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ICSE Class 10 Hindi Solutions साहित्य सागर – सूर के पद [कविता]

February 5, 2024

ICSE Class 10 Hindi Solutions साहित्य सागर – सूर के पद [कविता]

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

प्रश्न क-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जसोदा हरि पालने झुलावै।
हलरावै, दुलराइ मल्हावै, जोइ-सोइ कछु गावै॥
मेरे लाल कौं आउ निंदरिया, काहे न आनि सुवावै।
तू काहैं नहिं बेगहिं आवै, तोकौं कान्ह बुलावै॥
कबहुँ पलक हरि मूँदि लेत हैं, कबहुँ अधर फरकावै।
सोवत जानि मौन ह्वै कै रहि, करि-करि सैन बतावै॥
इहिं अंतर अकुलाइ उठे हरि, जसुमति मधुरै गावै।
जो सुख सूर अमर-मुनि दुरलभ, सो नँद-भामिनि पावै॥
कौन किसको सुलाने का प्रयास कर रहा है?

उत्तर:
प्रस्तुत पद्यांश में कवि ने माता यशोदा का कृष्ण के प्रति प्यार को प्रदर्शित किया है। यहाँ पर माता यशोदा कृष्ण को सुलाने का प्रयास कर रही है।

प्रश्न क-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जसोदा हरि पालने झुलावै।
हलरावै, दुलराइ मल्हावै, जोइ-सोइ कछु गावै॥
मेरे लाल कौं आउ निंदरिया, काहे न आनि सुवावै।
तू काहैं नहिं बेगहिं आवै, तोकौं कान्ह बुलावै॥
कबहुँ पलक हरि मूँदि लेत हैं, कबहुँ अधर फरकावै।
सोवत जानि मौन ह्वै कै रहि, करि-करि सैन बतावै॥
इहिं अंतर अकुलाइ उठे हरि, जसुमति मधुरै गावै।
जो सुख सूर अमर-मुनि दुरलभ, सो नँद-भामिनि पावै॥
यशोदा बालक कृष्ण को सुलाने के लिए क्या-क्या यत्न कर रही है?

उत्तर :
यशोदा जी बालक कृष्ण को सुलाने के लिए पलने में झुला रही हैं। कभी प्यार करके पुचकारती हैं और लोरी गाती रहती है।

प्रश्न क-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जसोदा हरि पालने झुलावै।
हलरावै, दुलराइ मल्हावै, जोइ-सोइ कछु गावै॥
मेरे लाल कौं आउ निंदरिया, काहे न आनि सुवावै।
तू काहैं नहिं बेगहिं आवै, तोकौं कान्ह बुलावै॥
कबहुँ पलक हरि मूँदि लेत हैं, कबहुँ अधर फरकावै।
सोवत जानि मौन ह्वै कै रहि, करि-करि सैन बतावै॥
इहिं अंतर अकुलाइ उठे हरि, जसुमति मधुरै गावै।
जो सुख सूर अमर-मुनि दुरलभ, सो नँद-भामिनि पावै॥
कृष्ण को सोता हुआ जानकर यशोदा क्या करती हैं?

उत्तर:
कृष्ण को सोते समझकर यशोदा माता चुप हो जाती हैं और दूसरी गोपियों को भी संकेत करके समझाती हैं कि वे सब भी चुप रहे।

प्रश्न क-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जसोदा हरि पालने झुलावै।
हलरावै, दुलराइ मल्हावै, जोइ-सोइ कछु गावै॥
मेरे लाल कौं आउ निंदरिया, काहे न आनि सुवावै।
तू काहैं नहिं बेगहिं आवै, तोकौं कान्ह बुलावै॥
कबहुँ पलक हरि मूँदि लेत हैं, कबहुँ अधर फरकावै।
सोवत जानि मौन ह्वै कै रहि, करि-करि सैन बतावै॥
इहिं अंतर अकुलाइ उठे हरि, जसुमति मधुरै गावै।
जो सुख सूर अमर-मुनि दुरलभ, सो नँद-भामिनि पावै॥
सूरदास के अनुसार यशोदा कौन-सा सुख पा रही हैं?

उत्तर:
सूरदास जी कहते हैं कि जो सुख देवताओं तथा मुनियों के लिये भी दुर्लभ है, वही श्याम को बालरूप में पाकर लालन-पालन तथा प्यार करने का सुख यशोदा प्राप्त कर रही हैं।

प्रश्न ख-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
खीजत जात माखन खात।
अरुन लोचन, भौंह टेढ़ी, बार-बार जँभात॥
कबहुँ रुनझुन चलत घुटुरुनि, धूरि धूसर गात।
कबहुँ झुक कै अलक खैँचत, नैन जल भरि जात॥
कबहुँ तोतरे बोल बोलत, कबहुँ बोलत तात।
सूर हरि की निरखि सोभा, निमिष तजत न मात॥
इस दोहे में सूरदास जी ने क्या वर्णन किया है?

उत्तर:
इस दोहे में सूरदास जी ने श्रीकृष्ण के अनुपम बाल सौन्दर्य का वर्णन किया है।

प्रश्न ख-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
खीजत जात माखन खात।
अरुन लोचन, भौंह टेढ़ी, बार-बार जँभात॥
कबहुँ रुनझुन चलत घुटुरुनि, धूरि धूसर गात।
कबहुँ झुक कै अलक खैँचत, नैन जल भरि जात॥
कबहुँ तोतरे बोल बोलत, कबहुँ बोलत तात।
सूर हरि की निरखि सोभा, निमिष तजत न मात॥
बाल कृष्ण कैसे चलते हैं?

उत्तर:
बाल कृष्ण घुटनों के बल चलते हैं। उनके पैरों में घुंघरुओं की आवाज़ आती है।

प्रश्न ख-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
खीजत जात माखन खात।
अरुन लोचन, भौंह टेढ़ी, बार-बार जँभात॥
कबहुँ रुनझुन चलत घुटुरुनि, धूरि धूसर गात।
कबहुँ झुक कै अलक खैँचत, नैन जल भरि जात॥
कबहुँ तोतरे बोल बोलत, कबहुँ बोलत तात।
सूर हरि की निरखि सोभा, निमिष तजत न मात॥
बाल कृष्ण के रूप सौंदर्य का वर्णन कीजिए।

उत्तर:
बाल कृष्ण बहुत सुंदर हैं। उनके नेत्र सुंदर हैं, भौंहें टेढ़ी हैं तथा वे बार-बार जम्हाई ले रहे हैं। उनका शरीर धूल में सना है।

प्रश्न ख-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
खीजत जात माखन खात।
अरुन लोचन, भौंह टेढ़ी, बार-बार जँभात॥
कबहुँ रुनझुन चलत घुटुरुनि, धूरि धूसर गात।
कबहुँ झुक कै अलक खैँचत, नैन जल भरि जात॥
कबहुँ तोतरे बोल बोलत, कबहुँ बोलत तात।
सूर हरि की निरखि सोभा, निमिष तजत न मात॥
बाल कृष्ण कैसी जबान में बोलते हैं?

उत्तर:
बाल कृष्ण तोतली जबान में बोलते हैं।

प्रश्न ग-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
मैया, मैं तौ चंद-खिलौना लैहौं।
जैहौं लोटि धरनि पर अबहीं, तेरी गोद न ऐहौं॥
सुरभी कौ पय पान न करिहौं, बेनी सिर न गुहैहौं।
ह्वै हौं पूत नंद बाबा को, तेरौ सुत न कहैहौं॥
आगैं आउ, बात सुनि मेरी, बलदेवहि न जनैहौं।
हँसि समुझावति, कहति जसोमति, नई दुलहिया दैहौं॥
तेरी सौ, मेरी सुनि मैया, अबहिं बियाहन जैहौं॥
सूरदास ह्वै कुटिल बराती, गीत सुमंगल गैहौं॥
उपर्युक्त पद का प्रसंग स्पष्ट कीजिए।

उत्तर:
उपर्युक्त पद महाकवि सूरदास द्वारा रचित है। इस पद में बाल कृष्ण अपनी यशोदा माता से चंद्रमा रूपी खिलौना लेने की हठ कर रहे हैं उसका वर्णन किया गया है।

प्रश्न ग-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
मैया, मैं तौ चंद-खिलौना लैहौं।
जैहौं लोटि धरनि पर अबहीं, तेरी गोद न ऐहौं॥
सुरभी कौ पय पान न करिहौं, बेनी सिर न गुहैहौं।
ह्वै हौं पूत नंद बाबा को, तेरौ सुत न कहैहौं॥
आगैं आउ, बात सुनि मेरी, बलदेवहि न जनैहौं।
हँसि समुझावति, कहति जसोमति, नई दुलहिया दैहौं॥
तेरी सौ, मेरी सुनि मैया, अबहिं बियाहन जैहौं॥
सूरदास ह्वै कुटिल बराती, गीत सुमंगल गैहौं॥
अपनी हठ पूरी न होने पर बाल कृष्ण अपनी माता को क्या-क्या कह रहे हैं?

उत्तर:
अपनी हठ पूरी न होने पर बाल कृष्ण अपनी माता को कहते हैं कि जब तक उन्हें चाँद रूपी खिलौना नहीं मिल जाता तब तक वह न तो भोजन ग्रहण करेंगे, न चोटी गुँथवाएगे, न मोतियों की माला पहनेंगे, न उनकी गोद में आएँगे, न ही नंद बाबा और यशोदा माता के बेटे कहलाएँगे।

प्रश्न ग-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
मैया, मैं तौ चंद-खिलौना लैहौं।
जैहौं लोटि धरनि पर अबहीं, तेरी गोद न ऐहौं॥
सुरभी कौ पय पान न करिहौं, बेनी सिर न गुहैहौं।
ह्वै हौं पूत नंद बाबा को, तेरौ सुत न कहैहौं॥
आगैं आउ, बात सुनि मेरी, बलदेवहि न जनैहौं।
हँसि समुझावति, कहति जसोमति, नई दुलहिया दैहौं॥
तेरी सौ, मेरी सुनि मैया, अबहिं बियाहन जैहौं॥
सूरदास ह्वै कुटिल बराती, गीत सुमंगल गैहौं॥
यशोदा माता श्रीकृष्ण को मनाने के लिए क्या कहती है?

उत्तर:
यशोदा माता श्रीकृष्ण को मनाने के लिए उनके कान में कहती है, तुम ध्यान से सुनो। कहीं बलराम न सुन ले। तुम तो मेरे चंदा हो और में तुम्हारे लिए सुंदर दुल्हन लाऊँगी।

प्रश्न ग-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
मैया, मैं तौ चंद-खिलौना लैहौं।
जैहौं लोटि धरनि पर अबहीं, तेरी गोद न ऐहौं॥
सुरभी कौ पय पान न करिहौं, बेनी सिर न गुहैहौं।
ह्वै हौं पूत नंद बाबा को, तेरौ सुत न कहैहौं॥
आगैं आउ, बात सुनि मेरी, बलदेवहि न जनैहौं।
हँसि समुझावति, कहति जसोमति, नई दुलहिया दैहौं॥
तेरी सौ, मेरी सुनि मैया, अबहिं बियाहन जैहौं॥
सूरदास ह्वै कुटिल बराती, गीत सुमंगल गैहौं॥
माँ यशोदा की बात सुनकर श्रीकृष्ण की क्या प्रतिक्रिया हुई?

उत्तर:
माँ यशोदा की बात सुनकर श्रीकृष्ण कहते हैं माता तुझको मेरी सौगन्ध। तुम मुझे अभी ब्याहने चलो।

ICSE Class 10 Hindi Solutions साहित्य सागर – नेता जी का चश्मा

February 5, 2024

ICSE Class 10 Hindi Solutions साहित्य सागर – नेता जी का चश्मा

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

प्रश्न क-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इसी नगरपालिका के उत्साही बोर्ड या प्रशासनिक अधिकारी ने एक बार ‘शहर’ के मुख्य चौराहे पर नेताजी सुभाषचंद्र बोस की एक संगमरमर की प्रतिमा लगवा दी यह कहानी उसी प्रतिमा के बारे में है,बल्कि उसके भी एक छोटे-से हिस्से के बारे में।
हालदार साहब कब और कहाँ-से क्यों गुजरते थे?

उत्तर:
हालदार साहब हर पंद्रहवें दिन कंपनी के काम के सिलसिले में एक कस्बे से गुजरते थे। जहाँ बाज़ार के मुख्य चौराहे पर नेताजी की मूर्ति लगी थी।

प्रश्न क-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इसी नगरपालिका के उत्साही बोर्ड या प्रशासनिक अधिकारी ने एक बार ‘शहर’ के मुख्य चौराहे पर नेताजी सुभाषचंद्र बोस की एक संगमरमर की प्रतिमा लगवा दी यह कहानी उसी प्रतिमा के बारे में है, बल्कि उसके भी एक छोटे-से हिस्से के बारे में।
कस्बे का वर्णन कीजिए।

उत्तर :
कस्बा बहुत बड़ा नहीं था। जिसे पक्का मकान कहा जा सके वैसे कुछ ही मकान और जिसे बाज़ार कहा जा सके वैसा एक ही बाज़ार था। कस्बे में एक लड़कों का स्कूल, एक लड़कियों का स्कूल, एक सीमेंट का कारखाना, दो ओपन एयर सिनेमाघर और एक नगरपालिका थी।

प्रश्न क-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इसी नगरपालिका के उत्साही बोर्ड या प्रशासनिक अधिकारी ने एक बार ‘शहर’ के मुख्य चौराहे पर नेताजी सुभाषचंद्र बोस की एक संगमरमर की प्रतिमा लगवा दी यह कहानी उसी प्रतिमा के बारे में है, बल्कि उसके भी एक छोटे-से हिस्से के बारे में।
नगरपालिका के कार्यों के बारे में बताइए।

उत्तर:
उस कस्बे नगरपालिका थी तो कुछ-न कुछ करती भी रहती थी। कभी कोई सड़क पक्की करवा दी, कभी कुछ पेशाबघर बनवा दिए, कभी कबूतरों की छतरी बनवा दी तो कभी कवि सम्मलेन करवा दिया।

प्रश्न क-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
इसी नगरपालिका के उत्साही बोर्ड या प्रशासनिक अधिकारी ने एक बार ‘शहर’ के मुख्य चौराहे पर नेताजी सुभाषचंद्र बोस की एक संगमरमर की प्रतिमा लगवा दी यह कहानी उसी प्रतिमा के बारे में है, बल्कि उसके भी एक छोटे-से हिस्से के बारे में।
शहर के मुख्य बाज़ार में प्रतिमा किसने लगवाईं थी और उस प्रतिमा की क्या विशेषता थी?

