of a gas in a liquid is exothermic process. Therefore in accordance with Le-Chatilier’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 12.
State Henry’s law and mention some important applications.
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 α P or m = KP, where K is a constant. We can understand Henry’s law by taking example of soda water bottle. Soda water contains carbon dioxide dissolved in water under pressure.

Applications of Henry’s law:

1. In the production of carbonated beverages : To increase the solubility of CO₂ in soft drinks, soda water, bear etc. the bottles are sealed at high pressure.

2. In exchange of gases in the blood : The partial pressure of O₂ is high inhaled air, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low therefore oxyhaemoglobin releases oxygen in order to carry out cellular activities.

3. In deep sea diving : Deep sea divers depend upon compressed air for breathing at high pressure under water. The compressed air contains N₂ in addition to O₂, which are not very soluble in blood at normal pressure. However, at great depths when the diver breathes in compressed air from the supply tank, more N₂ dissolve in the blood and in other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. When the divers come towards the surface at atmospheric pressure, this dissolve nitrogen bubbles out of the blood. These bubbles restrict blood flow, affect the transmission of nerve impulses. This causes a disease called bends or decompression sickness. To avoid bends, as well as toxic effects of high concentration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% He, 56.2% N₂ and 32.1% O₂).

4. At high altitudes: At high altitudes the partial pressure of O₂ is less than that at the ground level. This result in low concentration of oxygen in the blood and tissues of the people living at high altitudes or climbers. The low blood oxygen causes climbers to become weak and unable to think clearly known as anoxia.

5. Aquatic life : The dissolution of oxygen (from air) in water helps in the existence of aquatic life in various water bodies like : Lake, rivers and sea.

The partial pressure formula of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.

Question 13.
The partial pressure of ethane over a solution containing 6.56 x 10^-3 g of ethane is 1 bar. If the solution contains 5.00 x 10^-2 g of ethane, then what shall be the partial pressure of the gas ?
Solution:
According to Henry’s law m = K_P, 6.56 × 10^-3 g = K × 1 bar, K = 6.56 × 10^-3 g bar^-1
Now when m = 5 × 10^-2 g, P = ?
Applying m’ = K × P
5.00 × 10^-2g = 6.56 × 10^-3 g bar^-1 × P

Question 14.
What is meant by positive and negative deviations from Raoult’s law and how is the sign of ∆_mix H related to positive and negative deviations from Raoult’s law ?
Answer:
Positive deviation: When vapour pressure of the solution is greater than as expected on the basis of Raoult’s Law, it is known as positive deviation. For a solution formed by components A and B, if the A-B interactions in the solutions are weaker than the solute A-A and solvent B-B interactions in the two components, then the escaping tendency of A and B types of molecules from the solution becomes more than from pure liquids. As a result, each component of the solution has a partial vapour pressure greater than expected on the basis of Raoult’s Law.

Characteristics of solution representing positive deviation if:

Example:

• Ethyl alcohol and water
• Acetone and benzene.

Negative deviation: When vapour pressure of the solution is less than expected on the basis of Raoult’s Law, it is known as negative deviation. In this type of solution, A-B (solute-solvent) interaction is stronger than the interaction between A-A (solute-solute) and B-B (solvent-solvent). Thus, the escaping tendency of A and B types of molecules from the solution becomes less than from pure liquids. As a result, each component of the solution has a partial vapour pressure lesser than expected on the basis of Raoult’s Law.

Characteristics of solution representing negative deviation.if:

Example : (1) HNO₂ and water and
(2) Chloroform and Acetone.

Question 15.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1-004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Vapour pressure of pure water at the boiling point
P° = 1 atm = 1.013 bar
Vapour pressure of solution P_s = 1.004 bar
M₁ = 18 g mol^-1
M₂ = ?
Mass of solute = W₂ = 2 g
Mass of solution = 100 g
Mass of solvent W₁ = 98 g
Applying Raoult’s law for dilute solution

Question 16.
Heptane and octane form an ideal solution. At 373K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Solution:
Heptane (C₇H₁₆) Octane (C₈ H₁₈)
Mass = 26g Mass = 35g
Molar mass = 100 Molar mass =114

Question 17.
The vapour pressure of water is 12.3 kPa at 300K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Solution:
1 molal solution means 1 mol of the solute in 1 kg of the solvent (water)

Question 18.
Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Solution:
According to Raoult’s law, P = P°_A X_A …(1)
P° = 100 then P = 80
∴ From equation (1), X_A = 0.80

Question 19.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2-8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K.
Calculate:
(i) molar mass of the solute
(ii) vapour pressure of water at 298 K.
Solution:
Applying Raoult’s law


II^nd experiment:
Given, W_B = 30g, W_A = 90 + 18 = 108g, P_s = 2.9 kPa, M_A = 18 g mol^2-1
Substituting the value in equation (1)

Question 20.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Solution:
For cane sugar, ∆T_f = 273.15 – 271.0 = 2.15°C

Question 21.
Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20g of benzene (C₆H₆), lg of AB₂ lowers the freezing point by 2-3 Kwhereas 1.0g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Solution:
Applying the formula

Suppose atomic masses of A and B are ‘a’ and ‘b’ respectively.
Then, Molar mass of AB₂ = a + 2b = 110.87 g mol^-1 …(1)
Molar mass of AB₄ = a + 4b = 196.15 g mol^-1 …(2)
Equation (2) – (1) gives
2b = 85.28 or b = 42.64
Substituting in equation (1), we get
a + 2 × 42.64 = 110.87
or a = 25.59
Thus, atomic mass of A = 25.59 u
Atomic mass of B = 42.64u.

Question 22.
At 300 K, 36g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Solution:

Dividing equation (1) by equation (2), we get

Here, the volume of the solution is 1L. So, the conc, of the solution

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs :
(i) n-hexane and n-octane
(ii) I₂ and CCl₄
(iii) NaClO₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆O).
Answer:
(i) n-hexane and n-octane: Dispersion or London force.
(ii) I₂ and CCl₄ (Both non-polar): London or Dispersion forces.
(iii) NaClO₄ (Ionic) and water (Polar): Ion-dipole interaction also called hydration of ion.
(iv) Methanol (Polar) and acetone (Polar): Dipole-dipole interaction.
(v) Acetonitrile (Polar) and acetone (Polar): Dipole-dipole.

You can calculate the ionic strength of a solution by applying the Debye and Huckel formula. Alternatively, use an ionic strength calculator.

Question 24.
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 ‘like dissolves like’, n-octane is a non-polar solvent, hence non-polar compounds will be more soluble.
KCl < CH₃OH < CH₃CN < Cyclohexane.

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water :
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform and
(vi) pentanol.
Answer:
Highly soluble: Formic acid and ethylene glycol. They are able to form H- bonding with a water molecules.
Insoluble: Chloroform and toluene being non-polar are insoluble in a polar medium like water.
Partially soluble: Phenol and pentanol from weaker H-bonding with water hence, they are partially soluble.

Question 26.
If the density of some lake water is T25 g ml^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.
Solution:
Given W_B = 92g, M_B = 23, W_A = 1000g

Question 27.
If the solubility product of CuS is 6 × 10^-16, calculate the maximum molarity of CuS in an aqueous solution.
Solution:
Given K_sp of CuS = 6 × 10^-16
If ‘s’ is the solubility, then

Question 28.
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5g of C₉H₈O₄ is dissolved in 450g of CH₃CN.
Solution:

Question 29.
Nalorphine (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphine generally given is 1.5 mg. Calculate the mass of 1.5 × 10^-3 m aqueous solution required for the above dose.
Solution:

Question 30.
Calculate the amount of benzoic acid (C₆H₅CO-OH) required for preparing 250 ml of 0.15 M solution in methanol.
Solution:

Question 31.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:
Acetic acid < Trichloroacetic acid < Trifluoroacetic acid
Degree of ionisation, increases with the increase in the electron-withdrawing effect of the groups attached to the carboxylic group.

Increasing electron-withdrawing effect
Depression in freezing point is a colligative property. As more ions are produced by ionisation of trifluoroacetic acid, so depression in the freezing point is maximum.

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH₂CH₂CHClCOOH is added to 250g of water. Ka = 1.4 × 10^-3, K_f = 1.86 K kg mol^-1.
Solution:
Molar mass of solute CH₃CH₂CHClCOOH (MB) = 122.5 g mol^-1

If α is the degree of dissociation of CH₃CH₂ CHCl-COOH.

The only substantial evidence concerning the application of the dilution factor formula.

Question 33.
19-5g of CH₂FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1-0°C. Calculate the van’t Hoff factor and dissociation constant of fluoro acetic acid.
Solution:
Molecular mass of CH₂FCOOH (M_B) = 78, W_B = 19.5g, W_A = 500g

CH₂FCOOH dissociates as CH₂FCOO^– and H⁺ if α is the degree of dissociation.