उत्तर:
शहर के मुख्य बाज़ार के मुख्य चौराहे पर नगरपालिका के किसी उत्साही बोर्ड या प्रशासनिक अधिकारी ने नेताजी सुभाषचंद्र बोस की एक संगमरमर की प्रतिमा लगवा दी थी।
उस मूर्ति की विशेषता यह थी कि मूर्ति संगमरमर की थी। टोपी की नोक से कोट के दूसरे बटन तक कोई दो फुट ऊँची और सुंदर थी। नेताजी फौजी वर्दी में सुंदर लगते थे। मूर्ति को देखते ही ‘दिल्ली चलो’ और तुम मुझे खून दो… आदि याद आने लगते थे। केवल एक चीज की कसर थी जो देखते ही खटकती थी नेताजी की आँख पर संगमरमर चश्मा नहीं था बल्कि उसके स्थान पर सचमुच के चश्मे का चौड़ा काला फ्रेम मूर्ति को पहना दिया गया था।

प्रश्न ख-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वाह भाई! क्या आइडिया है। मूर्ति कपड़े नहीं बदल सकती, लेकिन चश्मा हर बार बदल कैसे जाता है?
प्रस्तुत कथन के वक्ता का परिचय दें।

उत्तर:
प्रस्तुत कथन के वक्ता हालदार साहब हैं। वे अत्यंत भावुक और संवेदनशील होने के साथ एक देशभक्त भी हैं। उन्हें देशभक्तों का मज़ाक उड़ाया जाना पसंद नहीं है। वे कैप्टन की देशभावना के प्रति सम्मान और सहानुभूति रखते हैं।

प्रश्न ख-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वाह भाई! क्या आइडिया है। मूर्ति कपड़े नहीं बदल सकती, लेकिन चश्मा हर बार बदल कैसे जाता है?
प्रस्तुत कथन के श्रोता का परिचय दें।

उत्तर:
प्रस्तुत कथन का श्रोता पानवाला है। पानवाला पूरी की पूरी पान की दुकान है, सड़क के चौराहे के किनारे उसकी पान की दुकान है। वह काला तथा मोटा है, उसकी तोंद भी निकली हुई है, उसके सिर पर गिने-चुने बाल ही बचे हैं। वह एक तरफ़ ग्राहक के लिए पान बना रहा है, वहीं दूसरी ओर उसका मुँह पान से भरा है। पान खाने के कारण उसके होंठ लाल तथा कहीं-कहीं काले पड़ गए हैं। स्वभाव से वह मजाकिया है। वह बातें बनाने में माहिर है।

प्रश्न ख-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वाह भाई! क्या आइडिया है। मूर्ति कपड़े नहीं बदल सकती, लेकिन चश्मा हर बार बदल कैसे जाता है?
कस्बे से गुजरते समय हालदार साहब को क्या आदत पड़ गई थी?

उत्तर:
कस्बे से गुजरते समय हालदार साहब को उस कस्बे के मुख्य बाज़ार के चौराहे पर रुकना, पान खाना और मूर्ति को ध्यान से देखने की आदत पड़ गई थी।

प्रश्न ख-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
वाह भाई! क्या आइडिया है। मूर्ति कपड़े नहीं बदल सकती, लेकिन चश्मा हर बार बदल कैसे जाता है?
मूर्ति का चश्मा हर-बार कौन और क्यों बदल देता था?

उत्तर:
मूर्ति का चश्मा हर-बार कैप्टन बदल देता था। कैप्टन असलियत में एक गरीब चश्मेवाला था। उसकी कोई दुकान नहीं थी। फेरी लगाकर वह अपने चश्मे बेचता था। जब उसका कोई ग्राहक नेताजी की मूर्ति पर लगे फ्रेम की माँग करता तो कैप्टन मूर्ति पर अन्य फ्रेम लगाकर वह फ्रेम अपने ग्राहक को बेच देता। इसी कारणवश मूर्ति पर कोई स्थाई फ्रेम नहीं रहता था।

प्रश्न ग-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
“लेकिन भाई! एक बात समझ नहीं आई।” हालदार साहब ने पानवाले से फिर पूछा, “नेताजी का ओरिजिनल चश्मा कहाँ गया?”
प्रस्तुत कथन में नेताजी का ओरिजिनल चश्मा से क्या तात्पर्य है?

उत्तर:
प्रस्तुत कथन में नेताजी का ओरिजिनल चश्मा से तात्पर्य नेताजी के बार-बार बदलने वाले फ्रेम से है। मूर्तिकार ने नेताजी की मूर्ति बनाते समय चश्मा नहीं बनाया था। नेताजी बिना चश्मे के यह बात एक गरीब देशभक्त चश्मेवाले कैप्टन को पसंद नहीं आती थी इसलिए वह नेताजी की मूर्ति पर उसके पास उपलब्ध फ्रेमों से एक फ्रेम लगा देता था।

प्रश्न ग-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
“लेकिन भाई! एक बात समझ नहीं आई।” हालदार साहब ने पानवाले से फिर पूछा, “नेताजी का ओरिजिनल चश्मा कहाँ गया?”
मूर्तिकार कौन था और उसने मूर्ति का चश्मा क्यों नहीं बनाया था?

उत्तर:
मूर्तिकार उसी कस्बे के स्थानीय विद्यालय का मास्टर मोतीलाल था। मूर्ति बनाने के बाद शायद वह यह तय नहीं कर पाया होगा कि पत्थर से पारदर्शी चश्मा कैसे बनाया जाये या फिर उसने पारदर्शी चश्मा बनाने की कोशिश की होगी मगर उसमें असफल रहा होगा।

प्रश्न ग-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
“लेकिन भाई! एक बात समझ नहीं आई।” हालदार साहब ने पानवाले से फिर पूछा, “नेताजी का ओरिजिनल चश्मा कहाँ गया?”
“वो लँगड़ा क्या जाएगा फ़ौज में। पागल है पागल!”कैप्टन के प्रति पानवाले की इस टिप्पणी पर अपनी प्रतिक्रिया लिखिए।

उत्तर:
पानवाले ने कैप्टन को लँगड़ा तथा पागल कहा है। जो कि अति गैर जिम्मेदाराना और दुर्भाग्यपूर्ण वक्तव्य है। कैप्टन में एक सच्चे देशभक्त के वे सभी गुण मौजूद हैं जो कि पानवाले में या समाज के अन्य किसी वर्ग में नहीं है। वह भले ही लँगड़ा है पर उसमें इतनी शक्ति है कि वह कभी भी नेताजी को बग़ैर चश्मे के नहीं रहने देता है। अत: कैप्टन पानवाले से अधिक सक्रिय तथा विवेकशील तथा देशभक्त है।

प्रश्न ग-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
“लेकिन भाई! एक बात समझ नहीं आई।” हालदार साहब ने पानवाले से फिर पूछा, “नेताजी का ओरिजिनल चश्मा कहाँ गया?”
सेनानी न होते हुए भी चश्मेवाले को लोग कैप्टन क्यों कहते थे?

उत्तर:
चश्मेवाला कभी सेनानी नहीं रहा परन्तु चश्मेवाला एक देशभक्त नागरिक था। उसके हृदय में देश के वीर जवानों के प्रति सम्मान था। वह अपनी ओर से एक चश्मा नेताजी की मूर्ति पर अवश्य लगाता था उसकी इसी भावना को देखकर लोग उसे कैप्टन कहते थे।

प्रश्न घ-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
हालदार साहब भावुक हैं। इतनी सी बात पर उनकी आँखें भर आईं।
हालदार साहब ने अपने ड्राईवर को चौराहे पर रुकने के लिए मना क्यों किया?

उत्तर:
करीब दो सालों तक हालदार साहब उस कस्बे से गुजरते रहे और नेताजी की मूर्ति में बदलते चश्मे को देखते रहे फिर एक बार ऐसा हुआ कि नेताजी के चेहरे पर कोई चश्मा नहीं था। पता लगाने पर हालदार साहब को पता चला कि मूर्ति पर चश्मा लगाने वाला कैप्टन मर गया और अब ऐसा उस कस्बे में कोई नहीं था जो नेताजी की मूर्ति पर चश्मा लगाता इसलिए हालदार साहब ने अपने ड्राईवर को चौराहे पर न रुकने का निर्देश दिया।

प्रश्न घ-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
हालदार साहब भावुक हैं। इतनी सी बात पर उनकी आँखें भर आईं।
हालदार साहब पहले मायूस क्यों हो गए थे?

उत्तर:
कैप्टन की मृत्यु के बाद हालदार साहब को लगा कि क्योंकि कैप्टन के समान अब ऐसा कोई अन्य देश प्रेमी बचा न था जो नेताजी के चश्मे के बारे में सोचता। हालदार साहब स्वयं देशभक्त थे और नेताजी जैसे देशभक्त के लिए उसके मन में सम्मान की भावना थी। यही सब सोचकर हालदार साहब पहले मायूस हो गए थे।

प्रश्न घ-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
हालदार साहब भावुक हैं। इतनी सी बात पर उनकी आँखें भर आईं।
मूर्ति पर सरकंडे का चश्मा क्या उम्मीद जगाता है?

उत्तर:
मूर्ति पर लगे सरकंडे का चश्मा इस बात का प्रतीक है कि आज भी देश की आने वाली पीढ़ी के मन में देशभक्तों के लिए सम्मान की भावना है। भले ही उनके पास साधन न हो परन्तु फिर भी सच्चे हृदय से बना वह सरकंडे का चश्मा भी भावनात्मक दृष्टि से मूल्यवान है। अतः उम्मीद है कि बच्चे गरीबी और साधनों के बिना भी देश के लिए कार्य करते रहेंगे।

प्रश्न घ-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
हालदार साहब भावुक हैं। इतनी सी बात पर उनकी आँखें भर आईं।
हालदार साहब इतनी-सी बात पर भावुक क्यों हो उठे?

उत्तर:
उचित साधन न होते हुए भी किसी बच्चे ने अपनी क्षमता के अनुसार नेताजी को सरकंडे का चश्मा पहनाया। यह बात उनके मन में आशा जगाती है कि आज भी देश में देश-भक्ति जीवित है भले ही बड़े लोगों के मन में देशभक्ति का अभाव हो परंतु वही देशभक्ति सरकंडे के चश्मे के माध्यम से एक बच्चे के मन में देखकर हालदार साहब भावुक हो गए।

ICSE Class 10 Hindi Solutions एकांकी-संचय – संस्कार और भावना

February 5, 2024

ICSE Class 10 Hindi Solutions एकांकी-संचय – संस्कार और भावना

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

प्रश्न क-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
अपराध और किसका है। सब मुझी को दोष देते हैं। मिसरानी कह रही थी बहू किसी की भी हो, पर अपने प्राण देकर उसने पति को बचा लिया।
यहाँ पर किसके कौन-से अपराध की बात हो रही है?

उत्तर:
यहाँ पर अतुल और अविनाश की माँ खुद के रुढ़िवादी विचारों तथा जात-पात के संस्कारों को मानने के अपराध की बात कर रही है।

प्रश्न क-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
अपराध और किसका है। सब मुझी को दोष देते हैं। मिसरानी कह रही थी बहू किसी की भी हो, पर अपने प्राण देकर उसने पति को बचा लिया।
माँ ने अविनाश की बहू को क्यों नहीं अपनाया? समझाकर लिखिए।

उत्तर :
माँ एक हिन्दू वृद्‌धा है। वे हिन्दू समाज की रूढ़िवादी संस्कारों से ग्रस्त हैं। वे संस्कारों की दास हैं। एक मध्यम परिवार में अपने पुराने संस्कारों की रक्षा करना धर्म माना जाता है। माँ भी वहीं करना चाहती थी। उसका बड़ा बेटा अविनाश अपनी माँ की इच्छा के विरुद्‌ध एक बंगाली लड़की से प्रेम-विवाह कर आया परन्तु माँ ने अपनी रूढ़िवादी मानसिकता के कारण विजातीय बहू को नहीं अपनाया।

प्रश्न क-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
अपराध और किसका है। सब मुझी को दोष देते हैं। मिसरानी कह रही थी बहू किसी की भी हो, पर अपने प्राण देकर उसने पति को बचा लिया।
बहू ने किसे और किस बीमारी से प्राण देकर बचा लिया?

उत्तर:
बहू ने अपने पति अविनाश को हैजे की बीमारी से प्राण देकर बचा लिया। हैजे की बीमारी को छुआ-छूत की बीमारी माना जाता है।

प्रश्न क-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
अपराध और किसका है। सब मुझी को दोष देते हैं। मिसरानी कह रही थी बहू किसी की भी हो, पर अपने प्राण देकर उसने पति को बचा लिया।
बहू किसकी, कौन और किस जाति की थी?

उत्तर:
बहू अविनाश की पत्नी थी जो की विजातीय (बंगाली) महिला थी।

प्रश्न ख-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
काश कि मैं निर्मम हो सकती, काश कि मैं संस्कारों की दासता से मुक्त हो पाती तो कुल धर्म और जाति का भूत मुझे तंग न करता और मैं अपने बेटे से न बिछुड़ती। स्वयं उसने मुझसे कहा था, संस्कारों की दासता सबसे भयंकर शत्रु है।
कौन संस्कारों की दासता से मुक्त होने में विफल रहा और क्यों?

उत्तर:
यहाँ पर अतुल और अविनाश की माँ हिन्दू समाज की रूढ़िवादी संस्कारों से ग्रस्त हैं। वे संस्कारों की दास हैं। एक मध्यम परिवार में अपने पुराने संस्कारों की रक्षा करना धर्म माना जाता है। इसलिए माँ संस्कारों की दासता से मुक्त होने में विफल रही।

प्रश्न ख-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
काश कि मैं निर्मम हो सकती, काश कि मैं संस्कारों की दासता से मुक्त हो पाती तो कुल धर्म और जाति का भूत मुझे तंग न करता और मैं अपने बेटे से न बिछुड़ती। स्वयं उसने मुझसे कहा था, संस्कारों की दासता सबसे भयंकर शत्रु है।
माँ ने अपने विचारों के प्रति क्या पश्चाताप किया है?

उत्तर:
माँ ने अपने रूढ़ीवादी विचारों के कारण अपने बेटे-बहू से बिछड़ने का पश्चाताप किया है।

प्रश्न ख-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
काश कि मैं निर्मम हो सकती, काश कि मैं संस्कारों की दासता से मुक्त हो पाती तो कुल धर्म और जाति का भूत मुझे तंग न करता और मैं अपने बेटे से न बिछुड़ती। स्वयं उसने मुझसे कहा था, संस्कारों की दासता सबसे भयंकर शत्रु है।
अतुल और उमा माँ के किस निर्णय से प्रसन्न हैं?

उत्तर:
जब माँ को अविनाश की पत्नी की बीमारी की सूचना मिलती है तब उसका हृदय मातृत्व की भावना से भर उठता है। उसे इस बात का आभास है कि यदि बहू को कुछ हो गया तो अविनाश नहीं बचेगा। माँ को पता है कि अविनाश को बचाने की शक्ति केवल उसी में है। इसलिए वह प्राचीन संस्कारों के बाँध को तोड़कर अपने बेटे के पास जाना चाहती है। इस प्रकार बेटे की घर वापसी के निर्णय से अतुल और उमा प्रसन्न हैं।

प्रश्न ख-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
काश कि मैं निर्मम हो सकती, काश कि मैं संस्कारों की दासता से मुक्त हो पाती तो कुल धर्म और जाति का भूत मुझे तंग न करता और मैं अपने बेटे से न बिछुड़ती। स्वयं उसने मुझसे कहा था, संस्कारों की दासता सबसे भयंकर शत्रु है।
अविनाश की वधू का चरित्र-चित्रण कीजिए।

उत्तर:
अविनाश की वधू बहुत भोली और प्यारी थी, जो उसे एक बार देख लेता उसके रूप पर मंत्रमुग्ध हो जाता। बड़ी-बड़ी काली आँखें उनमें शैशव की भोली मुस्कराहट उसके रूप तथा बड़ों के प्रति आदर के भाव ने अतुल और उमा को प्रभावित किया।

प्रश्न ग-i:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जिन बातों का हम प्राण देकर भी विरोध करने को तैयार रहते हैं। एक समय आता है, जब चाहे किसी कारण से भी हो, हम उन्हीं बातों को चुपचाप स्वीकार कर लेते हैं।
अविनाश का अपने परिवार वालों से किस बात पर विरोध था?

उत्तर:
अविनाश ने एक विजातीय (बंगाली) कन्या से विवाह किया था। किसी ने इस विवाह का समर्थन नहीं किया। अविनाश की माँ ने इसका सबसे ज्यादा विरोध किया और उसको घर से निकाल दिया।

प्रश्न ग-ii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जिन बातों का हम प्राण देकर भी विरोध करने को तैयार रहते हैं। एक समय आता है, जब चाहे किसी कारण से भी हो, हम उन्हीं बातों को चुपचाप स्वीकार कर लेते हैं।
माँ की मनोवृत्ति बदलने में अतुल और उमा ने क्या भूमिका निभाई?

उत्तर:
अविनाश ने एक विजातीय (बंगाली) कन्या से विवाह किया था। अविनाश की माँ ने इसका सबसे ज्यादा विरोध किया और उसको घर से निकाल दिया। माँ की इस रुढ़िवादी मनोवृत्ति को बदलने में अतुल और उमा ने भरपूर प्रयास किया। उन दोनों ने अविनाश की पत्नी के गुणों तथा विचारों से माँ को अवगत करवाया अतुल ने ही अपनी माँ को अविनाश की बहू को अपनाने के लिए प्रेरित किया। अतुल ने के द्वारा ही माँ को पता चलता है कि किस प्रकार उनकी बहू ने अपने प्राणों की परवाह न करके अविनाश की जान बचाई और अब बहू स्वयं बीमार है। इसलिए जब माँ को अविनाश की पत्नी की बीमारी की सूचना मिलती है तब उसका हृदय मातृत्व की भावना से भर उठता है। उसे इस बात का आभास है कि यदि बहू को कुछ हो गया तो अविनाश नहीं बचेगा।
इस प्रकार अतुल और उमा के सम्मिलित प्रयास से माँ अपनी बहू को अपना लेती है।

प्रश्न ग-iii:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जिन बातों का हम प्राण देकर भी विरोध करने को तैयार रहते हैं। एक समय आता है, जब चाहे किसी कारण से भी हो, हम उन्हीं बातों को चुपचाप स्वीकार कर लेते हैं। अतुल का चरित्र-चित्रण कीजिए।

उत्तर:
अतुल एकांकी का प्रमुख पुरुष पात्र है। वह माँ का छोटा पुत्र है। वह प्राचीन संस्कारों को मानते हुए आधुनिकता में यकीन रखने वाला एक प्रगतिशील नवयुवक है। वह माँ का आज्ञाकारी पुत्र होते हुए भी माँ की गलत बातों का विरोध भी करता है। वह अपनी माँ से अपने बड़े भाई को विजातीय स्त्री से विवाह करने पर न अपनाने का भी विरोध करता है। अतुल संयुक्त परिवार में विश्वास रखता है। उसमें भ्रातृत्व की भावना है। वह अपने बड़े भाई का सम्मान करता है।

प्रश्न ग-iv:
निम्नलिखित गद्यांश को पढ़कर नीचे दिए गए प्रश्नों के उत्तर लिखिए :
जिन बातों का हम प्राण देकर भी विरोध करने को तैयार रहते हैं। एक समय आता है, जब चाहे किसी कारण से भी हो, हम उन्हीं बातों को चुपचाप स्वीकार कर लेते हैं। एकांकी का सारांश लिखिए।

उत्तर:
विष्णु प्रभाकर द्‌वारा रचित “संस्कार और भावना” एकांकी में भारतीय हिंदू परिवार के पुराने संस्कारों से जकड़ी हुई रूढ़िवादिता तथा आधुनिक परिवेश में पले बड़े बच्चों के बीच संघर्ष की चेतना को चित्रित किया गया है।
अविनाश ने एक विजातीय (बंगाली) कन्या से विवाह किया था। किसी ने इस विवाह का समर्थन नहीं किया। अविनाश की माँ ने इसका सबसे ज्यादा विरोध किया और उसको घर से निकाल दिया। माँ अपने छोटे बेटे अतुल और उसकी पत्नी उमा के साथ रहती है पर बड़े बेटे से अलग रहना उसके मन को कष्ट पहुँचाता है।
एक बार जब माँ को पता चला कि अविनाश को प्राणघातक हैजे की बीमारी हुई थी और बहू ने अपने पति अविनाश को प्राण देकर बचा लिया। अब वह खुद बीमार है परंतु अविनाश में उसे बचाने की ताकत नहीं है। जब माँ को अविनाश की पत्नी की बीमारी की सूचना मिलती है तब उसका हृदय मातृत्व की भावना से भर उठता है। उसे इस बात का आभास है कि यदि बहू को कुछ हो गया तो अविनाश नहीं बचेगा। तब पुत्र-प्रेम की मानवीय भावना का प्रबल प्रवाह रूढ़िग्रस्त प्राचीन संस्कारों के जर्जर होते बाँध को तोड़ देता है। माँ अपने बेटे और बहू को अपनाने का निश्चय करती है।

ICSE Class 10 English Solutions Small Pain in my Chest [Poem]

February 5, 2024

ICSE Class 10 English Solutions Small Pain in my Chest [Poem]

Passage 1

Question 1.
Read the extract and answer the questions that follow.

The soldier boy was sitting calmly underneath that tree.
As I approached it, I could see him beckoning to me.
The battle had been long and hard and lasted through the night
And scores of figures on the ground lay still by morning’s light.

Why was the soldier sitting calmly under the tree?

Answer:
The war had ended by the break of day. This soldier was the only alive one among the hundreds of other who died fighting the war. He was sitting calmly under the tree as there are no more enemies to fight.

Question 2.
Read the extract and answer the questions that follow.

What does the phrase ‘sitting calmly’ signify?

Answer:
The phrase ‘sitting calmly’ indicates the stillness and silence which follows a storm. The soldier has no need to be worried or to be on his guard as there are no enemies left to fight, he is therefore calm It also stresses on the fact that his fellow soldiers are all lying dead around him and therefore, he has no one to talk to.

Question 3.
Read the extract and answer the questions that follow.

Why did the soldier call the narrator?

Answer:
The soldier was wounded and too tired to get water for himself. He called the narrator hoping that he would help him with some water so that he could quench his thirst and his wounded body could be relieved of some pain.

Question 4.
Read the extract and answer the questions that follow.

Explain the line ‘And scores of figures on the ground lay still by morning’s light’.

Answer:
The given lines explain how the soldiers fought hard all night. In the morning, the wounded soldier boy who sat under a tree saw all of his fellow soldiers lying dead beside him on the battle field.

Passage 2

Question 1.
Read the extract and answer the questions that follow.

“I wonder if you’d help me, sir”, he smiled as best he could.
“A sip of water on this morn would surely do me good.
We fought all day and fought all night with scarcely any rest –
A sip of water for I have a small pain in my chest.”

What help does the soldier need from the narrator?

Answer:
The soldier is wounded and wants the narrator to give him some water.

Question 2.
Read the extract and answer the questions that follow.

Why is the soldier smiling?

Answer:
Although the soldier is wounded, like a true fighter he has a smile on his face as he doesn’t believe in showing his pain. He is smiling to assure the narrator that he is just tired after the night’s fight and there is nothing else wrong with him.

Question 3.
Read the extract and answer the questions that follow.

Why doesn’t the soldier directly tell the narrator that he has been wounded by a bullet?

Answer:
The soldier has faced and suffered the devastating reality of a war and does not want a civilian to be acquainted with the language of guns and bullets. Therefore, he only tells the narrator that he is tired and feeling a slight pain in his chest instead of telling him how seriously injured he is.

Question 4.
Read the extract and answer the questions that follow.

Why do you think the soldier asks for a sip of water?

Answer:
The soldier asks for a sip of water to feel some relief from his pain. He perhaps knows that death is now imminent and wants to drink water for the last time before he falls into an eternal slumber of death.

Passage 3

Question 1.
Read the extract and answer the questions that follow.

As I looked at him I could see the large stain on his shirt
All reddish – brown from his warm blood mixed in with
Asian dirt.
“Not much”, said he. “I count myself more lucky than the rest.
They’re all gone while I just have a small pain in my chest.”

What does the narrator notice on the soldier’s shirt?

Answer:
The narrator notices a large reddish brown stain on the soldier’s shirt.

Question 2.
Read the extract and answer the questions that follow.

Why is the stain ‘all reddish brown’?

Answer:
The stain is all reddish brown because it is the soldier’s blood mixed with the mud of the Asian land where the Vietnam War was fought.

Question 3.
Read the extract and answer the questions that follow.

Why does the soldier count himself lucky? Do you think he is really lucky?

Answer:
The soldier counts himself as lucky because he is alive unlike his fellow mates, who have lost their lives while fighting the war. Unfortunately, he is not lucky as he has been hit with a bullet and will soon die like the other soldiers.

Passage 4

Question 1.
Read the extract and answer the questions that follow.

“Must be fatigue”, he weakly smiled. “I must be getting old.
I see the sun is shining bright and yet I’m feeling cold.
We climbed the hill, two hundred strong, but as we cleared the crest,
The night exploded and I felt this small pain in my chest.”

Who smiled weakly in the extract? Why?

Answer:
The soldier smiles weakly as he is seriously injured but doesn’t want to reveal his pain to a civilian (the narrator). His pain is excruciating but by smiling weakly, he tries to hide it.

Question 2.
Read the extract and answer the questions that follow.

Why is the soldier feeling cold?

Answer:
The soldier’s injury is causing his bodily functions to deteriorate as a result of which he is feeling cold even on a warm sunny morning.

Question 3.
Read the extract and answer the questions that follow.

Why does the soldier feel that he is getting old? Is the reason that he gives true?

Answer:
The young soldier feels that he is getting old because he is experiencing fatigue and weakness. However, the true reason for his weakness is the bullet in his chest that is pushing him towards death.

Question 4.
Read the extract and answer the questions that follow.

What does “the night exploded” refer to?

Answer:
The phrase ‘the night exploded’ refers to the sudden attack on the soldier’s army by the enemies which may have included bomb explosions along with the firing of bullets. The soldier along with the other mates had been fighting the entire night. The poet has personified the night to describe how gruesome the war had been.

Passage 5

Question 1.
Read the extract and answer the questions that follow.

“I looked around to get some aid – the only things I found
Were big, deep craters in the earth – bodies on the ground.
I kept on firing at them, sir. I tried to do my best,
But finally sat down with this small pain in my chest.”

Why is the soldier looking around?

Answer:
The soldier is injured and is looking around in the hope of getting some aid.

Question 2.
Read the extract and answer the questions that follow.

Does the soldier find help? What does he see around him?

Answer:
The soldier doesn’t find help, instead he sees deep craters in the earth and dead bodies of his fellow soldiers around him.

Question 3.
Read the extract and answer the questions that follow.

Why did the soldier have to stop firing?

Answer:
The soldier fought bravely until finally he was wounded by a shot and had to stop firing. .

Question 4.
Read the extract and answer the questions that follow.

What does the soldier want to justify when he says “I tried to do my best”?

Answer:
When the soldier tells the narrator that he tried to do his best, he is emphasising on the fact that he fought with all his might to bring down the enemies and keep his fellow mates safe.

Passage 6

Question 1.
Read the extract and answer the questions that follow.

“I’m grateful, sir”, he whispered, as I handed my canteen
And smiled a smile that was, U think, the brightest that I’ve seen.
“Seems silly that a man my size so full of vim and zest,
Could find himself defeated by a small pain in his chest.”

What according to the soldier is silly?

Answer:
According to him, it is silly for a soldier, who ought to be full of energy and vigour to be defeated by a small pain in the chest.

Question 2.
Read the extract and answer the questions that follow.

What is the actual condition of the solider?

Answer:
The soldier has been fatally wounded in the war and is currently in acute pain. However, he portrays exactly the opposite of his true condition. He tries to show that he is fine and that it is just a small pain in his chest. In reality, he is counting the last few minutes of his life.

Question 3.
Read the extract and answer the questions that follow.

What is the soldier grateful for?

Answer:
The soldier is grateful that the narrator gives him some water to sip in his last moments.

Question 4.
Read the extract and answer the questions that follow.

Why is the smile of the soldier described as the brightest by the narrator?

Answer:
The narrator understands that in order to hide his pain, the soldier is trying to force a smile on his face. Beneath the smiling face is a worn out body and a heart shredded apart by the death and destruction around him. Therefore, the narrator says that his is the brightest smile he has ever seen.

Passage 7

Question 1.
Read the extract and answer the questions that follow.

“What would my wife be thinking of her man so strong and grown,
If she could see me sitting here, too weak to stand alone?
Could my mother have imagined, as she held me to her breast,
That I’d be sitting HERE one day with this pain in my chest?”

What about his wife and mother is the soldier thinking?

Answer:
The soldier wonders what his wife would think of him if she knew that her husband, a strong soldier, could not even bear a small pain in his chest and stand alone. Then he thinks about his mother, he wondering how his mother will react after knowing that a boy that she held close to her heart is sitting in the battlefield with a fatal wound.

Question 2.
Read the extract and answer the questions that follow.

Why does the soldier always refer to his pain as small?

Answer:
It is a known fact that the injury caused by explosions or bullet wounds can be very serious, and at most times, fatal. However, the soldier constantly stresses on the fact that the pain in his chest is small or little. Through this he wishes to tell the narrator that the physical and mental trauma that soldiers suffer during the war is much severe than a bullet wound but they are trained to deal with it for the good of their nation.

Question 3.
Read the extract and answer the questions that follow.

Why do you think the soldier is thinking about his family?

Answer:
The soldier is thinking about his family as he realises that he is living the final moments of his life. In these moments, he tries to think about his life in retrospect, without any lament about the war that is taking him to his death. All that he does is talk to the narrator calmly as if wanting to be heard in his final minutes of being alive. He thinks about his family especially his wife and mother who he knows will wonder how a soldier like him couldn’t escape the explosion despite being trained to dodge life threatening circumstances like these. As he knows that he will not be able to talk to his family for the last time or listen to them, he ponders over what they might say if they see him in this condition.

Passage 8

Question 1.
Read the extract and answer the questions that follow.

“Can it be getting dark so soon?” He winced up at the sun.
“It’s growing dim and I thought that the day had just begun.
I think, before I travel on, I’ll get a little rest……
And, quietly, the boy died from that small pain in his chest.

I don’t recall what happened then. I think I must have cried;
I put my arms around him and I pulled him to my side
And, as I held him to me, I could feel our wounds were pressed
The large one in my heart against the small one in his chest.

What do the first two lines of the given extract tell about the soldier’s condition?

Answer:
The first two lines of the stanza are among the last few lines of the poem where the soldier succumbs to his injuries. In the given lines, the soldier winces up at the sun and tells the narrator that he feels as if it is getting dark soon although the day has just began.

Question 2.
Read the extract and answer the questions that follow.

Explain “before I travel on”.

Answer:
In the third line of the given extract, the soldier is finally seen telling the narrator that it is time for him to embrace death and ‘travel on’ but he must get a little rest before moving on.

Question 3.
Read the extract and answer the questions that follow.

What do you think are the feelings of the narrator in this stanza?

Answer:
The narrator has been a patient listener for the wounded soldier throughout their conversation. The soldier tells him how he fought the war all night along with the others, how his family will think about him after his death, and how he is luckier than his dead soldier friends to have lived till the break of this day.
At the end of this stanza, the narrator tells the readers that the soldier quietly dies from the small pain in his chest. The narrator feels helpless and remorseful to see the young soldier succumb to his injuries. In their short conversation the narrator has learned a lot about the soldier and cannot help but feel overwhelmed by the fact that the soldier, who put up a brave fight against death, has now humbly accepted it.

Question 4.
Read the extract and answer the questions that follow.

How does the narrator express his feelings for the dead soldier?

Answer:
When the soldier dies, a kind of numbness comes over the narrator; he mentions not being able to remember what happened next. He further says that he must have cried but does not seem to have a clear memory of it. He pulls the soldier to his side and as he holds him close to his chest, he feels a large pain in his own heart. The narrator says both their chests pressed to each other are wounded by this war.; His heart is emotionally wounded by the loss of the soldier’s life while the soldier’s chest is wounded the physical injury and trauma from the war.