Question 34.
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.
Solution:

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298K under 760 mm Hg.
Solution:
Here K_H = 4.27 × 10⁵ mm, P = 760 mm
Applying Henry’s law, P = K_HX

∴ Mole fraction of methane in benzene = 1.78 × 10^-3.

Question 36.
100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.
Solution:

Vapour pressure of a solution of two liquids A and B may be calculated as

Question 37.
Vapour pressures of pure acetone and chloroform at 328K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot P_total,P_chioroform and P_acetonc as a function of X_acetonc. The experimental data observed for different compositions of mixture is :

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
Answer:
From the question, we have the following data

It can be observed from the graph that the plot for the P_total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Question 38.
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.
Solution:
A → Benzene (C₆H₆); B → Toluene (C₇H_g)

Mole fraction of components in vapour phase may be calculated by using Dalton’s law,

Question 39.
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with the approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Solution:
The vapour pressure of air over water = 10 atm
The partial pressure of N₂ and O₂ are :

Question 40.
Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litres of water such that its osmotic pressure is 0.75 atm at 27°C.
Solution:

Question 41.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25° C, assuming that it is completely dissociated.
Solution:

Solutions Other Important Questions and Answers

Solutions Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
The elevation in boiling point of a solution of molal concentration of solute will be maximum if the solvent is :
(a) Ethyl alcohol
(b) Acetone
(c) Benzene
(d) Chloroform.

Question 2.
Solutions of similar osmotic pressure are known as :
(a) Hypotonic
(b) Hypertonic
(c) Isotonic
(d) Normal.

Question 3.
How many ml of 1 M H₂SO₄ is required to neutralize 10 ml of 1 N NaOH :
(a) 20 ml
(b) 2.5 ml
(c) 5 ml
(d) 10ml.

Question 4.
Which of the following solution does not show positive deviation from Raoult’s law:
(a) Benzene – chloroform
(b) Benzene – acetone
(c) Benzene – ethanol
(d) Benzene – CCl₄.

Question 5.
Molarity of solution of H₂SO₄ containing 9.8 gm H₂SO₄ dissolved in 2 litre water is:
(a) 0.1 M
(b) 0.05 M
(c) 0.01 M
(d) 0.2 M.

Question 6.
On dissolving common salt in water boiling point of water :
(a) Decreases
(b) Increases
(c) Does not change
(d) Cannot be said.

Question 7.
When blood cells are kept in high osmotic pressure solution than cellsap then:
(a) They contract
(b) They swell up
(c) Not affected
(d) First contract then swell.

Question 8.
All of the following form ideal solution, except one :
(a) C₂H₅Br and C₂H₅Cl
(b) C₆H₅Cl and C₆H₅Br
(c) C₆H₆ and C₆H₅CH₃
(d) C₂H₅I and C₂H₅OH

Question 9.
According to Raoult’s law relative lowering of vapour pressure of solution of non-volatile solute is equal to :
(a) Mole fraction of solvent
(b) Mole fraction of solute
(c) Mass percent of solvent
(d) Mass percent of solute.

Question 10.
For osmotic pressure (P), volume (V) and temperature (T) which of the follow¬ing statement is false:
(a) P α \(\frac { 1 }{ V } \) T’s constant
(b) P α T if T is constant
(c) p α V if T is constant
(d) PV is constant if T is constant.

Question 11.
Whose boiling point is highest at 1 atm. pressure :
(a) 0.1M glucose
(b) 0.1M BaCl₂
(c) 0.1M NaCl
(d) 0.1M urea.

Question 12.
Semipermeable membrane is chemically :
(a) Copper ferrocyanide
(b) Copper ferricyanide
(c) Copper sulphate
(d) Pottassium ferrocyanide.

Question 13.
Which among the following is a colligative property :
(a) Surface tension
(b) Viscosity
(c) Osmotic pressure
(d) Optical solution.

Question 14.
Experimental molecular mass of an electrolyte will ‘always be less than’ its calculated value because value of van’t Hoff factor i is :
(a) Less than 1
(b) More than 1
(c) Equal to 1
(d) Zero.

Question 15.
In molal solution, 1 mole of solute substance is dissolved in :
(a) In 1000 gm. solvent
(b) In 1 litre solution
(c) In 1 litre solvent
(d) In 224 litre solution.

Question 16.
If boiling point of solution is T₁ and boiling point of solvent is T₂, then elevation in boiling point will be :
(a) T₁ + T₂
(b) T₁ – T₂
(c) T₂ – T₁
(d) T₁T₂.

Question 17.
Colligative property is :
(a) Change in free energy
(b) Change in pressure
(c) Heat of vapourisation
(d) Osmotic pressure.

Question 18.
Gram molality of a solution is :
(a) Number of molecules of solute per 1000 ml solvent
(b) Number of molecules of solute per 1000 gm solvent
(c) Number of molecules of solute per 1000 ml solvent
(d) Number of gm equivalent of solute per 1000 ml solvent.

Question 19.
An Ideal solution is that:
(a) Which represents negative deviation towards Raoult’s law
(b) Which represents positive deviation towards Raoult’s law
(c) Is not related to Raoult’s law
(d) Obey’s Raoult’s law.

Question 20.
Order of osmotic pressure of BaCl₂, NaCl and glucose solutions of same molarity will be:
(a) BaCl₂ > NaCl > Glucose
(b) NaCl > BaCl₂ > Glucose
(c) Glucose > BaCl₂ > NaCl
(c) Glucose > NaCl > BaCl₂.

Question 21.
A solution contains 20 mole solute and total number of moles is 80. Mole frac-tion of solute will be :
(a) 2.5
(b) 0.25
(c) 1
(d) 0.75.

Question 22.
A solution contain 1 mole of water and 4 moles of ethanol. The mole fraction of water and ethanol in solution will be :
(a) 0.2 water + 0.8 ethanol
(b) 0.4 water + 0.6 ethanol
(c) 0.6 water + 0.8 ethanol
(d) 0.8 water + 0.2 ethanol.

Question 23.
Colligative properties of solution depends upon :
(a) Nature of solvent
(b) Nature of solute
(c) Number of solute particles present in solution
(d) None of these.

Question 24.
Molality of pure water is :
(a) 55.6
(b) 50
(c) 100
(d) 18.

Question 25.
Is not a colligative property :
(a) Osmotic pressure
(b) Vapour pressure depression
(c) Freezing point depression
(d) Boiling point elevation.

Question 26.
Formula for determining osmotic pressure :

Question 27.
Ratio of observed value of colligative property to that of theoritical value is known as:
(a) Colligative property
(b) van’t Hoff factor
(c) Solution constant
(d) Specific constant.

Question 28.
Which of the following does not show positive deviation from Raoult’s law :
(a) Benzene-chloroform
(b) Benzene-acetone
(c) Benzene-ethanol
(d) Benzene-CCl₄.

Question 29.
6 gm urea (mol. wt 60) dissolved in 180 gm of water. The mole fraction of urea will be:
(a) \(\frac { 10 }{ 10.1 } \)
(b) \(\frac { 10.1 }{ 10 } \)
(c) \(\frac { 0.1 }{ 10.1 } \)
(d) \(\frac { 10.1 }{ 0.1 } \)

Answers:
1. (c), 2. (c), 3. (c), 4. (a), 5. (b), 6. (b), 7. (a), 8. (d), 9. (b), 10. (c), 11. (b), 12. (a), 13. (c), 14. (b), 15. (a), 16. (b), 17. (d), 18. (b), 19. (d), 20. (a), 21. (b), 22. (a), 23. (c), 24. (a), 25. (b), 26. (c), 27. (b), 28. (d), 29 (c).

Question 2.
Fill in the blanks :

1. Value of van’t Hoff factor for a solute showing normal state in the solution will be …………………. one.
2. The mathematical expresssion for relative lowering in vapour pressure is ………………….
3. Number of moles of solute in 1000 gm solvent is known as ………………….
4. Liquid mixture which boil without any change in its composition is called ………………….
5. Through semipermeable membrane only …………………. molecules can pass through.
6. At high altitudes boiling point of water decreases because at high attitudes atmospheric pressure is ………………….
7. Molality of water is ………………….
8. Soda water is a solution of ………………….
9. Number of moles of solute present in one litre solution is known as ………………….
10. Non-ideal solution of 95.4% of H₂O + C₂H₅OH represents …………………. deviation
Answer:
1. Equal
2. \(\frac{P_{A}^{\circ}-P_{A}}{P_{A}^{\circ}}\)
3. Molality
4. Azeotropic liquid mixture
5. Solvent
6. Less
7. 55.6 m
8. Gas in liquid
9. Molality
10. Positive.

Question 3.
Match the following:

Answers:

1. (c)
2. (a)
3. (b)
4. (h)
5. (f)
6. (g)
7. (d)
8. (e).