Selina Concise Chemistry Class 10 ICSE Solutions Sulphuric Acid

February 4, 2024

Selina Concise Chemistry Class 10 ICSE Solutions Sulphuric Acid

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 11 Sulphuric Acid. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 11 Sulphuric Acid

Exercise 1

Solution 1.

(a) Sulphuric acid is called King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries.
(b) Sulphuric acid is referred to as Oil of vitriol as it was obtained as an oily viscous liquid by heating crystals of green vitriol.

Solution 2.

(a) Two balanced equations to obtain SO₂ is:
(i) 4FeS₂ + 11O₂ → 2Fe₂O₃ +8SO₂(ii) S +O₂ → SO₂

(b) The conditions for the oxidation of SO₂ are:
(i) The temperature should be as low as possible. The yield has been found to be maximum at about 410⁰C-450^oC
(ii) High pressure (2 atm) is favoured because the product formed has less volume than reactant.
(iii) Excess of oxygen increases the production of sulphur trioxide.
(iv) Vanadium pentoxide or platinised asbestos is used as catalyst.

(c) Sulphuric acid is not obtained directly by reacting SO₃ with water because the reaction is highly exothermic which produce the fine misty droplets of sulphuric acid that is not directly absorbed by water.

(d) The chemical used to dissolve SO₃ is concentrated sulphuric acid. The product formed is oleum.

(e) Main reactions of this process are:

selina-icse-solutions-class-10-chemistry-sulphuric-acid-2

Solution 3.

Water is not added to concentrated acid since it is an exothermic reaction. If water is added to the acid, there is a sudden increase in temperature and the acid being in bulk tends to spurt out with serious consequences.

Solution 4.

Impurity of ARSENIC poisons the catalyst [i.e. deactivates the catalyst]. So, it must be removed before passing the mixture of SO₂ air through the catalytic chamber.

Solution 5.

Balanced reactions are:

(a) Acidic nature:

(i) Dilute H₂SO₄ reacts with basic oxides to form sulphate and water.
2 NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(ii) CuO + H₂SO₄→ CuSO₄ + H₂O

(iii) It reacts with carbonate to produce CO₂.
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂ ↑

(b) Oxidising agent:

H₂SO₄ → H₂O +SO₂ + [O]
Nascent oxygen oxidizes non-metals, metals and inorganic compounds.
For example,
Carbon to carbon dioxide
C+H₂SO₄→ CO₂ +H₂O +2SO₂

Sulphur to sulphur dioxide
S +H₂SO₄→ 3SO₂ +2H₂O

It has great affinity for water. It readily absorbs moisture from atmospheric air.

selina-icse-solutions-class-10-chemistry-sulphuric-acid-5

(d) Non-volatile nature:

It has a high boiling point (356^oC) so it is considered to be non-volatile. Therefore, it is used for preparing volatile acids like hydrochloric acid, nitric acid from their salts by double decomposition reaction.

NaCl + H₂SO₄→ NaHSO₄ + HCl

KCl + H₂SO₄→ KHSO₄ + HCl

Solution 6.

(a) Bring a glass rod dipped in Ammonia solution near the mouth of each test tubes containing dil. Hel and dil. H₂SO₄each. selina-icse-solutions-class-10-chemistry-sulphuric-acid-6

(b)

Dilute sulphuric acid treated with zinc gives Hydrogen gas which bums with pop sound.
Concentrated H₂SO₄ gives SO₂ gas with zinc and the gas turns Acidified potassium dichromate paper green.
Barium chloride solution gives white ppt. with dilute H₂SO₄, This white ppt. is insoluble in all acids.
Concentrated H₂SO₄ and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

Solution 7.

selina-icse-solutions-class-10-chemistry-sulphuric-acid-7

Solution 8.

(a) Concentrated sulphuric acid is hygroscopic substance that absorbs moisture when exposed to air. Hence, it is stored in air tight bottles.

(b) Sulphuric acid is not a drying agent for H₂S because it reacts with H₂S to form sulphur.
H₂SO₄ + H₂S → 2H₂O + SO₂ + S

(c) Concentrated sulphuric acid has high boiling point (356^oC). So, it is considered to be non-volatile. Hence, it is used for preparing volatile acids like Hydrochloric acid and Nitric acids from their salts by double decomposition.
NaCl + H₂SO₄→ NaHSO₄ + HCl
NaNO₃ + H₂SO₄→ NaHSO₄ + HNO₃

Solution 9.

(a) Due to its reducing property. i.e, it is a non-volatile acid.

NaCl + H₂SO₄→ NaHSO₄ + HCl

selina-icse-solutions-class-10-chemistry-sulphuric-acid-9

(c) Magnesium is present above hydrogen in the reactivity series so sulphuric acid is able to liberate hydrogen gas by reacting with magnesium strip.

Mg + H₂SO₄→ MgSO₄ + H₂

(d) Due to its oxidizing character

Cu +H₂SO₄→ CuSO₄ +2H₂O +SO₂

(e) Due to its oxidizing property Hydrogen sulphide gas is passed through concentrated sulphuric acid to liberate sulphur dioxide and sulphur is formed.

H₂S + H₂SO₄→ S + 2H₂O + SO₂

Solution 10.

The name of the salt of
(a) Hydrogen sulphites and Sulphites.
(b) Sulphate and bisulphate.

Solution 11.

(a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic.

NaOH + H₂SO₄→ NaHSO₄ + H₂O

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

(b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours.

2KBr + 3H₂SO₄ → 2KHSO₄ + SO₂ + Br₂ ↑ + 2H₂O

(d) When sulphuric acid is added to sodium carbonate it liberates carbon dioxide which produces brisk effervescence.

Na₂CO₃ + H₂SO₄→ Na₂SO₄+ H₂O + CO₂ ↑

Solution 12.

Column 1 Substance reacted with acid	Column 2 Dilute or concentrated acid	Column 3 Gas
Substance reacted with acid	Dilute or concentrated sulphuric acid	Gas
Zinc	Dilute sulphuric acid	Hydrogen
Calcium carbonate	Concentrated sulphuric acid	Carbon dioxide
Bleaching power CaOCl₂	Dilute sulphuric acid	only chlorine

Solution 1 (2004).

Hydrogen sulphide (H₂S) can be oxidized to sulphur.

Solution 1 (2006).

(a) The process used for the large-scale manufacture of sulphuric acid is Contact process.

(b) Sulphuric acid has great affinity for water. It readily removes element of water from other compound. Thus it acts as a dehydrating agent.

(c) Concentrated acid is non-volatile thus it is used for the preparation of volatile acids:
NaCl + H₂SO₄→ NaHSO₄ + HCl
Concentrated acid act as an oxidizing agent:
C + 2H₂SO₄→ CO₂ + 2H₂O + 2SO₂

Solution 1 (2007).

(i) B
(ii) D
(iii) C
(iv) A
(v) A

Solution 1 (2008).

(a) Concentrated sulphuric acid is non-volatile; hence it is used for the preparation of higher volatile acids.
(b) Due to its dehydrating nature sugar turns black in the presence of concentrated sulphuric acid.

Solution 2 (2004).

When sodium sulphide is added to solution of HCl, Hydrogen sulphide gas is produced. It has rotten egg like smell.

Solution 2 (2007).

(a) The acid formed when sulphur dioxide dissolves in water is sulphurous acid.
(b) Carbondioxide gas is released when sodium carbonate is added to solution of sulphur dioxide.

Solution 3 (2004).

(a) The catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C is Vanadium pentoxide.
(b) The two steps for the conversion of sulphur trioxide to sulphuric acid is:
(i) SO₃+ H₂SO₄→ H₂S₂O₇(ii) H₂S₂O₇ + H₂O → 2H₂SO₄(b) The substance that will liberate sulphur dioxide in step E is dilute H₂SO_4.(c) The equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F is:
SO₂ + 2NaOH → Na₂SO₃ + H₂O
Or Na₂O + SO₂ → Na₂SO₃

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Selina Concise Biology Class 10 ICSE Solutions Transpiration

February 4, 2024

Selina Concise Biology Class 10 ICSE Solutions Transpiration

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 5 Transpiration. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 5 Transpiration

Exercise 2

Solution A.1.
(a) Open stomata, dry atmosphere and moist soil

Solution A.2.
(a) increase

Solution A.3.
(b) temperature is high

Solution A.4.
(c) sunken stomata

Solution A.5.
(d) hydathodes

Solution A.6.
(d) transpiration

Solution A.7.
(d) hot, dry and windy

Solution A.8.
(b) Lenticels

Solution A.9.
(b) evaporation of water from the aerial surfaces of a plant

Solution B.1.
(a) Lenticels
(b) Guttation
(c) Potometer
(d) Nerium
(e) Ganong’s photometer
(f) Stomata and cuticle
(g) Hydathodes
(h) Guttation

Solution B.2.
(a) vapour, aerial
(b) stomata, transpiration
(c) suction, water (heat)

Solution C.1.
(a) guttation
(b) protection and reduced transpiration
(c) transpiration
(d) reduced transpiration

Solution C.2.
(i) False
(ii) True
(iii) True
(iv) False
(v) Most transpiration occurs at mid-day.
(vi) Potometer is an instrument used for measuring the rate of transpiration in green plants.

Solution C.3.
(a) Transpiration increases with the velocity of wind. If the wind blows faster, the water vapours released during transpiration are removed faster and the area surrounding the transpiring leaf does not get saturated with water vapour.

(b) When the rate of transpiration far exceeds the rate of absorption of water by roots, the cells lose their turgidity. Hence, excessive transpiration results in wilting of the leaves.

(c) Plants absorb water continuously through their roots, which is then conducted upwards to all the aerial parts of the plant, including the leaves. Only a small quantity of this water i.e. about 0.02% is used for the photosynthesis and other activities. The rest of the water is transpired as water vapour. Hence water transpired is the water absorbed.

(d) There are more stomatal openings on the lower surface of a dorsiventral leaf. More the number of stomata, higher is the rate of transpiration. Hence more transpiration occurs from the lower surface.

(e) Cork and Bark of trees are tissues of old woody stems. Bark is thick with outermost layer made of dead cells and the cork is hydrophobic in nature. These properties make them water-proof and hence they prevent transpiration.

(f) In both perspiration and transpiration, water is lost by evapouration from the body of the organism as water vapour. This evaporation reduces the temperature of the body surface and brings about cooling in the body of the organism.

(g) On a bright sunny day, the rate of transpiration is much higher than any other days. The leaves of certain plants roll up on a bright sunny day to reduce the exposed surface and thus reduce the rate of transpiration.

Solution C.4.
(a) False
Reason: Potometer is used to measure the rate of transpiration in a plant. Demonstration of transpiration occurring from the lower surface of a leaf is done by analyzing the changes in colour of pieces of dry cobalt chloride paper attached (and held in place) to the two surfaces of a leaf.

(b) True
Reason: Transpiration carried out by the large number of trees in a forest. This increases the moisture in the atmosphere and brings rain.

(c) False
Reason: Hydathodes are special pores present on the ends of leaf veins through which guttation occurs and water droplets are given out. Their openings cannot be regulated. Stomata on the other hand are minute openings in the epidermal layer of leaves through which exchange of gases as well as transpiration occurs. Water is given out as water vapour. Stomatal opening is regulated by guard cells.

(d) False
Reason: Transpiration is reduced during high atmospheric humidity. High humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

(e) True
Reason: Desert plants need to reduce transpiration as much as possible so as to survive in the hot and dry environment. Hence some of them have sunken stomata as an adaptation to curtail transpiration.

(f) True
Reason: During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. During mid-day, the outside temperature is higher, due to which there is more evaporation of water from the leaves. Therefore more transpiration occurs during mid-day.

Solution C.5.

Guttation	Bleeding
It is the removal of excess of water from the plants because of excess water buildup in the plant.	It is the removal of water from the plant because of injury.
Water escapes from specialisedstructures called hydathodes.	Water escapes in the form of sap from the injured part of the plant.

Solution D.1.
Wilting refers to the loss of cellular turgidity in plants which results in the drooping of leaves or plant as a whole because of lack of water.
During noon the rate of transpiration exceeds the rate of absorption of water by roots. Due to the excessive transpiration, the cells of leaves lose their turgidity and wilt.

Solution D.2.
The lower surface of leaf is sheltered from direct sunlight. If more stomata are on the upper surface of a leaf, then excessive transpiration would occur, resulting in quick wilting of the plant. Hence most plants have more numerous stomata on the lower surface of a leaf to control the rate of transpiration.

Solution D.3.
Take the small potted rose plant and cover it with a transparent polythene bag. Tie its mouth around the base of the stem. Leave the plant in sunlight for an hour or two.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 1
Drops of water will soon appear on the inner side of the bag due to the saturation of water vapour given out by the leaves. A similar empty polythene bag with its mouth tied and kept in sunlight will show no drops of water. This is the control to show that plants transpire water in the form of water. If tested with dry cobalt chloride paper, the drops will be confirmed as water only.

Solution D.4.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 2
Potometer is a device that measures the rate of water intake by a plant. This water intake is almost equal to the water lost through transpiration. Potometers do not measure the water lost due to transpiration but measure the water uptake by the shoot.

Solution D.5.

Transpiration occurring through lenticels i.e. minute openings on the surface of old stems is called lenticular transpiration.
Stomatal transpiration is controlled by the plant by altering the size of the stoma, where as this does not happen in case of lenticular transpiration. This is because the lenticels never close, but remain open all the time.
The amount of stomatal transpiration is much more than the amount of lenticular transpiration.

Solution D.6.
The factors that accelerate the rate of transpiration are:

High intensity of sunlight
High temperature
Higher wind velocity
Decrease in atmospheric pressure
(Any three)

Solution D.7.
Forests have large number of plants especially trees. Each plant loses water in the form of water vapour everyday into the atmosphere through transpiration. A large apple tree loses as much as 30 litres of water per day. So huge amount of water is escaped into the atmosphere by forests. This increases the moisture in the atmosphere and brings more frequent rains.

Solution D.8.
The advantages of transpiration to the plants are:

Transpiration brings about a cooling effect to the plant body since evaporation of water reduces the temperature of leaf surface.
Transpiration helps in the ascent of sap by producing a suction force acting from the top of the plant.
Transpiration helps in distributing water and mineral salts throughout the plant body.
Transpiration helps in eliminating excess water.

Solution D.9.

If the water content of the leaves decreases due any reason, the guard cells turn flaccid, thereby closing the stomatal opening and transpiration stops.
Some plants have sunken stomata whereas others have reduced number of stomata to reduce transpiration.
In some plants, leaves may be dropped or may be absent or changed into spines as an adaptation to reduce transpiration.
The leaves may be covered by thick cuticle such as in Banyan tree, so as to reduce transpiration.

Solution D.10.
No, they are not dew drops.

This is water given out by the plant body through guttation. Since the banana plant is growing in humid environment, transpiration is hampered. But the roots continue to absorb water from the soil. This builds up a huge hydrostatic pressure within the plant and forces out the excess water from the hydathodes, which are pores present at the tips of veins in the leaf. This is observed especially during the mornings.

Solution D.11.
(a) Intensity of light – During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. At night they are closed. Hence more transpiration occurs during the day. During cloudy days, the stomata are partially closed and the transpiration is reduced.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 3
(b) Humidity of the atmosphere – When the air is humid; it can receive very less water vapour. Thus, high humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

Solution E.1.

(i) The leaf D would become most limp. This is because water would be lost through transpiration from upper as well as the lower surface of leaf D since it is uncoated.

(ii) The least limping would be shown by leaf C since its upper and lower surfaces have been coated with vaseline. So no water is lost from the leaf through transpiration since the stomatal openings get blocked by vaseline.