Question 4.
Answer in one word / sentence :

1. Write the formula to determine normality.
2. What is the unit of molality ?
3. Write the formula which states the relation between relative lowering in vapour pres¬sure and mass of solute.
4. What is the property of a dilute solution which depend on the number of solute present in it ?
5. What is Raoult’s law ?
6. Give an example of non-ideal solution showing positive deviation.
7. Give an example of non-ideal solution showing negative deviation.
8. Give an example of antifreeze compound.
9. Write the unit of representing pollution.
10. Write the example of minimum boiling constant solution.
Answers:
1.

2. Mole per kilogram,
3. \(\frac{P_{A}^{0}-P_{A}}{P_{A}^{\circ}}=\frac{W_{B}}{M_{B}} \times \frac{M_{A}}{W_{A}}\)
4. Colligative properties
5. At a definite tempearture for a solution of a non-volatile solute relative lowering in vapour pressure is equal to the mole fraction of solute
6. CH₃COCH₃ + C₆H₆
7. CHCl₃ + CH₃COCH₃
8. Ethylene glycol
9. ppm
10. 96.4% C₂H₅OH + 4.5% H₂O.

Solutions Very Short Answer Type Questions

Question 1.
Write formula of van’t Hoff factor T.
Answer:

Question 2.
Write down van’t Hoff equation. Give formula used for calculating molecular mass with its help.
Answer:

Where, W is mass of solute, R is solution constant, T is temperature, n is osmotic pressure and V is volume of solution.

Question 3.
6.3 gm oxalic acid (Eqv. wt 63) is dissolved in 500 ml of solution. Find out the normality of solution.
Solution:

Question 4.
Define formality and write its formula.
Answer:
Formality is defined as number of gram formula mass of substance dissolved in a litre of solvent. It is denoted by F. It is used for solutions in which solute associates.

Question 5.
Determine the molarity of a solution of 4#0 gram per litre concentration of NaOH.
Solution:

Question 6.
Give two-two example of solution showing negative deviation.
Answer:
(i) CHCl₃ + CH₃COCH₃
(ii) CHCl₃ + C₂H₅OC₂H₅.

Question 7.
Give two examples of non-ideal solutions showing positive deviation.
Answer:
Examples of non-ideal solutions showing positive deviation are :
(i) CCl₄ and CHCl₃
(ii) CCl₄ and C₆H₅CH₃ (Toluene).

Question 8.
Define Normality.
Answer:
Number of gram equivalent of solute present in one litre of solution is called normality. It is represented by N.

Normality of a solution changes with temperature as it is based on mass-volume relationship and volume changes with change in temperature.

Question 9.
If 2 gm NaOH is present in 250 ml solution, then determine the normality of the solution.
Solution:
Equivalent mass of sodium hydroxide (NaOH) = 40
Amount dissolved in 250 ml NaOH solution = 2 gm
∴ 1000 ml solution of NaOH contains

Question 10.
Differentiate between Molarity and Molality.
Answer:
Differences between Molarity and Molality :

Molarity (M)

1. Molarity involves the total volume of solution.
2. In molarity, gram moles of solute are dissolved in 1 litre of solution.
3. Molarity changes with temperature because volume changes with tempe-rature.

Molality (m)

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

Question 11.
Explain the following term : Parts per million.
Answer:
(i) 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.

Where, ppm. is the concentration of component A in parts per million.

Question 12.
Write two examples of non-ideal solution showing negative deviation.
Answer:

1. Chloroform and acetone
2. Water and hydrochloric acid.

Question 13.
On which factor colligative properties of a solution depend ?
Answer:
Number of solute particles.

Question 14.
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 324, then it starts decreasing. Thus, 324°C is the transition temperature of sodium sulphate.

Question 15.
Give an example of such a solid solution in which solute is a gas.
Answer:
Hydrogen (solute) in Palladium (solvent).

Question 16.
Sprinkling of salt help in clearing the snow covered roads in hilly areas. Why?
Answer:
On sprinkling salts like CaCl₂ or NaCl over snow covered roads, the freezing point of water lowers to such an extent that water does not freeze to form ice and as a result the snow starts melting from the surface and therefore it helps in clearing the roads.

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 proportional to the mole fraction of the solute,
Mathematically,

Where, P°_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 ?
Ans.
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 Raoult’s law i.e. their vapour pressure is high thus their boiling point is low are known as low boiling azeotropic mixture.
Example : (i) CS₂ + Acetone, (ii) C₂H₅OH + 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 : (i) Acetone + Chloroform, (ii) Ether + Chloroform.

Question. 3.
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 W_A gram of solvent and the mollecular mass of non-volatile solute is MB gram. Then, molality, m will be

Question 4.
Derive the expression for molecular mass of solute by relative lowering in vapour pressure.
Answer:
Determination of molecular mass of solute by relative lowering in vapour pressure : Suppose a known mass(WB)of solute is dissolved in known mass (WA) of solvent to give a dilute solution and the relative lowering of vapour pressure of the solution is equal to mole fraction of solute.

It is determined by experiment, when the molecular mass of solvent (M_A) is known, the molecular mass of the solute (M_B) can be calculated as given below :



If relative lowering of vapour pressure \(\left[\frac{P_{A}^{0}-P_{A}}{P_{A}^{0}}\right]\) is known and WA,WB,MA are also known then molecular mass of solute MB can be calculated from the eqn. (5).

Question 5.
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 :

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:

Question 6.
Differentiate between Diffusion and Osmosis.
Answer:
Differences between Diffusion and Osmosis :

Diffusion:

1. Molecules move from a region of high concentration to lower concentration.
2. Semipermeable membrane is not required.
3. This process takes place in gases and in liquids.
4. Molecule of both solute and solvent move.
5. It cannot be stopped by applying pressure from opposite direction.

Osmosis:

1. Molecules of solvent move from solution of low concentration to solution of high concentration.
2. Semipermeable membrane is required.
3. This process takes place only in solution.
4. Only molecules of solvent move.
5. It can be stopped by applying pressure from opposite direction.

Question 7.
Write four examples of colligative properties of solutions.
Answer:
Physical properties of solution which depends upon number of solute particles dissolved in solution, are called colligative properties.

Colligative properties are:

1. Lowering of vapour pressure
2. Elevation in boiling point
3. Depression in freezing point
4. Osmotic pressure.

Value of ail colligative properties increases with increase in concentration of solute and decreases with decrease in concentration.

Question 8.
Establish van’t Hoff solution equation.
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 α T
Derivation : Osmotic pressure n of a solution is directly proportional to a molar concentration.

Question 9.
Define the following :
(i) Molal elevation boiling point constant
(ii) Molal freezing point depression constant.
Answer:
(i) 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

Where Kb = Molal boiling point elevation constant

(ii) 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 1 gm mole of solute is dissolved in 1000 gm of solvent.”
∴ Depression in freezing point

Question 10.
(a) What is osmotic pressure ?
(b) Solution of urea is prepared by dissolving 6 gm urea in 1 litre. Determine the osmotic pressure of that urea solution at 300 K. (R = 0.0821 L atom K-1 mol^-1) (Mo-lecular mass of urea = 60)
Answer:
(a) Osmotic Pressure: Osmotic pressure is the excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semipermeable membrane. It is denoted by π.

Solutions Long Answer Type Questions

Question 1.
What are constant boiling mixture ? Write three differences in Ideal solution and Non-ideal solution.
Answer:
Constant boiling mixtures or azeotropic mixture. A solution which distils without change in composition is called azeotropic mixture.
Differences between Ideal and Non-ideal solution

Question 2.
(a) What is meant by depression in freezing point ?
(b) Solution is prepared by dissolving 1 gm NaCl in 100 gm water. If molal de-pression constant for water is 1-85 K kg mol^-1 then determine the extent of dissociation of NaCl. Depression in freezing point for NaCl solution is 0-604 K.
Answer:
(a) Freezing point of a substance is the temperature at which its solid and liquid phases have the same vapour pressure. If non-volatile solute is dissolved in pure liquid to constitute a solution its freezing point decreases, this decrease in freezing point is called depression of freezing point and it is denoted by ∆T_f.
(b) Observed molecular mass of NaCl can be calculated by the following formula:

Thus, observed molecular mass = 30.6 and normal molecular mass of sodium chloride = 58.5.

Question 3.
Write five differences in solution having Positive deviation and Negative deviation.
Answer:
Differences between Positive deviation and Negative deviation :

Question 4.
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 ferro-cyanide 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 5.
(a) What is molal elevation boiling point constant ?
(b) On dissolving phenol in benzene, two of its molecule associate to form a bigger molecules. When Z gm phenol is dissolved in 100 gm benzene, then its freezing point decreases by 0.69°C. Determine the extent of association of phenol. (K_f = 512 K kg mol^-1).
Answer:
(a) Molal boiling elevation constant: It is defined as the elevation in boiling point when 1 gm of non-volatile solute is dissolved in 1000 gm of the solvent.
We know that, ∆T_b α m or ∆T_b = K_bm
Where, K_b is a molal elevation boiling point constant.
Elevation in boiling point is directly proportional to molality of the solution.

Normal molecular mass of phenol = 6 × 12 + 1 × 5 + 16 + 1 = 94

Question 6.
What do you understand by van’t Hoff’s factor ? Write its importance.
Answer:
van’t Hoff’s factor (i) : This factor expresses the extent of association or dissociation of solutes in solution. It is defined as the ratio of the observed value of colligative property to the theoretical value of colligative property, i.e

If association of solute in solution takes place, number of particles decreases. Whereas, in case of dissociation number of particles increases. Since, colligative property depends upon the actual number of particles in solution and is inversely proportional to molecular mass, thus, observed value may be more or less due to association or dissociation.

After introducing the van’t Hoff’s factor (i), the modified equations for colligative properties may be written as :

From the value of ‘i’ degree of dissociation or degree of association of a solute can be calculated.
Value of i : (i) If it is l, then it represents neither association nor dissociation.
(ii) If value of i is less than l, then it expresses association, e.g., solution of benzoic acid in benzene.
(iii) If it is more than 1, then dissociation takes place, e.g., solution of NaCl in water.

Assuming no association or dissociation.

Question 7.
Molecular weight obtained on the basis of colligative property is sometimes different from the actual molecular weight Explain.
Answer:
When value of observed molar mass for a solution is more or less than values of normal molar mass.Then they are known as abnormal molar mass. Abnormal molar mass primarily due to :
(1) Association of solute molecule and (2) Dissociation of solute molecule.
(1) Association of solute molecule: This leads to decrease in the number of molecular particles on dissolving in a solvent. Due to association, there is decrease in the values of colligative properties. Hence, higher values are obtained for the molecular mass of solutes compound to the normal values.
Example : When acetic acid is dissolved in benzene it shows a molecular mass of 120. While the normal molecular mass is 60, these are due to dimer formation as a result of hydrogen bonding.

(2) Dissociation of solute molecules : In electrolytic solution molecules of electro-lytes dissociate to give two or more particles. Since the number of solute particles in solution of such substance is more than the expected value, these solution give higher value of colligative properties. The value of colligative properties are inversely proportional to mo-lecular masses, so the calculated values of molecular mass will be less than normal values.
Example: KCl dissociates into K⁺ and Cl^– ions when dissolved in water so the number of solute particles in solution would be double, the number of particle if no dissociation had been take place. So on the basis of colligative properties the expected value of molecular
mass is half of its normal molecular mass i.e. = \(\frac { 74.5 }{ 2 } \) = 37.25

Question 8.
An aqueous solution freezes at – 0.385°C
if K_f= 3.85 K kg/mol, Kb = 0.712 K kg/mol
Then, determine the elevation in its boiling point.
Solution:

Question 9.
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_b°). The difference T_b – T_b° 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_b° the vapour pressure of the solvent becomes equal to the atmospheric pressure hence it boils at T_b°. 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 Tb the solution boils. Thus, the boiling point is now elevated from T_b° 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_b°.

Question 10.
Prove that the relative lowering in vapour pressure of a solution is equal to mole fraction of solute present in the solution.
Or, 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 deter-mined with its help ?
Answer:
Raoult’s law : For a solution in which solute is non-volatile, the Raoult’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 (P_A).

P°_A – P_A is lowering in vapour pressure and \(\frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}-\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}\) 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 11.
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 deter-mined 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 Ky i.e.,

Thus, depression in freezing point is proportional to molality of solution. Molality is directly proportional to number or molecules of solute substrance. Therefore, depression in freezing point is a colligative property.
Calculation of molecular mass of solute : By determination of depression in freez-ing point, the Molecular mass of non-volatile solute can be determined.
For a solution of non-volatile solute,
∆T_f = K_f × m ….(1)
Let W_B gram non-volatile solute is dissolved in W_A gram solvent and molecular mass of solute is M_B.

From eqn. (3), molecular mass of solute (MB) can be calculated.

Solutions Numerical Questions

Question 1.
1.325 gram sodium carbonate is dissolved in 250 ml solution. Determine the concentration of solution in gram/litre.
Solution:

Mass of Na₂CO₃ = 1.325 gram
Volume of solution = 250 gram
Concentration of sodium carbonate in gram per litre \(\frac { 1.325 }{ 250 } \) × 1000 = 5.3.

Question 2.
4 gm caustic soda (NaOH) is dissolved in 500 ml aqueous solution. Deter-mine the normality of the solution.
Solution:
∵ In 500 ml. solution 4 gm NaOH is dissolved

Question 3.
Determine the osmotic pressure of 5% glucose solution at 25°C. Molecular mass of glucose = 180, R = 0-0821 litre atmosphere.
Solution:
∵ 5 gm glucose is dissolved in 100 ml.
∴ 180 gm glucose will be dissolved in \(\frac { 100 }{ 5 } \) × 180
= 3600 ml = 3.6 litre
We know that,

Question 4.
12.5 gm of urea dissolved in 170 gm of water. The elevation in boiling point was found to be 0.63 K. If Kb for water = 0.52 Km^-1, calculate the molecular mass of urea.
Solution:

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 4 Chemical Kinetics

Chemical Kinetics NCERT Intext Exercises

Question 1.
For the reaction R → P, the concentration of reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:
For the reaction R → P

Question 2.
In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval ?
Answer:

Question 3.
For a reaction, A + B → Product; the rate law is given by, r = k [A]^1/2 [B]². What is the order of the reaction?
Answer:
r = k[A]^1/2 [B]²
k = \(\frac{\mathrm{r}}{[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{2}}\) = \(\frac{\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}}{\left[\mathrm{~mol} \mathrm{~L}^{-1}\right]^{1 / 2}\left[\mathrm{~mol} \mathrm{~L}^{-1}\right]^{2}}\) = 0.05 mol L^-1 min^-1

Question 4.
The conversion of molecules A; toy follows second order kinetics. If concentration of JC is increased to three times how will it affect the rate of formation of y²?
Answer:
For the reaction x → y
Rate of reaction (r) = k[x]² …(1)
If concentration of x is increased three times, now
Rate of reaction (r)¹ = K[3x]² = k[9x²]
Dividing equation (2) by eq. (1)
\(\frac{r^{1}}{r}=\frac{k\left[9 x^{2}\right]}{k\left[x^{2}\right]}=9\)
Thus, rate of reactions will become 9 times.

nuclear equation calculator is balancing calcualter.

Question 5.
A first-order reaction has a rate constant of 1.15 × 10^-3 s^-1. How long will 5g of this reactant take to reduce to 3g?
Solution:
According to the question,

Applying first-order kinetic equation

Question 6.
The time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first-order reaction, calculate the rate constant of the reaction.
Solution:
For a first-order reaction,

Question 7.
What is the effect of the rate constant on temperature? How can this effect of temperature be measured quantitatively?
Answer:
Rate constant increases with an increase in temperature.

Question 8.
The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea.
Solution:
Given: k₂ = 2k₁, T₁ = 298K, T₂ = 308K, R = 8.314 JK^-1 mol^-1
We know that

Question 9.
The activation energy for the reaction
2HI_(g) → H_2(g) + I_2(g)
is 209.5kJ mol^-1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Solution:
According to the question, E_a = 209.5kJ mol^-1 = 209.5 × 103 J mol^-1, T = 581K Fraction of molecules with energy equal to or greater than activation energy is given by [R = 8.314J]

Chemical Kinetics NCERT Textbook Exercises

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Solutions:

Question 2.
For the reaction :
2A + B → A2B
the rate= k [A][B]² with k = 2.0 × 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1
Solution:
Given, Rate = k [A][B]²
Initial rate = 2 × 10-6 × [0.1] × [0.2]²
= 8 × 10^-9 mol L^-1 s^-1.
Decrease in concentration of A
i.e. ∆[A] = 0.1 – 0.06 = 0.04 mol L^-1
Decrease in concentration of B
i.e. ∆[B] = \(\frac { 1 }{ 2 } \) × 0.04 = 002 mol L^-1
Remaining concentration of B
= 0.2 – 0.02 = 0.18M
Rate = k[A] [B]²
= 2 × 10-6 × 0.06 × [0.18]²
= 3.89 × 10^-9 mol L^-1 s^-1.

Question 3.
The decomposition of NH₃ on the platinum surface is a zero-order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 L s^-1?
Solution:
2NH₃ → N₂ + 3H₂

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H_2, and CO, and the reaction rate is given by Rate = k [CH₃OCH₃]^3/2. The rate of reaction is followed by an increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k(P_CH3OCH3)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Solution:
The rate law of reaction is
Rate = k [P_CH3OCH3]^3/2
Rate = Pressure change/Time change
Unit of rate = bar min^-1

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Answer:

1. Concentration of reactants
2. Temperature
3. Nature of reactants and products
4. Exposure to light (Radiation)
5. Presence of catalysts

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is :
(i) doubled
(ii) reduced to half?
Solution:
Let the reaction A → B is a second-order reaction, for A

(i) When the concentration of [A] is doubled
A = 2 a
The new rate of reaction

The rate of reaction will become four times when concentration is doubled.