Solution E.2.
(a)
Guard Cell
Inner wall of the Guard Cell
Stoma/Stomatal Aperture
(b) Open state
(c) The structure of stoma remains same in monocots as well as in dicots. Hence, the stoma from the diagram can be of a monocot leaf or of a dicot leaf.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 4

Solution E.3.
(a) Transpiration
(b) Oil is put on the surface of water to prevent loss of water by evaporation.
(c) Yes, the transpiration rate will increase. Transpiration would occur faster. The observable changes will occur in less time.
(d) The spring balance progressively measures the change in weight of the set-up. This because as the plant transpires, it creates the suction force in plant which allows roots to absorb more water from the test tube. Hence, the water in the test will get reduced. Thus, the weight of the entire set will decrease.

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Selina Concise Chemistry Class 10 ICSE Solutions Metallurgy

February 4, 2024

Selina Concise Chemistry Class 10 ICSE Solutions Metallurgy

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 7 Metallurgy. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 7 Metallurgy

Exercise 7(A)

Solution 1.

Three classes in which elements are classified are:
Metals, Non-metals and Metalloids
Copper was the first metal used by man.

Solution 2.

(a) The metal which is a constituent of blood pigment is Iron (Fe)
(b) The metal which is a constituent of plant pigment is Magnesium (Mg).

Solution 3.

(a) Nitrogen: It is used to preserve food.
(b) Hydrogen: It is used in the hydrogenation of vegetable oils to make ghee.
(c) Carbon: It is essential for the growth and development of living beings.

Solution 4.

The metal which is present in abundance in earths crust is aluminium.
The non-metal which is present in abundance in the earth crust is oxygen.

Solution 5.

Metals are defined as the elements which form positive ions by the loss of electrons.
Non-metals are the elements which form negative ions by the gain of electrons.

Solution 6.

(a) Alkali metals: They are placed in IA group, the first column on the left of the periodic table.
(b) Alkaline earth metal: They are placed in IIA group, the second column on the left of the periodic table.
(c) Iron and Zinc: Fe is placed in VIII group and Cu is placed in IB group.
(d) Aluminium: It is placed in IIIA group present on the right of periodic table.

Solution 7.

(a) Alkali metals:-
(i) Bonding: All alkali metal salts are ionic in nature.
(ii) Action of air: The react rapidly with oxygen and water vapour in the air.
(iii) Action of water: They react violently with water and produce hydrogen gas.
2M + 2H₂O → 2MOH + H₂(iv) Action of acid: They react violently with dil. HCl and dil. H₂SO₄ to produce hydrogen gas.
2M + 2HCl → 2MCl + H₂

(b) Alkaline earth metal:-
(i) Bonding: All alkaline earth metal salts except beryllium are ionic compounds.
(ii) Action of air: They are less reactive than alkali metals.
(iii) Action of water: They react with water to produce hydrogen gas.
M + 2H₂O → M(OH)₂ + H₂(iv) Action of acid: They react with dilute HCl and dil. H₂SO₄ to produce hydrogen gas.
M + 2HCl → MCl₂ + H₂

Solution 8.

Elements which show properties of both metals and non-metals are called metalloids.
For example: Silicon, Germanium.

Solution 9.

Hydrogen is placed with alkali metals as it has one electron similar to the alkali metals.

Solution 10.

(a) Bromine
(b) Lead
(c) Gallium
(d) Carbon
(e) Sodium
(f) Sodium
(g) Tungsten
(h) Carbon fibre
(i) Carbon
(j) Mercury

Solution 11.

(i) Ion formation: Metals form positive ions by loss of electrons whereas non-metals form negative ions by gain of electrons.

(ii) Discharge of ions: Metals are discharged at the cathode during electrolysis whereas non-metals are liberated at the anode during electrolysis.

(iii) Nature of oxide formed: Oxides of metals are usually basic. Soluble basic oxides dissolve in water forming an alkaline solution whereas oxides of non-metals are usually acidic. Soluble acidic oxides dissolve in water forming an acidic solution.

(iv) Oxidizing and reducing property: Metals ionize by loss of electrons and hence are reducing agents whereas non-metals ionize by gain of electrons and hence are oxidizing agents.

(v) Reaction with acids: Metals above hydrogen in activity series usually replace hydrogen from dilute non-oxidising acids whereas non-metals do not react with dilute hydrochloric acid or sulphuric acid.

Solution 12.

(a) Na – e^– → Na⁺(b) N + 3e^– → N^3-(c) Cl + e^– → Cl^–(d) Mg – 2e^–→ Mg²⁺(e) M + 2HCl → MCl₂ + H₂(f) Mg + H₂SO₄ → MgSO₄ + H₂

Solution 13.

(a) Fe₂O₃(b) PbO
(c) Mn₂O₇(d) NO

Solution 14.

Exercise 7(B)

Solution 1.

(a) Mercury and gallium
(b) Sodium and potassium
(c) Mercury
(d) Iodine
(e) Graphite
(f) Zinc
(g) Neon , Argon
(h) CrO₃ , Mn₂O₇(i) Al₂O₃,PbO
(j) Potassium , sodium
(k) Basic copper(II) sulphate
(l) Aluminium , Oxygen
(m) Hydrogen
(n) Carbon
(o) Iron

Solution 2.

(a) Occurrence of metals: The metals placed at the top of activity series are most reactive, so they always exist in the combined state whereas the metals placed below the activity series are least reactive, so they can be found in the isolated state also.

(b) Tendency to corrosion: The metals lying above the hydrogen in activity series can easily react with moisture and air and corrode easily whereas the metals such as gold and platinum do not corrode easily.

Potassium and sodium reacts with cold water whereas magnesium reacts with warm water and aluminium, zinc and iron reacts with steam.

(d) Reaction with acids: All the metals above hydrogen, in the activity series, reduce hydrogen ions from dil. hydrochloric or sulphuric acid and give out hydrogen gas. The rate of reaction decreases on moving down the series.

Solution 3.

selina-icse-solutions-class-10-chemistry-metallurgy-7b-3

Solution 4.

The metals placed higher in the activity series (i.e. Na and K) are stable to heat and soluble in water.
Whereas metals like Ca, Mg, Al, Zn, Fe, Pb, Cu decompose on heating with decreasing vigour to form metal oxide and carbon dioxide.
The metals which lie below in the activity series (i.e. Hg, Ag) decompose on heating to form metal, oxygen and carbon dioxide.

Solution 5.

(a) Alkali metals like sodium and potassium are kept in kerosene as they react with moisture and air.
(b) (i) Basic lead carbonate is a mixture of lead hydroxide and lead carbonate.
(ii) Brown powder is mainly hydrated iron(III) oxide (Fe₂O₃.xH₂O)

Solution 6.

Oxides of metals like Na, K, Ca, Mg, Al are stable to heat and so can be reduced only by electrolysis.
Zinc oxide can be reduced by coke only.
Oxides of iron, lead and copper are reduced by C, CO, H₂ and NH₃.
Oxides of mercury and silver decompose to give metal and oxygen.

Solution 7.

Metal A is more reactive than Metal B.
(a) Metal A is Na (Sodium). Metal B is Ca (Calcium).
Reaction with HCl:

selina-icse-solutions-class-10-chemistry-metallurgy-7b-7

(b) (i) Oxides: Sodium and calcium oxides are stable to heat.
(ii) Hydroxides: Sodium hydroxide is stable to heat whereas calcium hydroxide decomposes on heating to metal oxide and water vapour.
(iii) Carbonates: Sodium carbonate is stable to heat whereas calcium carbonates decompose on heating to form calcium oxide and carbon dioxide.
(iv) Nitrates: Sodium nitrate on heating form nitrite and oxygen whereas calcium nitrate decomposes on heating to form calcium oxide, nitrogen dioxide and oxygen.

Solution 8.

(a)

Metals	Non-metals
(i) Good conductors of heat (ii) Malleable (iii) Form positive ions (iv) Form basic oxides	Poor conductors of heat Non-Malleable Forms negative ions Form acidic oxides

(b) Valence electrons present in :
(i) Metals have 1, 2 or 3 valence electrons.
(ii) Non-metals have 5, 6 or 7 valence electrons.

Solution 9.

When the surface of metal is attacked by air, moisture or any other substance around it, the metal is said to corrode and the phenomenon is known as corrosion.
Necessary conditions for corrosion are:

Presence of oxygen and moisture.
Metals which are placed higher in the activity series corrode more easily.

Solution 10.

Conditions for increase of corrosion are:

Presence of oxygen and moisture.
Metals which are placed higher in activity series corrode more easily
Dissolved salts in water act as electrolyte and enhance the rate of corrosion.
The presence of pollutants like NO₂and CO₂increases rusting.

Solution 11.

Corrosion of metals is an advantage as it prevents the metal underneath from further damage. For example: On exposure to air, the surface of metal like aluminium and Zinc forms layers of their oxides which are very sticky and impervious in nature and hence act as protective layer. This layer protects the metal from further damage.

Solution 12.

Rusting is the slow oxidation of iron by atmospheric oxygen in the presence of water.
Equation:
4Fe + 3O₂ + 2x H₂O → 2Fe₂O₃.xH₂O

Solution 13.

Two conditions necessary for rusting of iron are:

Air
Water

Solution 14.

By painting an iron object, the iron do not come in contact with atmospheric reagents .This prevents rusting.

Solution 15.

Galvanisation is the process of applying a protective zinc coating to steel or iron, in order to prevent rusting.
The zinc coating does not allow iron to come in contact with air and moisture and thus protects it from rusting.

Solution 16.

Silver gets tarnished when exposed to the atmosphere which contains pollutant H₂S and forms a black coating of Ag₂S.
Copper forms a green deposit on its surface when exposed to moist air. This is usually basic copper (II) sulphate.

Solution 17.

Aluminium forms white colour oxide on exposure to the atmosphere. This white colour oxide prevents it from further corrosion whereas iron reacts with air to form hydrated oxide called rust. So, iron undergoes corrosion to greater extent.

Solution 18.

The noble metals such as gold and platinum do not corrode easily.

Solution 19.

Gold is the most unreactive metal so it does not react with air or water and other gases in atmosphere. So gold does not corrode. That is why gold look new after several years of use.

Exercise 7(C)

Solution 1.

The process used for the extraction of metals in their pure form from their ores is referred to as Metallurgy.
The processes involved in Metallurgy are:

Crushing and Grinding
Concentration
Roasting andcalcination
Reduction
Refining

Solution 2.

(a) A metal which occurs as sulphide is lead.
(b) A metal which occurs as halide is silver.
(c) A metal which occurs as carbonate is zinc.
(d) A metal which occurs as oxide is iron.

Solution 3.

(a) Minerals are naturally occurring compounds of metals which are generally present with other matter such as soil, sand, limestone and rocks. Ores are those minerals from which the metals are extracted commercially at low cost and comfortably. All ores are minerals, but all minerals are not necessarily ores.

(b) Ores are those minerals from which the metals are extracted commercially at low cost and with minimum effort. A metallic compound is a compound that contains one or more metal elements. Examples: AgNO₃ – Silver nitrate is a metallic compound.

Solution 4.

The metals that can be extracted from the following ores are:
(a) Bauxite- Aluminium
(b) Calamine- Zinc
(c) Haematite- Iron

Solution 5.

Three objectives achieved during the roasting of ores is:

It removes moisture from ores.
It makes the ore porous and more reactive.
It expels volatile impurities.
It convertssulphideores into oxides.

Solution 6.

(a) Hydraulic washing: The difference in the densities of the ore and the gangue is the main criterion.
(b) Forth floatation: This process depends on the preferential wettability of the ore with oil and the gangue particles by water.
(c) Electromagnetic separation: Magnetic properties of the ores.

Solution 7.

(a) The processes involved in

(i) Processes involved in concentration are:

Hydrolytic method
Magnetic Separation
Froth floatation
Leaching

(ii) Processes involved in Refining of ores are:

Distillation
Liquation
Oxidation
Electro- refining

(b) Potassium and sodium oxides cannot be reduced by carbon, carbon monoxide and hydrogen.

Solution 8.

(a) Flux: A flux is a substance that is added to the charge in a furnace to remove the gangue.
(b) Gangue: Earthly impurities including silica, mud etc., associated with the ore are called gangue.
(c) Slag: It is the fusible product formed when flux reacts with impurities during the extraction of metals.
(d) Smelting: Smelting is the process of reducing the roasted oxide ore and removing the gangue with the help of an appropriate flux added with the ore.

Solution 9.

Iron and zinc are quite reactive and hence they do not occur in the free state. The compounds of metals found in nature are their oxides, carbonate and sulphides.

Solution 10.

Solution 11.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-11

Solution 12.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-12

Solution 13.

Oxides of highly active metals like potassium, sodium, calcium, magnesium and aluminium have great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide or hydrogen.

Metals in the middle of activity series (iron, zinc, lead, copper) are moderately reactive and are not found in oxide form. These are found in nature as sulphides or carbonate. These are first converted into oxides and can be reduced by C, CO or H₂.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-13
Metals low in the activity series is very less reactive and oxides of these metals are reduced to metals by heating alone.

Solution 14.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-14

Solution 15.

Aluminium has a great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide. So it is extracted from its oxide by electrolysis.
Metals like copper, lead and iron are placed in the middle of the activity series and re moderately reactive and their oxides can be reduced by carbon, CO and hydrogen.
Mercury and silver are less reactive and are placed lower in the reactivity series. The oxides of these metals are reduced to metals by heating their oxides.

Solution 16.

The process used for the concentration of the ore is froth floatation process.

Solution 17.

(a) The purification depends upon:

Nature of metal.
Nature of impurities present in the metal.
Purpose for which metal is to be used

(b) Methods used for purification are:

Distillation
Liquation
Oxidation
Electro-refining

(c) The impure metal is made the anode, while a thin sheet of pure metal is made the cathode. Electrolyte used is a salt solution of a metal which is to be refined. Pure metal deposits at the cathode and impurities settle down forming anode mud.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-17

Solution 18.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-18

Exercise 7(D)

Solution 1.

Position in the Periodic Table: Period 3,Group IIIA(13)

Solution 2.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-2

Solution 3.

Bauxite ore contains approximately 60% aluminium oxide. The rest being sand, ferric oxide and titanium oxide.

Solution 4.

Red mud consists of ferric oxide, sand etc. left after bauxite dissolves in NaOH forming sodium aluminate and is removed by filtration.

Solution 5.

As aluminium has great affinity for oxygen, so it is stable compound. It is not easily reduced by common reducing agents like carbon, carbon monoxide or hydrogen. Hence, electrolytic reduction is chosen as the method for reducing alumina.

Solution 6.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-6

Solution 7.

(a) The process by which refining of aluminium is done is called Hoope’s electrolytic process.
(b) Molten impure aluminium forms the bottom layer. The bottom layer has carbon lining and serves as anode.
Pure molten aluminium with carbon electrodes serves as cathode in top layer.
(c) Reactions at the two electrodes are:
Anode: Al -3e^– → Al³⁺Cathode: Al³⁺ + 3e^– → Al

Solution 8.

Reaction of aluminium:

(a) Air: Aluminium forms oxide at room temperature.
Aluminium powder burns in air at about 800⁰C forming its oxide and nitride with a bright light.
4Al + 3O₂ → 2Al₂O₃2Al + N₂ → 2AlN

(b) Water: Water has no action on aluminium due to layer of oxide on it.
When steam is passed over pure heated aluminium, hydrogen is produced.
2Al + 3H₂O → Al₂O₃+ 3H₂

(c)Acid: It reacts with acids to produce salt and hydrogen.
2Al + 6HCl → 2AlCl₃ + 3H₂Dilute sulphuric acid reacts with metal to liberate hydrogen.
2Al + 3H₂SO₄ (dilute) → Al₂(SO₄)₃ + 3H₂

Concentrated sulphuric acid reacts with aluminium to produce sulphur dioxide.
2Al + 6H₂SO₄ → Al₂(SO₄)₃ + 6H₂O + 3SO₂Dilute and concentrated nitric acid does not attack the metal aluminium.