(ii) When the concentration of [A] is 1/2.
A = \(\frac { a }{ 2 } \)
So new rate of reaction

Therefore, the rate of the reaction could be reduced to l/4^th.

Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Answer:
The rate constant of the reaction increases with temperature, it is independent of whether the reaction is exothermic or endothermic. The Arrhenius equation shows the dependence of rate constants on temperature.

Question 8.
In pseudo-first-order hydrolysis of ester in water, the following results were obtained:

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo-first-order rate constant for the hydrolysis of the ester.
Solution:

Where [R]0 = 0.55 M is initial concentration of ester (t = 0) and [R] is the concentration of ester at time ‘t’

Question 9.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected by increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Solution:

Let [A] = x and [B] = y
Rate (r₁) = kxy² ….(i)

(ii) When [B] = 3y
r₂ = k x (3y)² ….(ii)
On dividing equation (ii) by (i),
r₂ = 9r₁
i. e. rate increases 9 times.

(iii) When both [A] and [B] are doubled.
r₃ = k (2x) (2y)² ….(iii)
Dividing equation (iii) by (i),
r₃ = 8r₁
i.e. rate increases 8 times.

Question 10.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B ?
Solution:
Assuming that the order of reaction w.r.t. A is x and w.r.t. B is y.
Rate = k [A]^x [B]^y
Rate₁ = 1(0.20)^x (0.30)^y = 5.07 × 10^-5 ….(i)
Rate₂ = k(0.20)^x (0.10)^y = 5.07 × 10^-5 …..(ii)

Order of reaction w.r.t A = 1.5
Order of reaction w.r.t B = 0.

Question 11.
The following results have been obtained during the kinetic studies of the reaction :
2A + B → C + D

Determine the rate law and the rate constant for the reaction.
Solution:
Comparing experiments I and IV and substituting the values.
As the concentration of ‘B’ remains constant therefore we get order w.r.t. ‘A’.
(Rate)₁ = k (0.1)^x (0.1)^y = 6.0 × 10^-3 ….(i)
(Rate)₂ = k (0.4)^x (0.1)^y = 2.40 × 10^-2 ….(ii)
Dividing equation (ii) by (i),

Similarly, by comparing experiments II and III, we get the order w.r.t. ‘B’.
(Rate)₂ = k (0.3)^x (0.2)^y = 7.2 × 10^-2 ….(iii)
(Rate)₃ = k (0.3)^x (0.4)^y = 2.88 × 10^-1 …(iv)
Dividing equation (iv) by (iii),

Question 12.
The reaction between A and B is first order with respect to A and zero-order with respect to B. Fill in the blanks in the following table:

Solution:
rate law = K [A]
from I 2.0 × 10^-2 = K 0.1
⇒ K = 2.0 × 10^-1min^-1
from II 4 .0 × 10^-2 = 2 .0 × 10^-31 x
⇒ K = 2 .0 × 10^-1 mol L^-1
from III y = 2.0 × 10^-1 × 0.4
⇒ y = 8 × 10^-2 mol 1L
from IV 2.0 × 10^-2 = 2.0 × 10^-1 × z
⇒ z = 0.1 mol L^-1.

Question 13.
Calculate the half-life of a first-order reaction from their rate constants given below:
(i) 200 s^-1
(ii) 2 min^-1
(iii) 4 years^-1.
Solution:

Question 14.
The half-life for the radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Solution:
Radioactive decay follows first-order kinetics,

Question 15.
The experimental data for the decomposition of N₂O₅
[2N₂O_s → 4NO₂ + O₂]
In gas phase at 318K are given below:

(i) Plot [N₂O₅] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N₂O₅] and t.
(iv) What is the rate law?
(v) Calculate the half-life period from k and compare it with (ii).
Solution:
(i) The plot of [N₂O₅] versus time is shown below :

(ii) Initial concentration of N₂O₅ = 1.63 × 10^-2 M
Half of initial concentration = 1.63 × 10^-2 \(\frac { 1 }{ 2 } \) × = 0.815 × 10^-2 M
Time corresponding to half of initial conc. (t_1/2) from the plot = 1440 sec.

(iii) The plot of log [N₂O₅] Versus time:

(iv) Since the graph between log [N₂O₅] vs time is a straight line, the reaction is of 1st order.
Rate = k [N₂O₅]

Question 16.
The rate constant for a first-order reaction is 60s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?
Solution:

Question 17.
During a nuclear explosion, one of the products is ⁹⁰Sr with a half-life of 28.1 years. If µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Solution:

Question 18.
For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.
Solution:

Question 19.
A first-order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Solution:
30% decomposition means that x = 30% of a = 0.30
As reaction is of first order

Question 20.
For the decomposition of azo isopropanol to hexane and nitrogen at 543 K, the following data are obtained :

Calculate the rate constant.
Solution:

Question 21.
The following data were obtained during the first-order thermal decomposition of SO₂Cl₂ at a constant volume.
SO₂Cl₂(g) → SO₂(g) + cl₂(g)

Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution:

For the first-order reaction in terms of pressure

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30°C and 50°C.
Solution:
To draw the plot of log k versus 1/T, we can rewrite the given data as follows:

Draw the graph as shown in the ahead figure :


Compare it with y = mx + c
log A = Value of intercept on y axis i.e„ on logk axis
[y₂ – y₁ = -1 – (-7.2)] = (-1 + 7.2) = 6.2
log A = 6.2
A = Antilog 6.2
= 1.585 × 10⁶ s^-1
The value of rate constant k can be found from graph as follows :

We can also calculate the value of k from the following formula :

Question 23.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-s s^-1 at 546 K. If the energy of activation is 179-9 KJ/mol, what will be the value of the pre-exponential factor.
Solution:
According to the question, Rate constant (k) = 2.418 × 10^-5 s^-1, Temperature (T) = 546K, Activation energy (E_a) = 179.9 kJ mol^-1

Question 24.
Consider a certain reaction A → Products, with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.
Solution:
Unit of k is s^-1.

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first-order rate law, with t_1/2 = 3.00 hours. What fraction of the sample of sucrose remains after 8 hours?
Solution:

Let initial concentration of sucrose (a) = 1M
Concentration after 8hrs = (a – x) = (1 – x).
Where ‘x’ is the amount of sucrose decomposed.

Question 26.
The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s^-1) e^– 28000 K/T Calculate E_a.
Solution:
k = (4.5 × 10¹¹ s^-1) e^-28000K/T
Comparing the equation with Arrhenius equation
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)

Question 27.
The rate constant for the first-order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 × 10⁴ K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:
(i) We know that
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Taking logarithm on both sides of equation (i), we get

Question 28.
The decomposition of A into the product has a value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kj mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1?
Solution:
Given k₁ = 4.5 × 10³ s^-1, k₂ = 1.5 × 10⁴ s^-1, T₁ =10°C = 10 + 273 = 283K, E_a = 60kJ mol^-1 = 60000J mol^-1
Applying the Arrhenius equation and substituting the values, we get

Question 29.
The time required for 10% completion of a first-order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s^-1. Calculate k at 318 K and Ea.
Solution:
For 10% completion of the reaction

Question 30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:

Chemical Kinetics Other Important Questions and Answers

Chemical Kinetics Objective Type Questions

Choose the correct answer.

Question 1.
For most of the reactions, the value of the temperature coefficient lies in between:
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4.

Question 2.
For first-order reaction value of t_1/2 is :

Question 3.
A first order reaction gets completed to 75% in 32 minutes. How much time would have been required for 50% completion :
(a) 24 minute
(b) 16 minute
(c) 8 minute
(d) 4 minute.

Question 4.
For reaction H_2(g) + Br_2(g) → 2HBr_(g) the experimental values indicate that:
Rate of reaction = k [H₂] [Br₂]^1/2 Molecularity and order of reaction is :
(a) 2, \(\frac { 3 }{ 2 } \)
(b) \(\frac { 3 }{ 2 } \),\(\frac { 3 }{ 2 } \)
(c) 1,1
(d) 1,\(\frac { 1 }{ 2 } \)

Question 5.
The reaction 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄ is an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) None of these.

Question 6.
In the reaction 2A + B → A₂B, the rate of consumption of reactant A is:
(a) Half of the consumption rate of B
(b) Equal to the consumption rate of B
(c) Twice to the consumption rate of B
(d) Equal to the rate of formation of A₂B.

Question 7.
Hydrolysis of sucrose to glucose and fructose is :
C₁₂H₂₂O₁₁ + H₂O > C₆H₁₂O₆ + C₆H₁₂O₆
sucrose  glucose  fructose
an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) Zero order reaction.

Question 8.
At a given temperature the rate of reaction becomes slow when :
(a) Energy of activation becomes high
(b) Energy of activation becomes low
(c) Entropy changes
(d) Initial concentration of reactants remains constant.

Question 9.
Plants prepare starch in the process of:
(a) Flash photolysis
(b) Photolysis
(c) Photosynthesis
(d) None of these.

Question 10.
For first order reaction the specific reaction constant depends upon:
(a) Concentration of reactants
(b) Concentration of products
(c) Time
(d) Temperature.

Question 11.
In the reaction between A and B to form C, A represents first order and B rep¬resents second order. Rate equation will be written as :
(a) Rate = k [A]² [B]
(b) Rate = k [A] [B]²
(c) Rate = k [A]^1/2 [B]
(d) Rate = k [A] [B]^1/2.

Question 12.
Molecularity of reaction of Inversion of sugar :
(a) 3
(b) 2
(c) 1
(d) 0.

Question 13.
Minimum energy required for molecules to react is called :
(a) Potential energy
(b) Kinetic energy
(c) Nuclear energy
(d) Activation energy.

Question 14.
2A + B → A₂B, if concentration A is doubled and concentration of B is being half, the rate of reaction will:
(a) Increase 4 times
(b) Decrease 2 times
(c) Increase 2 times
(d) Remains unchanged.

Question 15.
Half life for second order reaction :
(a) Is proportional to initial concentration
(b) Does not depend upon initial concentration
(c) Inversely proportional to initial concentration
(d) Inversely proportional to square root of intial concentration.

Question 16.
The reaction 2H₂O₂ → 2H₂O + O₂, r = k[H₂O₂] is:
(a) Zero order reaction
(b) First order reaction
(c) Second order reaction
(d) Third order reaction.

Question 17.
Thermal decomposition of a compound is first order reaction. If a sample of compound decomposes 50% in 120 minutes, how much time it will take for 90% decomposition:
(a) About 240 minutes
(b) About 480 minutes
(c) About 450 minutes
(d) About 400 minutes.

Question 18.
Arrhenius equation is :

Question 19.
When temperature is raised, rate of reaction increases. This is due to :
(a) Decrease in activation energy
(b) Increase in number of collisions
(c) Decrease in number of active molecules
(d) Decrease in number of collisions.

Question 20.
Unit of rate constant of second order reaction is :
(a) mol^-1 litre^-1 second^-1
(b) mol litre^-1 second^-1
(c) mol litre second
(d) mol^-1 litre second^-1.

Question 21.
Reaction

(a) Bimolecular and second order
(b) Unimolecular and first order
(c) Bimolecular and first order
(d) Bimolecular and zero order.

Question 22.
Unit of rate constant of first order reaction is :
(a) mol L^-1 S^-1
(b) mol^-1 LS^-1
(c) S^-1
(d) mol L^-1 S.

Question 23.
Time required for the completion of 90% reaction of first order is :
(a) 1-1 times that of half life
(b) 2-2 times that of half life
(c) 3-3 times that of half life
(d) 4-4 times that of half life.

Question 24.
The rate of chemical reaction depends upon :
(a) Active mass
(b) Atomic mass
(c) Equivalent weight
(d) Molecular mass.

Question 25.
Unit of reaction rate is :
(a) mol litre^-1 second^-1
(b) mol^-1 litre second^-1
(c) mol^-1 litre^-1 second
(d) mol litre second.

Answer:
1. (b), 2. (a), 3. (b), 4. (a), 5. (d), 6(b), 7. (a), 8. (a), 9. (c), 10. (d), 11. (b), 12. (c), 13. (d), 14. (d), 15. (c), 16. (b), 17. (d), 18. (d), 19. (b), 20. (d), 21. (c), 22. (c), 23. (c), 24. (a), 25. (a).

Question 2.
Fill in the blanks :

1. The rate of a reaction does not depends on the concentration of the reacting species, then the reaction is of …………………
2. Half life period of a radioactive element is 140 days. On taking 1 gm element initially the amount left after 560 days will be …………………
3. Fast reactions are completed in less than ………………… seconds.
4. In the mechanism of a reaction, the slowest step is called …………………
5. Study of fast reactions is done by …………………
6. Unit of rate constant for third order reaction is …………………
7. Difference in the minimum and maximum energy state of reactants is called …………………
8. Reactions which take place by the absorption of radiations are called …………………
9. The total number of molecules which participate in a reaction is called …………………
10. ………………… gives the idea about the mechanism of the reaction.
11. Hydrolysis of Ethyl acetate in acidic medium is an example of ………………… order reaction.
12. Molecularity is always a …………………
13. For the excitation of one mole reactant ………………… photons are required.
14. Rate of change in concentration of reactant at a specific instant of time is known as ………………… rate.
15. Rate of reaction is ………………… to the concentration of reactant.
16. cm-1 is the unit of …………………
Answer:
1. Zero
2. \(\frac { 1 }{ 16 } \) gm
3. 10^-9
4. Rate determining step
5. Flash photolysis
6. mol^-2 Litre² second^-1
7. Activation energy
8. Photochemical reactions
9. Molecularity
10. Order of reaction
11. Pseudo unimolecular
12. whole number
13. one mole
14. Instantaneous
15. Proportional
16. cell constant.

Question 3.
Match the following :


Answers:

1. (c)
2. (e)
3. (a)
4. (b)
5. (d)
6. (g)
7. (f).

Question 4.
Answer in one word / sentence :

1. What is the relation between threshold energy and activation energy ?
2. What is the expression of rate constant for first order reaction ?
3. Write alternative form of Arrhenius equation.
4. If reaction A + B → C is a zero order reaction. Write rate law expression for it.
5. If rate equation for a reaction is Rate = k [NO₂]² [Cl₂] then state the order of reaction with respect to Cl₂ and order with respect to NO₂ and total order of reaction.
6. Write rate equations on the basis of concentration of reactants and products for the following reactions.
2NO₂ → 2NO + O₂
7. What is rate determining step ?
8. Write an example of zero order reaction.
9. What is Quantum Efficiency ?
10. Write Arrhenius Equation.
11. Write the expression of half life period for first order reaction.
12. What is the effect of surface area of reactant on the rate of reaction ?
13. What is instantaneous rate ?
14. What is half-life period of a reaction ?
15. What is the unit of k for zero order reaction ?
16. Explain Threshold energy.
17. What are fast reactions ?
18. For a zero order reaction tm is proportional to what ?
Answer:
1. Activation energy = Threshold energy – Energy of molecules in normal state
2.

3.

4.

5. Order with respect to cl₂ = l, Order with respect to NO₂ = 2, Thus, Total order = 1 + 2 = 3
6.

7. The slowest step is the rate determining step
8.

9. Quantum Efficiency

10.

11.

12. Larger surface area increases the rate of reaction
13. Rate of change in concentration of reactant or product at a specific instant of time is known as instantaneous rate
14. Half-life period of a reaction is the time in which concentration of reactant is reduced to one half of the initial concentration
15. Litre mol^-1 sec^-1
16. The minimum amount of energy which the reactant molecules should possess for effective collision
17. Reactions which gets completed in 10-9 second or even lesser time interval are called fast reactions
18. Initial concentration of reactant [A]

Chemical Kinetics Very Short Answer Type Questions

Question 1.
What is first order reaction ?
Answer:
If the rate of reaction depends on the first power of concentration of reactant then the order of reaction is said to be first order reaction.

Question 2.
What is Energy Barrier ?
Answer:
The minimum energy achieved by the reactant only after which it can be converted to product is known as energy barrier. Reactant molecules cannot form activated complex till they reach this height (activation energy) and cannot be converted to form product.

Question 3.
Explain the rate determining step.
Answer:
Some chemical reactions complete in one or more steps. Rate of reaction is determined by the slowest step which is known as rate determining step.

Question 4.
Write the characteristics of photochemical reactions. Any three.
Answer:
Characteristics :

1. Absorption of magnetic radiations is necessary for these reactions.
2. These reactions are unaffected by temperature but intensity of radiations affects them.
3. Light is necessary for activating the reactants.

Question 5.
What are the applications of Arrhenius equation ? (Any two)
Answer:

1. In calculating activation energy.
2. In determining the rate constant at a temperature by the help of rate constant at another temperature of the reaction.

Question 6.
What is photochemical reaction ? Give an example.
Answer:
Such reactions which are induced by light or other electromagnetic radiations are known as photochemical reactions. Wavelength of these radiations is from 2000Å to 8000Å.

Question 7.
What is specific reaction rate ?
Answer:
Specific reaction rate of a reaction at a given temperature is equal to that rate of a reaction when concentration of each reactant is unity.

Rate of reaction = k where k = Specific reaction rate constant.

Question 8.
When is the average rate of the reaction equal to its instantaneous rate ?
Answer:
When value of time interval is nearly zero or when time by infinite form is minute then the average rate of reaction is comparable to its instantaneous rate.

Question 9.
What are pseudo unimolecular reactions ?
Answer:
There are some reactions in which molecularity is more than one i.e. two or more molecules are present, but in the chemical reaction concentration of only one reactant molecule is changed and it is only responsible for rate of reaction. Thus, order is one. These reactions are called pseudo unimolecular reactions.