(d) Base: Aluminium reacts with boiling and dilute alkalies to produce meta aluminate while with fused alkali produce aluminate.
2Al+ 2NaOH +2H₂O → 2NaAlO₂ +3H₂(Sodium meta aluminate)
2Al + 6NaOH → 2NaAlO₃ +3H₂
(Sodium aluminate)

Solution 9.

The role of cryolite in the electrolytic reduction of alumina in Hall’s process is :

Lowers the fusion temperature from 2050⁰C to 950⁰C and enhances conductivity.
Increases its conductivity since pure alumina is almost a non-conductor of electricity.
Cryoliteacts as a solvent for the electrolytic mixture.

Solution 10.

(a) Aluminium is more active metal but it gets oxidized and forms a thin protective layer on its surface which prevents further corrosion.
(b) Aluminium vessels should not be cleaned with powders containing alkalis because it results in the formation of meta aluminates and hydrogen.
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

Solution 11.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-11-1

(c) A layer of aluminium is formed on iron at high temperature during cooking and food becomes deficient in iron.

Solution 12.

(a) In the electrolytic reduction of alumina, the graphite (anode) is oxidized by oxygen to CO and further forms CO₂, so it is consumed and has to be replaced from time to time.
2C + O₂ → 2CO
2CO + O₂ → 2CO₂

(b) Roasting provides oxygen to convert metallic sulphides into metallic oxide and SO₂ which takes place when heated in excess of air.
Carbonate is converted into oxide by loss of CO₂ which takes place in the absence of air and when heated strongly.

(c) Aluminium has a great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide or hydrogen whereas lead oxide can be easily reduced to metal lead by carbon.
PbO + C → Pb + CO

Solution 13.

(a) Flux combines with the gangue to form a fusible mass called slag.
(b) It forms slag[CaSiO₃] with silica.
(c) It is removed from upper outlet, slag being lighter float on molten iron.

Solution 14.

(a) Froth flotation process: Zinc blende[ZnS]
(b) Magnetic Separation: Haematite[Fe₂O₃]

Solution 15.

Electrolytic Reduction

It is removal of oxide or halide from a metal.
Oxides of highly active metals like Na,K,Ca,Mg,Al are reduced by electrolytic reduction of their fused salts.
Oxides of these metals have great affinity for oxygen than carbon and cannot be reduced by carbon or CO or hydrogen.

Electrolytic refining of metals is the separation of residual impurities like Si and phosphorus.

Presence of other metals and non-metals like Si and phosphorus.
Unreduced oxides and sulphides of metals.

It depends upon:

Nature of metal
Purpose for which metal is to be obtained.
Nature of impurities present.
Impure metal is made anode while a thin sheet of pure metal is made cathode and electrolyte used is a salt of solution of a metal to be refined.

Solution 16.

The three ways in which metal zinc differs from the non-metal carbon is:

Zinc has avalency2 and carbon has valency 4.
Zinc does not form hydride but carbon does (CH₄).
Oxides of zinc areamphoteric(ZnO) whereas oxides of carbon are acidic (CO₂) and neutral (CO).

Exercise 7(E)

Solution 1.

(a) To prevent from rusting.
(b) Due to strong electropositive nature, it easily forms Zn⁺² ions.
(c) Antiseptic in face creams.

Solution 2.

(a) Aluminium:

Being a strong, light and corrosion resistant metal, it is used in alloys.
Aluminium is light, it has high tensile strength, is resistant to corrosion, good conductor of heat, unaffected by organic acids and has attractive appearance. So it is used for making cooking utensils, in building and construction work.
Aluminium has a strong affinity for oxygen so it is used as a deoxidizer in the manufacture of steel.

(b) Zinc:

Zinc has a strong electropositive character, so it is used for coating iron and steel sheets to prevent them from rusting and this process is known as galvanization.
Due to strong electropositive nature, it forms Zn⁺²ions, so it is used to make dry cell containers which act as negative electrode.
Zinc act as a reducing agent for many organic reductions and these reductions are employed in manufacturing drugs, dyes.

Solution 3.

(a) Zinc is electropositive metal than iron, gets oxidized and saves iron. Also zinc forms protective layer of ZnO on iron. This layer is sticky and impervious in nature and protects the iron metal underneath from rusting.

(b) A neutral gas other than oxygen which is formed at anode during electrolysis of fused alumina is carbon monoxide.

(c) Nitric acid can be stored in aluminium containers as the dilute and conc. nitric acid does not react with aluminum. It renders aluminium passive due to the formation of an oxide film on its surface.

Solution 4.

(a) Cast iron: It is used in drain pipes, gutter covers, weights and railings.
(b) Wrought iron: It is used in chains, horse shoes and electromagnets.
(c) Mild steel: It is to manufacture nuts, bolts etc.
(d) Hard steel: It is used to make tools.

Solution 5.

(a) Galvanized iron sheets
(b) Zinc
(c) Zinc

Solution 6.

(a) Aluminium being strong, light and corrosion resistant metal is used for making alloy.
(b) Aluminium is light, malleable and does not rust so it is used for wrapping chocolates.
(c) To prevent them from rusting.
(d) It is used in aluminothermy as it is a good reducing agent.
(e) As aluminium forms a film of aluminium oxide, it protects the ships from corrosion. So it is used for making ships.

Solution 7.

(a) A mixture of 3 parts of ferric oxide (Fe₂O₃) and one part of aluminium powder (Al).
(b) A mixture of Potassium chlorate and magnesium powder is the ignition mixture.
(c) Fe₂O₃ + 2Al → Al₂O₃ + 2Fe + heat

Solution 8.

Alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements.
The properties of alloys are often greatly different from those of the components.
For example: Gold is too soft to be used without small percentage of copper.
A low percentage of molybdenum improves the toughness and wear resistance of steel.
Bell metal is more sonorous than copper or tin.
Alnico an alloy of aluminium, nickel and cobalt can lift 60 times its own mass.
These added elements improve hardness, wear resistance, toughness and other properties.

Solution 9.

Alloy’s name	Composition	Uses
1. Stainless steel	73% Fe, 18%Cr,8%Ni,1%C	Used for making utensils, cutlery, ornamental pieces and surgical instruments.
2. Manganese steel	85% Fe,1%C ,14%Mn	Used for making rock drills and armour plates.
3. Tungsten steel	84%Fe, 5%W, 1%C	Used for cutting tools for high speed lathes.

Solution 10.

The other element in Brass is Zinc.
The other elements in Bronze are Tin and Zinc.

Solution 11.

(a) Duralumin
(b) Solder
(c) Brass
(d) Zinc amalgam

Solution 12.

A mixture or an alloy of mercury with a number of metals or an alloy such as sodium, zinc, gold and silver as well as with some non-metals is known as amalgam.
Dental amalgam is a mixture of mercury and a silver tin alloy.

Solution 13.

(a) Two properties of brass that make it more useful than its components are:

It is malleable and ductile.
It resists corrosion.
Can be easily cast.

(b) A metal which forms a liquid alloy at ordinary temperature is sodium.

Solution 14.

Magnalium is an alloy of aluminium with composition 90-95% and magnesium with composition 10-5%. It is used for making aircrafts.

Solution 15.

The constituents of
(a) Duralumin are aluminium (95%), copper (4%), magnesium (0.5%) and manganese (0.5%).
(b) Solder are lead (50%) and tin (50%).
(c) Bronze are copper (80%), tin (18%) and zinc (2%).
(d) Invar are iron (63%), nickel (36%) and carbon (1%).

Miscellaneous Exercise

Solution 1.

(a) Bauxite: Aluminium is extracted from its main ore bauxite Al₂O₃.2H₂O. It contains 60% Al₂O_3.(b) Sodium hydroxide: Sodium hydroxide dissolves bauxite to form sodium meta aluminate, removes insoluble impurities from Al₂O₃ by forming red mud.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-1

Solution 2.

(a) Copper
(b) Iron
(c) Zinc
(d) Magnesium

Solution 3.

Arrangement of metal in decreasing order of reactivity are:
Sodium > Magnesium > Zinc > Iron > copper

Solution 4.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-4
(iii) In electrolytic process, the graphite acts as anode. The anode has to be replaced from time to time as it gets oxidized by the oxygen evolved at the anode.
(iv) The reaction that occurs at cathode is:
4Al³⁺ + 12e^–→ Al
(d) In construction the alloy of aluminium -duralumin is used because it is hard and resistant to corrosion.

Solution 5.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-5
(d) In Aluminium thermite welding, the reduction with aluminium is highly exothermic and heat generated is sufficient to melt the metal.
Fe2O3 + 2Al → 2Fe + Al₂O₃ + Heat

Solution 1 (2005).

(i)

Zinc: Froth Flotation, ZincBlende,Coke
Aluminium: Bauxite,Cryolite, Sodium hydroxide solution

(ii)

Sodium hydroxide.
Cryolite

(iii) The formula of Cryolite is Na₃AlF_6.

Solution 1 (2006).

(a) Mercury
(b) Roasting
(c) CaSiO₃(d) Cryolite
(e) Graphite

Solution 1 (2007).

Acidic oxide(D)
Discharged at anode (F)
Covalent chlorides (I)
5,6,7 valence electrons (L)
Brittle(C)

Solution 1 (2008).

(i) A is made of carbon and B is thick graphite rod.
A → Cathode
B → Anode
(ii) Aluminium is formed at electrode A.
(iii) The two aluminium compound in the electrolyte C is Na₃AlF₆, Al₂O₃.
(iv) It is necessary to continuously replace electrode B from time to time as it gets oxidized by the oxygen evolved.

Solution 2 (2005).

(a) Stainless steel : Iron, Chromium
(b) Brass: Copper , Zinc

Solution 2 (2007).

selina-icse-solutions-class-10-chemistry-metallurgy-mis-2-2007

Solution 2 (2008).

Brass is an alloy of copper and Zinc.

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Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

February 4, 2024

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses Of Ammonium Hydroxide And Sodium Hydroxide. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE Solutions Selina ICSE Solutions

Selina ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

Exercise 1

Solution 1.

(a) Ferrous salts : Light green
(b) Ammonium salts : Colourless
(c) Cupric salts : Blue
(d) Calcium salts : Colourless
(e) Aluminium salts : Colourless

Solution 2.

(a) Cu(OH)₂(b) ZnO
(c) NaOH
(d) NH₄OH
(e) Na⁺, Ca²⁺(f) Fe²⁺, Mn²⁺(g) Aluminium
(h) Zn(OH)₂ and Al(OH)₃(i) PbO
(j) Ammonium ion

Solution 3.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry Uses of Ammonium Hydroxide And Sodium Hydroxide img 1

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 2

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 3

Solution 6.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 4

Solution 7.

(a) ZnCl₂(b) Zn(OH)₂

Solution 8.

(a) PbO
(b) ZnO
(c) K₂ZnO₂

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 5

Solution 10.

When freshly precipitated aluminum hydroxide reacts with caustic soda solution, whitesalt of sodium meta aluminate is obtained.

Solution 11.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 7

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 8
With excess of NaOH, white gelatinous ppt. of Zn (OH)₂ is soluble. So, these two cannot be distinguished by NaOH alone. However white ppt. of Pb(OH)₂is readily soluble in acetic acid also.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 9
(i) On addition of NH₄OH to calcium salts no precipitation of Ca(OH)₂occurs even with addition of excess of NH₄OH because the concentration of OH^–ions from ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.
Pb(NO₃)₂+2 NH₄OHPb(OH)₂+2NH₄NO₃
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 10
Solution 13.

Lead carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is insoluble in excess.
Zinc carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is soluble in excess.

Solution 14.

Reagent bottles A and B can identified by using calcium salts such as Ca(NO₃)₂.

On adding NaOH to Ca (NO₃)₂, Ca (OH)₂ is precipitated as white precipitate which is sparingly soluble in excess of NaOH.
Ca(NO₃)₂+2NaOH → Ca(OH)₂+ 2NaNO₃

Whereas, on addition of NH₄OH to calcium salts, no precipitation of Ca(OH)₂occurs even with addition of excess of NH₄OH because the concentration of OH^–ions from the ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.
So the reagent bottle which gives white precipitate is NaOH and the other is NH₄OH.

Intext Exercise

Solution 1.

(i) Analysis: The determination of chemical components in a given sample is called analysis.
(ii) Qualitative analysis: The analysis which involves the identification of the unknown substances in a given sample is called qualitative analysis.
(iii) Reagent: A reagent is a substance that reacts with another substance.
(iv) Precipitation: It is the process of formation of an insoluble solid when solutions are mixed. The solid thus formed is called precipitate.

Solution 2.

(i) Yellow
(ii) Colourless
(iii) PaleGreen
(iv) Colourless
(v) Colourless

Solution 3.

(i) Fe³⁺(ii) Cu²⁺(iii) Cu⁺²(iv) Mn²⁺

Solution 4.

(i) Ca(OH)₂(ii) Fe(OH)₂ and Cu(OH)₂(iii) Zn(OH)₂ and Pb(OH)₂

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 11

Solution 6.

NH₄OH and NaOH can be distinguished by using calcium salts.

For example on adding NaOH to Ca(NO₃)₂, Ca(OH)₂ is obtained as white precipitate which is sparingly soluble in excess of NaOH.

Ca(NO₃)₂ + 2NaOH → Ca(OH)₂ + 2NaNO₃

On addition of NH₄OH to calcium salts, no precipitation of Ca(OH)₂ occurs even with the addition of excess of NH₄OH.This is because the concentration of OH^– ions from the ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.

Solution 7.

(i) Fe(OH)₂and Pb(OH)₂(ii) Cu(OH)₂andZn(OH)₂

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Hydrogen Chloride

February 4, 2024

Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Hydrogen Chloride

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 8 Study of Compounds Hydrogen Chloride. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 8 Study of Compounds – Hydrogen Chloride

Exercise 1

Solution 1.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-1

Solution 2.

(a) Hydrogen chloride is dried by passing through conc. Sulphuric acid.
(b) Phosphorous pentoxide and CaO cannot be used to dry HCl because they react with HCl.
2P₂O₅+ 3HCl → POCl₃ + 3HPO₃CaO + 2HCl → CaCl₂ + H₂O

Solution 3.

(a) Anhydrous HCl is poor conductor due to the absence of ions in it whereas aqueous HCl is excellent conductor since it contains ions.

(b) When the stopper is opened HCl gas comes in contact with water vapors of air and gives white fumes due to the formation of hydrochloric acid.

(c) A solution of HCl in water gives hydronium ions and conducts electricity, but HCl is also soluble in dry toluene, but in that case it neither (i) turns blue litmus red (ii) nor does conducts electricity. This indicates the absence of H⁺ ions in toluene showing thereby that hydrogen chloride is a covalent compound.

(d) When ammonium hydroxide is brought near the mouth of HCl, dense white fumes are formed due to the formation of ammonium chloride.
HCl + NH₄OH → NH₄Cl + H₂O

(e) Dry hydrogen chloride is not acidic whereas moist Hydrogen chloride is acidic. In presence of a drop of water HCl gas dissolves in water and forms hydrochloric acid which turns blue litmus paper red.

(f) Hydrogen chloride is not collected over water as it is highly soluble in water.

Solution 4.

Difference between Hydrogen chloride gas and Hydrochloric acid is:

Hydrogen chloride gas

Hydrochloric acid

1. Dry hydrogen chloride gas does not turn blue litmus red due to non-acidic character.

2. Hydrogen chloride gas does not conduct electricity.

1. Being acidic it turns blue litmus red.

2. Hydrochloric acid is a good conductor of electricity.

Solution 5.