Above reaction is bimolecular but concentration of water does not affect the rate of reaction, thus, rate of reaction is proportional to the concentration of sucrose only.
Rate = k[C₁₂H₂₂O₁₁]
Thus, inversion of sucrose is a first order reaction. It is known as pseudo unimoleciilar reaction.

Question 10.
What is temperature coefficient ?
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10 C, velocity of reaction is increased up to 2-3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10 C, the ratio of these two constant is called temperature coefficient, i.e.,

Thus, at various time the ratio of rate of reaction which differ by 10°C is known as temperature coefficient.

Chemical Kinetics Short Answer Type Questions

Question 1.
Write down the expression representing the rate of reaction.
Answer:
Rate of reaction is the rate of change in concentration of reactant or concentration of product in unit time interval.

Unit of rate of reaction depend on the units of concentration and time. If concentration is represented in mole per litre and time in second, then unit of rate of reaction is mol per litre per second. If time is expressed in minute then unit of reaction rate is mole per litre per minute.

Question 2.
What is the meaning of instantaneous rate of reaction ?
Answer:
The rate of reaction does not remain constant during the whole time interval because rate of reaction depends upon the concentration of reactants. As the concentration of reactants decreases with time, the rate of reaction also decreases with time.

In order to express the reaction rate as accurately as possible, the instantaneous rate of reaction is expressed. For this the time interval (∆t) is taken as small as possible.

Where, d [A] is a change in concentration of reactant A.

Question 3.
What do you know about molecularity of any reaction ?
Answer:
Molecularity of Reaction : ‘Number of moles of reactant participating in elementary step of chemical reaction is called molecularity of the reaction.’

There are many reactions which proceed through the number of steps and each step being independent is called elementary step. The rate of reaction is determined by slowest step and it is known as rate determining step. Here molecularity can be defined as “Number of molecules/atoms or ions participating in rate determining step is called molecularity.”

Example : (i) Unimolecular reaction :
O₃ → O₂ + O
(ii) Bimolecular reaction:
NO + O₃ → NO₂ + O₂

Question 4.
Differentiate molecularity and order of reaction.
Answer:
Differences between Molecularity and Order of reaction :

Molecularity:

1. It is the total number of molecules which participate in the reaction.
2. It is a theoretical concept.
3. It is always a whole number.
4. Its value is never zero.
5. It does not provide any information about mechanism of reaction.

Order of reaction

1. It is the number of molecules which participate in reaction and whose concentration is changed.
2. Order of reaction is determined by experimentally.
3. Fractional values are also possible.
4. Zero value is possible.
5. It provides information about mechanisms of reaction.

Question 5.
Write any four factors which affects rate of a chemical reaction.
Answer:
The rate of reaction depends upon the following factors :
(i) Concentration of reactants: At constant temperature, the rate of a reaction increases by increasing the concentration of the reactants.

(ii) Temperature of the system : If the concentration of the reactants are constant then the rate of reaction increases by increasing temperature. For 10 degree rise of temperature, the reaction rate becomes double or triple.

(iii) Presence of catalyst: Positive catalyst increases the rate of reaction and negative catalyst decreases the reaction rate.

(iv) Nature of the reactants: Nature of reactants also affect the reaction rate. In any chemical reaction some old bonds are broken and new bonds are formed. Thus, in case of more simple molecules the lesser is the number of bond breaking and the rate of reaction increase whereas in case of complex molecules more bonds are broken and rate decreases.

(v) Exposure to radiations: The rate of some reactions increases due to some special radiations.

Question 6.
Write characteristics of rate constant
Answer:

1. At fixed temperature, the value of k is constant.
2. For a particular reaction, k is independent of concentration but depends on temperature.
3. The value of k is different for different reactions.
4. It is a measure of intrinsic rate of reaction i. e., larger the value of k, faster will be the reaction and viceversa.

Question 7.
What do you understand by order of reaction ? Give example.
Answer:
Order of reaction : The order of a reaction is defined as the sum of all the powers to which concentration terms in the rate law are raised to express the observed rate of the reaction. Suppose there is a general reaction,
aA + bB + cC → product
For which the rate law is
Rate = – \(\frac { dx }{ dt } \) = k[A]^p [B]^q [C]^r
Then the order of the reaction n = p + q + r, where, p, q and r are the orders with respect to individual reactants and overall order is the sum of the exponents i.e.,p + q + r.
When n = 1 the reaction is of first order, if n = 2 the reaction is of second order and so on.
For example : Decomposition of ammonium nitrite occurs as follows :
NH₄NO₂ → N₂ + 2H₂O
Rate of reaction = – \(\frac { dx }{ dt } \) = k[NH₄NO₂]
Thus, order of this reaction will be 1.

Question 8.
What do you mean by zero order reaction ? Give one example,
Answer:
In some reactions, rate does not depends, upon the concentration of reactants. This type of reactions are called zero order reaction.
Example : In contact of Au or Pt, the ammonia molecule dissociate and rate of this reaction is independent of concentration of ammonia.

Question 9.
How does rate of any reaction depend on temperature ? Explain.
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2-3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac{k_{t+10}}{k_{t}} \approx 2 \text { to } 3\)
Arrhenius provide the following relation for showing the effect of temperature on velocity constant.
i.e., \(k=\mathbf{A} \cdot e^{-\mathrm{E}_{\alpha} / \mathrm{RT}}\)

Question 10.
What is activation energy ?
Answer:
According to Arrhenius, any chemical reaction is only possible when reacting . molecules are activated with minimum energy which is called threshold energy. Kinetic energy of most of the molecules are less than this minimum energy. The excess energy which is required to activate reactant molecules, is called activation energy.

Activation energy can be determined by the use of Arrhenius equation.

Question 11.
Write four differences between Rate of reaction and Rate constant.
Answer:
Differences between Rate of reaction and Rate constant:

Rate of reaction:

1. It is expressed in terms of consumption of reactants or formation of product per unit time.
2. It depends on concentration of reactant at particular moment.
3. It generally decreases with the progress of reaction.
4. Its unit is mol L^-1 cm^-1.

Rate constant:

1. It is proportionality constant in differential form in rate law or rate equation.
2. It is independent of concentration of reactant.
3. It does not depend on the progress of reaction.
4. It changes according to order of reaction.

Question 12.
Prove that half-life period of zero order reaction is proportional to initial concentration of reactant.
Answer:
If rate of reaction does not depend on the concentration of reactants, then it is known as zero order reaction.

For a zero order reaction the relation between rate constant and concentration is expressed by the following equation :

Thus, half-life period of zero order reaction is proportional to initial concentration of reactant.

Question 13.
Prove that half-life period is independent of the initial concentration for the first order reaction.
Answer:
Half-life period for the reaction is that period in which the initial concentration of reactant is reduced to half.
For first order reaction :

Thus, the half-life period is independent of the initial concentration for the first order reaction.

Question 14.
What is Integrated rate law method ? Write a note.
Answer:
Integrated rate equation for first order reaction is :

By this equation, order of reaction can be calculated. By knowing the initial concen¬tration (a) of reactant, concentration at a definite time (a – x) can be known.

This way, by substituting the values of t and (a – x), value of k is calculated. If value of k comes to be constant, the reaction is of first order. For, first order reaction, on plotting a graph between concentration log)0 (a – x) and t, a straight line is obtained. For other orders of reaction suitable equations are used and orders are determined.

Question 15.
A first order reaction is 90% complete in 40 minutes. Calculate its half- life period. (log 2 = 0.3010).
Solution:
Let initial cone, of reactant (a) = 100, t = 40 minutes
90% reaction gets completed in 40 minutes i.e.,

Question 16.
Show that time required for completing 99.9% of a first order reaction is 10 times of its half-life period.
Solution:
If initial concentration of reactant is a, then t = ? for x = 0.999 a
We know that for a first order reaction

Question 17.
Write unit of rate constant k for the zero order, first order and second order reaction
Solution:

Question 18.
(1) 2N₂O_s 4NO₂ + O₂
(2) H²₂ + I₂ → 2HI
Reaction (1) is first order reaction and (2) is second order reaction. Why ?
Answer:
(1) In reaction 2N₂O₅ → 4NO₂ + O₂, when graph is plotted between rate of reaction and then again graph is plotted between rate of reaction and [N₂O₅]², it is shown that in the first graph straight line is obtained i.e.
rate ∝ [N₂O⁵]
or rate = k[N₂O⁵]
Therefore, 2N₂O₅ → 4NO₂ + O₂ is first order reaction.

(2) In reaction H₂ + I₂ → 2HI when graph is plotted between rate of reaction and (H₂) (I₂), it is seen that straight line is obtained.
Therefore rate ∝ [H₂][I₂]
Hence, this reaction is of 2^nd order reaction.

Chemical Kinetics Long Answer Type Questions

Question 1.
Determine the expression for zero order reaction.
Answer:
Reactions in which rate of reaction does not depend on the concentration of the reactants are called zero order reactions. Consider a zero order reaction
A → B
Where, A and B are concentration of reactants and products. Since in this type of reaction, rate of reaction does not depend on the concentration of reactant therefore, rate of change in concentration of reactant remains constant.
Rate of reaction = Constant.
Let initial cone, of reactants be a moles /It and after time ‘t’ x moles are converted into product. Then cone, of A after time t will be (a – x) mol /It.

Thus, eqn. (7) is the velocity equation for zero order reaction.

Question 2.
Write a note on Arrhenius equation.
Answer:
Arrhenius equation : Arrhenius gave a relation between rate constant of reaction and temperature which is known as Arrhenius equation i.e.,
\(k=\mathbf{A} e^{-\Delta \mathbf{E}_{a} / \mathbf{R} \mathbf{T}}\)
Where, k = Rate constant of reaction, A = Frequency factor, Ea = Activation energy, R = Gas constant, T = Absolute temperature.

Taking logarithm both sides of above equation we get

E_a and A can be determined by measuring rate constants of the reaction at two different temperature. Let k₁ and k₂ are the rate constant for the reaction at two temperatures T₁ and T₂ respectively. Then from eqn. (1)

Application : Activation energy AE_a may be calculated easily from this equation.

Question 3.
Write the Arrhenius equation in the form of equation of straight line. What will be the slope of a graph using this equation ? Calculate the activation energy for the decomposition in a decomposition reaction in which the value of slope obtained is – 9920 when log k is plotted against \(\frac { 1 }{ T } \).
Answer:
Arrhenius equation : Arrhenius gave a relation between rate constant of reac¬tion and temperature which is known as Arrhenius equation, i.e.,
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Where, k = Rate constant of reaction, A = Frequency factor, Ea = Activation energy, R = Gas constant and T = Absolute temperature.
Taking logarithm both sides of above equation, we get

This is a straight line equation. If log₁₀ A and \(\frac { 1 }{ T } \) is plotted at different temperatures, a straight line is obtained whose slope = \(\frac{-\mathrm{E}_{a}}{2 \cdot 303 \mathrm{R}}\) From the graph, value of slope = \(\frac{\mathrm{E}_{a}}{2 \cdot 303 \mathrm{R}}\) can be calculated and activation energy (E_a) can be calculated.

Alternatively, E_a and A can be determined by measuring rate constant of the reaction at two different temperatures. Let k₁ and k₂ are the rate constants for the reaction at two temperatures T₁ and T₂ respectively. Then, from eqn. (1),

Calculation of activation energy :
E_a = -2.303 × slope × R = -2.303 × (-9920) × 1.986
∴ E_a = 45334 cal/gm/mol.

Question 4.
Derive an expression for velocity constant of first order reaction.
Answer:
The reaction in which velocity of reaction depends upon the concentration of one mole, are called first order reaction.
Let this reaction is
A → Product
Suppose intial concentration of A is a gram mole and after t second x mole consumed and remaining concentration is (a – x) gram mole.
So after t time, the rate of reaction will be proportional to (a – x)

Putting the value of c from eqn. (4), into eqn. (3).
– ln(a – x) = kt – In a
or In a – ln(a – x) = kt

It is the desired expression for first order of reaction.

MP Board Class 12th Chemistry Solutions

In this article, we share MP Board Class 10th Maths Book Solutions Chapter 1 वास्तविक संख्याएँ Ex 1.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 10th Maths Solutions Chapter 1 वास्तविक संख्याएँ Ex 1.2

प्रश्न 1.
निम्नलिखित संख्याओं को अभाज्य गुणनखण्डों के गुणनफल के रूप में व्यक्त कीजिए :

1. 140
2. 156
3. 3825
4. 5005
5. 7429

हल :

1. 140 = 2 × 2 × 5 × 7 = 2² × 5¹ × 7¹ उत्तर
2. 156 = 2 × 2 × 3 × 13 = 2² × 3¹ × 13¹ उत्तर
3. 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17¹ उत्तर
4. 5005 = 5 × 7 × 11 × 13 = 5¹ × 7¹ × 11¹ × 13¹ उत्तर
5. 7429 = 17 × 19 × 23 = 17¹ × 19¹ × 23¹ उत्तर

प्रश्न 2.
पूर्णांकों के निम्नलिखित युग्मों के HCF और LCM ज्ञात कीजिए तथा इसकी जाँच कीजिए कि दो संख्याओं का गुणनफल = HCF × LCM है :
(i) 26 और 91
(ii)510 और 92
(iii) 336 और 54
हल:
(i) 26 = 2 × 13
91 = 7 × 13
HCF = 13
उत्तर LCM = 2 × 7 × 13 = 182
उत्तर अब HCF (26,91) × LCM (26,91)= 13 × 182 = 2366
एवं 26 × 91 = 2366
अत: HCF (26, 91) × LCM (26,91) = 26 × 91 सत्यापित

(ii) 510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2 उत्तर
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460 उत्तर
अब HCF (510, 92) × LCM (510, 92)= 2 × 23460 = 46920
एवं 510 × 92 = 46920
अत:, HCF (510, 92) × LCM (510, 92) = 510 × 92 सत्यापित

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6 उत्तर
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024 उत्तर
अब HCF (336,54) × LCM (336,54) = 6 × 3024 = 18144
एवं 336 × 54 = 18144
अतः HCF (336,54) × LCM (336,54) = 336 × 54 सत्यापित

365 / 838 = 0.436.

प्रश्न 3.
अभाज्य गुणनखण्ड विधि द्वारा निम्नलिखित पूर्णांकों के HCF और LCM ज्ञात कीजिए :
(i) 12, 15 और 21
(ii) 17, 23 और 29
(iii) 8,9 और 25
हल :
(i) 12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420
अतः, अभीष्ट HCF = 3 एवं LCM = 420

(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
अतः अभीष्ट HCF = 1 एवं LCM = 11339 उत्तर

(iii) 8 = 1 × 2 × 2 × 2
9 = 1 × 3 × 3
25 = 1 × 5 × 5
HCF = 1
LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
अतः, अभीष्ट HCF = 1 एवं LCM = 1800 उत्तर

प्रश्न 4.
HCF (306, 657) = 9 दिया है। LCM (306, 657) ज्ञात कीजिए।
हल :
LCM (306, 657) × HCF (306, 657) = 306 × 657
⇒ LCM (306, 657) = \(\frac{306 \times 657}{9}\)
[∵ HCF (306, 657) = 9 दिया है।
⇒ LCM (306, 657)= \(\frac { 201042 }{ 9 } \) = 22338
अतः, अभीष्ट LCM (306, 657) = 22338 उत्तर

प्रश्न 5.
जाँच कीजिए कि क्या किसी प्राकृत संख्या n के लिए संख्या 6n अंक 0 पर समाप्त हो सकती है?
हल :
हम जानते हैं कि 6ⁿ = 2ⁿ × 3ⁿ का गुणनखण्ड 5 नहीं है, अतः किसी भी प्राकृत संख्या n के लिए 6n संख्या अंक 0 पर समाप्त नहीं होगी क्योंकि 0 पर समाप्त होने वाली संख्याएँ 5 से विभाज्य होती हैं और यह संख्या 5 से विभाज्य नहीं है।
अतः, ऐसी कोई संख्या n नहीं है जिसके लिए 6ⁿ अंक 0 पर समाप्त होगी। उत्तर

प्रश्न 6.
व्याख्या कीजिए कि 7 × 11 × 13 + 13 और 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 भाज्य संख्याएँ क्यों हैं।
हल :
7 × 11 × 13 + 13 = 13 (7 × 11 + 1) = 13 × 78
जो एक भाज्य संख्या है।
एवं 7 × 6 × 5 × 4 × 3× 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1) = 5 × 1009
जो एक भाज्य संख्या है।
अतः, दी हुई दोनों संख्याएँ भाज्य संख्याएँ हैं। उत्तर

प्रश्न 7.
किसी खेल के मैदान के चारों ओर एक वृत्ताकार पथ है। इस मैदान का एक चक्कर लगाने में सोनिया को 18 मिनट लगते हैं, जबकि इसी मैदान का एक चक्कर लगाने में रवि को 12 मिनट लगते हैं। मान लीजिए वे दोनों एक ही स्थान और एक ही समय पर चलना प्रारम्भ करके एक ही दिशा में चलते हैं। कितने समय बाद वे पुनः प्रारम्भिक स्थान पर मिलेंगे?
हल :
18 = 2 × 3 × 3 = 2¹ × 3²
12 = 2 × 2 × 3 = 2² × 3¹
LCM (18, 12) = 2² × 3² = 2 × 2 × 3 × 3 = 36
अतः, वे पुन: 36 मिनट बाद प्रारम्भिक स्थान पर मिलेंगे।