(a) Hydrochloric acid is prepared by this method.
(b) The reactants are sodium chloride and Sulphuric acid.
(c) The empty flask acts as Anti-Suction device. In case the back suction occurs the water will collect in it and will not reach the generating flask.
(d) The drying agent is Conc. Sulphuric acid. Sulphuric acid is chosen as drying agent because it does not react with HCl.
(e) The Inverted funnel :
Prevents or minimizes back suction of water.
Provides a large surface area for absorption of HCl gas.

Solution 6.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-6

Solution 7.

(a) Chlorine.
The compound formed which is strongly acidic in water, is HCl.
H₂ + Cl₂ → 2HCl

(b) A dilute aqueous solution of hydrochloric acid gets gradually concentrated on distillation, till the concentration of the acid reaches 22.2% HCl by weight which boils at 110⁰C.When this concentration is reached, no further increase in concentration of the acid becomes possible by boiling. This is because vapours evolved before 110⁰C are vapours of water but at temperature above 110⁰C vapours consist mostly of molecules of HCl.

Solution 8.

We can prove that hydrochloric acid contains both hydrogen and chlorine by the following experiment.
Take a voltameter used for electrolysis of water, fitted with platinum cathode and graphite anode.
Into the voltameter pour 4 molar HCl and pass direct current.
It is seen that a colourless gas is evolved at cathode and a greenish gas is evolved at anode.
When a burning splinter is brought near a colourless gas, it bursts into flame thereby proving that it is hydrogen gas.
When moist starch iodide paper is held in the greenish yellow gas, it turns blue black, thereby proving that the gas is chlorine.
2HCl → H₂ + Cl₂This experiment proves that hydrochloric acid contains both hydrogen and chlorine.

Solution 9.

(a) Manganese dioxide
(b) Hydrogen chloride and ammonia
(c) Hydrogen and oxygen
(d) AgCl(Silver chloride)
(e) Aqua regia
(f) Fountain experiment
(g) Hydrogen chloride gas

Solution 10.

(a) An aqueous solution of chlorine is acidic as it dissolves in water to form hydrochloric and hypochlorous acids.
(b) Silver nitrate reacts with hydrochloric acid to form thick curdy white ppt. of silver chloride whereas silver nitrate does not react with nitric acid.
AgNO₃ + HCl → AgCl + HNO₃(White ppt.)

Solution 11.

A is Silver nitrate
B is Hydrochloric acid
C is Silver chloride

Solution 12.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-12

Solution 13.

a. Sodium carbonate on treating withdil.HCl results in the formation of sodium chloride with the liberation of carbon dioxide gas.
Na₂CO₃+ 2HCl → 2NaCl + H₂O + CO₂ ↑
Sodium sulphite on treating with dil.HCl results in the formation of sodium chloride with the liberation of sulphur dioxide gas.
Na₂SO₃+ 2HCl → 2NaCl + H₂O + SO₂ ↑

b. Sodiumthiosulphate reacts with dil. HCl to produce sulphur dioxide gas and precipitates yellow sulphur.
Na₂S₂O₃ + 2HCl → 2NaCl + H₂O + SO₂ + S↓
Sulphur is not precipitated when sulphites are treated with dil.HCl.

Solution 14.

Three tests are:

HCl gas gives thick white fumes of ammonium chloride when glass rod dipped in ammonia solution is held near the vapours of the acid.
NH₃ + HCl NH₄Cl
With silver nitrate HCl gives white precipitate of silver chloride. The precipitate is insoluble in nitric acid but soluble in ammonium hydroxide.
AgNO₃ + HCl AgCl + HNO₃
A greenish yellow gas is liberated when concentrated hydrochloric acid is heated with oxidizing agent like manganese dioxide.
MnO₂ + 4HCl MnCl₂ +2H₂O + Cl₂

Solution 15.

MnO₂, PbO₂ and red lead react with conc. HCl acid to liberate Cl₂. This shows that hydrochloric acid is oxidized to chlorine by oxidizing agents.

Solution 16.

HCl dissolves both in water and toluene, when HCl dissolves in water it ionizes and forms hydronium and chloride ions. Whereas this ionization is not observed in toluene hence a solution of HCl in water can be used as an electrolyte.

Solution 17.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-17

Solution 18.

A mixture having three parts of conc. Hydrochloric acid and one part of conc. Nitric acid is called aqua-regia.
Nitric acid acts as oxidizing agent.

Solution 19.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-19

Solution 20.

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-20

Solution 1 (2004).

Solution 1 (2005).

(a) (i) CuO +2HCl CuCl₂ + H₂O
(ii) MnO₂+ 4HCl MnCl₂ +2H₂O +Cl₂

(b) (i) The experiment is called Fountain Experiment.
(ii) This experiment shows that hydrogen chloride is highly soluble in water.
(iii) Red

Solution 1 (2007).

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-1-2007

Solution 1 (2008).

When hydrogen chloride is collected by downward delivery or upward displacement, it shows that it is heavier than air.

Solution 2 (2008).

Hydrogen chloride is not collected over water as it is soluble in water.

Solution 3 (2008).

selina-icse-solutions-class-10-chemistry-study-compounds-hydrogen-chloride-3-2008

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Selina Concise Physics Class 10 ICSE Solutions Radioactivity

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Selina Concise Physics Class 10 ICSE Solutions Radioactivity

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 12 Radioactivity. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 12 Radioactivity

Exercise 12(A)

Solution 1.

Three constituent of an atom are:

Electrons: mass is 9.1 x 10^-31kg, charge is -1.6 x 10^-19C
Neutron: mass is 1.6749 x 10^-27 kg, charge is zero.
Protons: mass is 1.6726 x 10^-27 kg, charge is +1.6 x 10^-19 C

Solution 2.

Atomic number – the number of protons in the nucleus is called atomic number.
Mass number – the total number of nucleons in the nucleus is called mass number.

Solution 3.

The nucleus at the centre of atom, whose size is of the order of 10^-15 m to 10^-14 m.
The size of a nucleus is 10^-5 to 10^-4 times the size of an atom. It consists of protons and neutrons.
If Z is the atomic number and A is the mass number of an atom, then the atom contains Z number of electrons; Z number of protons and A – Z number of neutrons.
The atom is specified by the symbol _ZX^Awhere X is the chemical symbol for the element.

Solution 4.

Atomic number Z = 11
Mass number A = 23
Number of neutrons A – Z = 12

Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 1

Solution 5.

Isotopes: the atoms of the same element which have the same atomic number Z but differ in their mass number A are called isotopes.

Solution 6.

Isobars: the atoms of different elements which have the same mass number A, but differ in their atomic number Z are called isobars.

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 4

Solution 8.

Radioactivity: Radioactivity is a nuclear phenomenon. It is the process of spontaneous emission of α or β and γ radiations from the nuclei of atoms during their decay.
Example: uranium, radium.

Solution 9.

There will be no change in the nature of radioactivity. This is because radioactivity is a nuclear phenomenon.

Solution 10.

(a) Three types of radiations: Alpha, beta and gamma.
(b) Alpha and beta radiations
(c) Gamma radiations
(d) Gamma radiations
(e) Alpha radiations
(f) Beta radiations

Solution 11.

(a) Gamma radiations have zero mass.
(b) Gamma radiations have the lowest ionizing power.
(c) Alpha particles have lowest penetrating power.
(d) Alpha particle has positive charge equal to 3.2 x 10^-19C and rest mass equal to 4 times the mass of proton i.e. 6.68 x 10^-27 kg.
(e) The gas is Helium.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 5
(f) These radiations come from nucleus of the atom.

Solution 12.

Radiations labeled A, B and C are α, β and γ respectively.
Radiation labeled A is gamma radiation because they have no charge and hence under action of magnetic field they go undeflected.
Radiation B is alpha radiation because its mass is large and it would be deflected less in comparison to beta radiation. The direction of deflection is given by Fleming’s left hand rule. Also directions of deflection of alpha and beta radiations are opposite as they have opposite charge.

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 6

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 7
(b) The radioactive substances are kept in thick lead containers with a very narrow opening, so as to stop radiations coming out from other directions because they may cause biological damage.

Solution 15.

This is because alpha and beta particles are charged particles, but gamma rays are neutral particles.

Solution 16.

No, it is not possible to deflect gamma radiation in a way similar to alpha and beta particles, using the electric or magnetic field because they are neutral and hence do not deflected under the action of electric or magnetic field.

Solution 17.

Property	α-particle	β-particle	γ-particle
Nature	Stream of positively charged particles, i.e. helium nuclei.	Stream of negatively charged particles, i.e. energetic electrons.	Highly energetic electromagnetic radiation.
Charge	Positive charge (Two times that of a proton) = + 3.2 x 10^-19 C (or +2e)	Negative charge = – 1.6 x 10^-19 C (or -e)	No charge
Mass	Four times the mass of proton i.e., 6.68 x 10^-27 kg	Equal to the mass of electron, i.e. 9.1 x 10^-31 kg	No mass (Rest mass is zero)
Effect of electric field	Less deflected	More deflected than alpha particles but in direction opposite to those of α particles	Unaffected

Solution 18.

Ionizing power of alpha radiation is maximum i.e., 10000 times of gamma radiation while beta particles have lesser ionizing power i.e., 100 times of gamma radiation and gamma radiation have least ionizing power.
Penetration power is least for alpha particle and maximum for gamma radiation.

Solution 19.

Speed of α radiation is nearly 10⁷ m/s.
Speed of β radiation is about 90% of the speed of light or 2.7 x 10⁸ m/s.
Speed of γ radiation is 3 x 10⁸ m/s in vacuum.

Solution 20.

Alpha radiations are composed two protons and two neutrons.
Beta particles are fast moving electrons.
Gamma radiations are photons or electromagnetic waves like X rays.
Alpha radiations have the least penetrating power.

Solution 21.

Gamma radiation are produced when a nucleus is in a state of excitation (i.e., it has an excess of energy). This extra energy is released in the form of gamma radiation.
Gamma radiations like light are not deflected by the electric and magnetic field.
Gamma radiations have the same speed as that of light.

Solution 22.

It will become singly ionized helium He⁺.

Solution 23.

Any physical changes (such as change in pressure and temperature) or chemical changes (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) do not alter the rate of decay of the radioactive substance. This clearly shows that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. The radioactivity should therefore be the property of the nucleus. Thus radioactivity is a nuclear phenomenon.

Solution 24.

On emitting a β particle, the number of nucleons in the nucleus (i.e. protons and neutrons) remains same, but the number of neutrons is decreased by one and the number of protons is increased by one.

If a radioactive nucleus P with mass number A and atomic number Z emits a beta particle to form a daughter nucleus Q with mass number A and atomic number Z+1, then the change can be represented as follows:

(a) Atomic number ‘Z’ is not conserved. It is increased by 1.
(b) Mass number A is conserved.

Solution 25.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 9

Solution 26.

(a) Atomic number decreases by 2.
(b) Atomic number increases by 1.
(c) Atomic number does not change.

Solution 27.

(a) After emitting an alpha particle the daughter element occupies two places to the left of the parent element in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of α-decay, then the α-decay can be represented as:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 10
Thus, the resulting nucleus has an atomic number equal to (Z-2). Hence, it shifts two places to the left of the parent element in the periodic table.

(b) After emitting a -particle, the daughter element occupies one place to the right of the parent element in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of β-decay, then the β-decay can be represented as:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 11
Thus, the resulting nucleus has an atomic number equal to (Z+1). Hence, it shifts one place to the right of the parent element in the periodic table.

(c) After emitting -radiation, the element occupies the same position in the periodic table.
Reason: If a parent nucleus X becomes a new daughter nucleus Y as a result of γ-decay, then the γ-decay can be represented as:
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 12
Thus, the resulting nucleus has atomic number equal to Z. Hence, it occupies the same position as the parent element in the periodic table.

Solution 28.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 13

Solution 29.

(a) The composition of B – 82 protons and 126 neutrons.
(b) The composition of C – 83 protons and 125 neutrons.
(c) The mass number of nucleus A = no. of protons +no. of neurons = 84+128=212.
(d) Their will be no change in the composition of nucleus C.

Solution 30.

(a) The alpha particle was emitted.
(b) This is because the atomic number has decreased by 2 and mass number has decreased by 4.

Solution 31.

(a) This is allowed.
(b) This is not allowed because mass number is not conserved.

Solution 32.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 15

Solution 33.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 16

Solution 34.

The atomic number of P decreases by 2 and mass no. decreases by 4 due to the emission of one alpha particle and then increases by 1 due to the emission of each beta particle, so the atomic number of Q formed after the emission of one alpha and two beta particles is same as that of P. Hence P and Q are the isotopes.

Solution 36.

(a) The mass number (A) of an element is not changed when it emits beta and gamma radiations.
(b) The atomic number of a radioactive element is not changed when it emits gamma radiations.
(c) During the emission of a beta particle, the mass number remains same.

Solution 37.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 17

Solution 38.

Radio isotopes: The isotopes of some elements with atomic number Z
Example: carbon (Z=6, A=14).
Radio isotopes are used in medical and scientific and industrial fields. Radio isotopes such as ₉₂U²³²are used as fuel for atomic energy reactors.

Solution 39.

Because they cannot penetrate the human skin.

Solution 40.

Gamma radiations have very high penetration power and can easily pass through the human body. Therefore they are used as radioactive tracers in medical science.

Solution 41.

When the number of neutrons exceeds much than the number of protons in a nuclei, it become unstable or radioactive.

Solution 42.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 18

Solution 43.

Many diseases such as leukemia, cancer, etc., are cured by radiation therapy. Radiations from cobalt -60 are used to treat cancer by killing the cells in the malignant tumor of the patient.
The salt of weak radioactive isotopes such as radio-sodium chloride, radio-iron and radio-iodine are used for diagnosis. Such radio isotopes are called the tracers.

Solution 44.

a < β < γ

An α-particle rapidly loses its energy as it moves through a medium and therefore its penetrating power is quite small. It can penetrate only through 3 – 8 cm in air. It can easily be stopped by a thin card sheet or a thick paper.

The penetrating power of β-particles is more than that of the α-particles. They can pass through nearly 5 m in air, through thin card sheet, and even through thin aluminium foil, but a 5 mm thick aluminium sheet can stop them.

Whereas, the penetrating power of γ-rays is high. It is about 10⁴ times that of α-particles and 10² times that of β-particles. They can pass through 500 m in air or through 30 cm thick sheet of iron. Thick sheet of lead is required to stop them.

Solution 45.

Two main sources of nuclear radiations are:

Radioactive fallout from nuclear plants and other sources.
Disposal of nuclear waste.

These radiations are harmful because when these radiations falls on the human body, they kill the human living tissues and cause radiation burns.

Solution 46.

The following safety measures must be taken in a nuclear power plant:

The nuclear reactor must be shielded with lead and steel walls so as to stop radiations from escaping out to the environment during its normal operation.
The nuclear reactor must be housed in an airtightbuilding of strong concrete structure which can withstand earthquakes, fires and explosion.
There must be back up cooling system for the reactor core, so that in case of failure of one system, the other cooling system could take its place and the core is saved from overheating and melting.

Solution 47.

The radioactive material after its use is known as nuclear waste.
It must be buried in the specially constructed deep underground stores made quite far from the populated area.

Solution 48.

Three safety precautions that we would take while handling the radioactive substances are:

Put on special lead lined aprons and lead gloves.
Handle the radioactive materials with long lead tongs.
Keep the radioactive substances in thick lead containers with a very narrow opening, so as to stop radiations coming out from other directions.

Solution 49.

Radioactive substance should not be touched by hands because these radiations are harmful; when radiation falls on the human body, they kill the human living tissues and cause radiation burns.

Solution 50.

Background radiation: These are the radioactive radiations to which we all are exposed even in the absence of an actual visible radioactive source.
There are two sources of background radiation:

Internal source: potassium, carbon and radium are present inside our body.
External sources: cosmic rays, naturally occurring radioactive elements such as radon-222 and solar radiation.

It is not possible for us to keep ourselves away from the background radiations.

Solution 1 (MCQ).

α or β
Hint: In a single radioactive decay, α and β particles are never emitted simultaneously. There will be either an α-emission or a α β-emission, which may be accompanied by γ emission.

Solution 2 (MCQ).

The nucleus of the atom.
Hint: Radioactivity is a nuclear phenomenon. Hence, electrons come out from the nucleus. Electron is created as a result of decay of one neutron into a proton inside the nucleus and it is not possible for the electron to stay inside the nucleus; thus, it is spontaneously emitted.

Solution 3 (MCQ).

(a) α – particles

An α – particle rapidly loses its energy as it moves through a medium and therefore its penetrating power is quite small. It can penetrate only through 3 – 8 cm in air. It can easily be stopped by a thin card sheet or a thick paper.

Solution 4 (MCQ).

(b) β-particles

β-particles are negatively charged, so they get deflected by the electric and magnetic fields. The deflection of β-particle is more than that of a-particle since a β-particle is lighter than the α-particle. Whereas, gamma radiations are not deflected by the electric and magnetic fields since they are not charged particles.

Exercise 12(B)

Solution 1.
Energy released by combining of nuclei of an atom or by decay of an unstable radioactive nucleus during a nuclear reaction i.e., during fusion or fission is known as nuclear energy.

Solution 1 (MCQ).
(d) neutron
A neutron is used in nuclear fission for bombardment.

Solution 1 (Num).
1 a.m.u. = 1.66 × 10^-27 kg
→ 0.2 a.m.u. = 0.2 × 1.66 × 10^-27 kg
Δm = 0.332 Δ 10^-27 kg
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 19

Solution 2.
Einstein’s mass-energy equivalence relation : E = Δmc²Where E is the energy released due to the loss in the mass Δm and c is the speed of light.

Solution 2 (MCQ).
(d) 10⁷K
To make the fusion possible, a high temperature of approximately 10⁷ K and high pressure is required.

Solution 2 (Num).
Given that Δm = 0.0265 a.m.u.
1 a.m.u. liberates 931.5 MeV of energy. Thus, energy liberated equivalent to 0.0265 a.m.u. is
= 0.0265 a.m.u. × 931.5 MeV
= 24.7 meV

Solution 3.
(a) The mass of atomic particles is expressed in terms of atomic mass unit (a.m.u.). 1 a.m.u. of mass is equivalent to 931 MeV of energy.
(b) Mass of proton = 1.00727 a.m.u.
Mass of neutron = 1.00865 a.m.u.
Mass of electron = 0.00055 a.m.u.

Solution 4.
Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.

Solution 5.
(a) \(_{ 92 }^{ 235 }{ U }\) and \(_{ 92 }^{ 235 }{ U }\)
(b) Experimentally it is found that isotope of \(_{ 92 }^{ 235 }{ U }\) is more easily fissionable because the fission of
is \(_{ 92 }^{ 235 }{ U }\) possible by sloe neutron unlike \(_{ 92 }^{ 238 }{ U }\) where fission is possible only by the fast neutrons.
(c) Slow and fast both.

Solution 6.
Nearly 190 MeV of energy is released due to fission of one nucleus of \(_{ 92 }^{ 235 }{ U }\) The cause of emission of this energy is the loss in mass i.e., the sum of masses of product nuclei is less than the sum of mass of the parent nucleus and neutron.

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 21

Solution 8.
A chain reaction is a series of nuclear fissions whereby the neutrons produced in each fission cause additional fissions, releasing enormous amount of energy.
It is controlled by absorbing some of the neutrons emitted in the fission process by means of moderators like graphite, heavy water, etc. then the energy obtained in fission can be utilized for the constructive purposes

Solution 9.
(i) It is used in a nuclear bomb.
(ii) It is used in a nuclear reactor where the rate of release of energy is slow and controlled which is used to generate electric power.

Solution 10.

Radioactive decay	Nuclear Fission
It is a self process.	It does not occur by itself. Neutrons are bombarded on a heavy nucleus.
The nucleus emits either the a or b particles with the emission of energy in form of g rays which is not very large.	A tremendous amount of energy is released when a heavy nucleus is bombarded with neutrons and the nucleus splits in two nearly equal fragments.
The rate of radioactive decay cannot be controlled.	The rate of nuclear fission can be controlled.

Solution 11.
Nuclear fission is the process in which a heavy nucleus is splits into two light nuclei nearly of the same size by bombarding it with slow neutrons.
When uranium with Z = 92 is bombarded with neutron, it splits into two fragments namely barium (Z = 56) and krypton (Z = 36) and a large amount of energy is released which appears due to decrease in the mass.

Nuclear fusion is also known as thermo-nuclear reaction. This is because nuclear fusion takes place at very high temperature.

Solution 12.
When two nuclei approach each other, due to their positive charge, the electrostatic force of repulsion between them becomes too strong that they do not fuse. Thus, nuclear fusion is not possible at ordinary temperature and ordinary pressure.
Hence to make the fusion possible, a high temperature of approximately 10⁷ K and high pressure is required. At such a high temperature, due to thermal agitations both nuclei acquire sufficient kinetic energy so as to overcome the force of repulsion between them when they approach each other, and so they get fused.

Solution 13.

Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 23
(b) In all three deuterium nuclei fuse to form a helium nucleus with a release of 21·6 MeV energy.
(c) When two deuterium nuclei (\(_{ 1 }^{ 2 }{ H }\)) fuse, nucleus of helium isotope (\(_{ 3 }^{ 2 }{ He }\) ) is formed and 3·3 MeVenergy is released. This helium isotope again gets fused with one deuterium nucleus to form a helium nucleus (\(_{ 4 }^{ 2 }{ He }\)
) and 18·3 MeV of energy is released in this process.

Solution 14.
(a)
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 24

Solution 15.
(a) Nuclear fusion
(b) Nuclear fission

Solution 16.
Both fission and fusion create release of neutrons and large amount of energy.
Nuclear fission: A heavy nucleus splits in two nearly equal light fragments when bombarded with neutrons. It is possible at very ordinary temperature and pressureNuclear fusion: Two light nuclei combine to form a heavy nucleus at very high temperature and high pressure. Possible only at a very high temperature (≈10⁷ K) and a very high pressure.

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Radioactivity img 25

Solution 18.
The source of energy in the Sun and stars is the nucleus fusion of light nuclei such as hydrogen present in them in their inner part. This takes place at a very high temperature and high pressure due to which helium nucleus is formed with the release of high amount of energy.

Solution 19.
(a) Nuclear fission
(b) Nuclear fusion

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Study of Compounds – Ammonia

February 4, 2024

Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Study of Compounds – Ammonia

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 9 Study of Compounds Ammonia. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 9 Study of Compounds – Ammonia

Exercise Intext 1

Solution 1.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-1

Solution 2.

The different forms of ammonia:

Gaseous ammonia(dry ammonia gas)
Liquid ammonia
Liquor ammonia fortis
Laboratory bench reagent

Solution 3.

Formula of liquid ammonia is: NH₃.
Liquid ammonia is liquefied ammonia and is basic in nature. It dissolves in water to give ammonium hydroxide which ionizes to give hydroxyl ions.
selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-3

Solution 4.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-4

Solution 5.

(a) Lab preparation of ammonia:
2NH₄Cl + Ca(OH)₂ → CaCl₂ +2H₂O +2NH₃(b) The ammonia gas is dried by passing through a drying tower containing lumps of quicklime (CaO).
(c) Ammonia is highly soluble in water and therefore it cannot be collected over water.

Solution 6.

The drying agent used is CaO in case of ammonia.
Other drying agents like P₂O₅ and CaCl₂ are not used. As ammonia being basic reacts with them.
6NH₃ + P₂O₅ + 3H₂O → 2(NH₄)₃PO₄CaCl₂ +4NH₃ → CaCl₂.4NH₃

Solution 7.

The substance A is Ammonium chloride and ‘B’ is Ammonia.
Reaction:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃

Solution 8.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-8

Solution 9.

(a) Ammonium compounds being highly soluble in water do not occur as minerals.
(b) Ammonium nitrate is not used in the preparation of ammonia as it is explosive in nature and it decomposes forming nitrous oxide and water vapours.
(c) Conc. H₂SO₄ is not used to dry ammonia, as ammonia being basic reacts with them.
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Solution 10.

Preparation of Aqueous Ammonia: An aqueous solution of ammonia is prepared by dissolving ammonia in water. The rate of dissolution of ammonia in water is very high; therefore, back suction of water is possible. To avoid this, a funnel is attached to the outer end of the delivery tube with rubber tubing.

Procedure: Water is taken in a container and only a small portion of the mouth of funnel is dipped in water.

As ammonia dissolves in water at a higher rate than its production in the flask, the pressure in the funnel above water level decreases for a moment and water rushes into the funnel. As a result, the rim of the funnel loses its contact with water. Since, ammonia produced pushes the water down, the funnel comes in contact with water again. In this way, ammonia dissolves in water without back suction of water.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-10

Exercise 1

Solution 1.

Physical properties of ammonia are:

Color: Colourless
Odour: Strong, Pungent chocking smell
Taste: Slightly bitter (alkaline ) taste
Physiological action: Non-Poisonous
Density: V.D = 8.5 Lighter than air
Nature: Alkaline
Liquefaction: easily liquefied at 10^oC by compressing it at 6 atm. Pressure
Boiling Point: Liquid ammonia boils at -33.5^oC
Freezing Point: Solid NH₃ melts at -77.7^oC
Solubility: Highly soluble in water, 1vol of water dissolves about 702 vols. of ammonia at 20^oC and 1 atm. pressure.
Reaction:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂ + 2NH₃

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-1

Solution 2.

Ammonia is less dense than air. By Fountain Experiment we demonstrate the high solubility of ammonia gas in water.
Balanced equation for the reaction between ammonia and sulphuric acid is:
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Solution 3.

(a) Ammonia is basic in nature.
(b) Copper oxide because CuO is less reactive can be reduced by C, CO or by hydrogen whereas Al₂O₃, Na₂O, MgO are reduced by electrolysis.

Solution 4.

(a) The formula of the compound is Mg₃N_2.(b) Balanced equation :
Mg₃N₂ + 6 H₂O → 3Mg(OH)₂ + 2 NH₃(c) Ammonia is a reducing agent and reduces less active metal oxide to its respective metal.

Solution 5.

Reducing property.

Solution 6.

When a piece of moist red litmus paper is placed in a gas jar of ammonia it turns blue.

Solution 7.

(a) The gas is ammonia.
(b) The formula is NH_3.(c) Uses of ammonia:

It is used in the industrial preparation of nitric acid by Ostwald process.
It is used in the manufacture of fertilizers such as ammonium sulphate, ammonium nitrate, ammonium phosphate.
It is used in the manufacture sodium carbonate by Solvay process.
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ + NH₄Cl

Solution 8.

Equation:
CuSO₄ + 2NH₄OH → Cu(OH)₂↓ + NH₄]₂SO₄pale blue

Ammonia solution in water gives a blue precipitate when it combines with a solution of copper salt.

The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue(deep blue)soluble complex salt.
Cu(OH)₂ +(NH₄)₂SO₄ +2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O

Solution 9.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-9

Solution 10.

(a) Liquid ammonia takes a lot of energy to vaporize .This heat is taken from the surrounding bodies which are consequently cooled down. Thus it is used as a refrigerant in ice plant.
(b) Ammonia emulsifies or dissolves fats, grease so it is used to remove grease from woolen clothes.
(c) Aqueous solution of ammonia gives pungent smell because of the presence of ammonia.
(d) Aqueous ammonia when dissolved in water breaks into ions which help in the conductance of electricity.

Solution 11.

(a) AlN + 3H₂O → Al(OH)₃ +NH₃(b) 2NH₃ + 3PbO → 3Pb + 3H₂O + N₂(c) 8NH₃ +3Cl₂→ N₂ + 6NH₄Cl
(d) 2NH₃ + CO₂ → NH₂CONH₂ + H₂O
(i) Ammonia act as reducing agent is explained by equation (c).
(ii) Urea the nitrogenous fertilizer is prepared from equation (d).

Solution 12.

(a) A Dirty green precipitate of Fe(OH)₂ is obtained when ammonium hydroxide is added to ferrous sulphate.
(b) Liquid ammonia is liquefied ammonia.
(c) Finely divided Iron is used in Haber process.
(d) Quicklime is a drying agent for NH₃.
(e) Ammonium salts when heated with caustic alkali.

Solution 13.

(a) Dirty green ppt. of Ferrous hydroxide is formed which is insoluble in excess of NH₄OH.
FeSO₄+ 2NH₄OH → [NH₄]₂SO₄ + Fe(OH)₂

(b) Reddish brown ppt. of ferric hydroxide is formed which is insoluble in ammonium hydroxide.
FeCl₃ + 3NH₄OH → 3NH₄Cl + Fe(OH)₃

(d) White gelatinous ppt. of Zinc hydroxide is formed which is soluble in NH₄OH.
Zn(NO₃)₂ + 2NH₄OH → 2NH₄NO₃ + Zn(OH)₂

Solution 14.

When correct amount of ammonium hydroxide is added drop wise to solutions of the metallic salts, ppts. (coloured generally) are formed. They help us to identify their metal ions.
Two equations:
FeSO₄ +2NH₄OH → (NH₄)₂SO₄ + Fe (OH)₂(Green) (Dirty green)
shows the presence of Fe⁺² ion.
FeCl₃ + 3NH₄OH → 3NH₄Cl + Fe (OH)₃(Brown) (Reddish brown)
shows the presence of Fe⁺³ ion.

Solution 15.

NH₄Cl on strong heating sublimes to form dense white fumes which condense to white powdery mass on cooler parts of the tube whereas no white fumes on heating NaCl.

(b) When ammonium hydroxide is added drop wise to solution to be tested.
Ferrous salt gives dirty green ppt.
Ferric salt gives reddish brown ppt of their hydroxides.

Solution 16.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-16

Solution 17.

(a) In the presence of Platinum at 800^oC, ammonia reacts with oxygen to give nitric oxide and water vapour.
Procedure:
Pass dry ammonia gas and oxygen gas through inlets over heated platinum placed in the combustion tube, which in the heated state emits reddish glow.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-17
Observation:
Reddish brown vapours of nitrogen dioxide are seen in the flask due to the oxidation of nitric oxide.
The platinum continues to glow even after the heating is discontinued since the catalytic oxidation of ammonia is exothermic.

(b) Two reactions to show reducing property of ammonia are:
8NH₃ +3Cl₂→ N₂ + 6NH₄Cl
2NH₃ +3CuO → 3Cu + 3H₂O +N₂

Solution 18.

(i) Neutralization
NH₃ + HCl → NH₄Cl

(ii) Thermal dissociation
NH₄Cl → NH₃ + HCl

(iii) Ammonia
NH₄Cl + NaOH → NH₃ + NaCl + H₂O

Solution 19.

(a) Ammonia
(b) Hydrogen chloride and chlorine gas.

(d) Acidic gas: HCl
Basic gas: Ammonia
Neutral gas: NH₄Cl

(e) Silver chloride
(f) Nitrogen
(g) Magnesium nitride
(h) Lead oxide
(i) Ammonium chloride

Solution 20.

CuSO₄ + 2NH₄OH → (NH₄)₂SO₄ + Cu(OH)₂ [Pale blue]
The cation present in solution B is Copper (Cu⁺²).
The colour of solution B is Blue.
The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue(deep blue) soluble complex salt.
Cu(OH)₂ + (NH₄)₂SO₄ +2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O

Solution 21.

Three ways in which ammonia gas can be identified is:

It has a sharp characteristic odour.
When a glass rod dipped in HCl is brought in contact with the gas white colour fumes of ammonium chloride are formed.
It turns moist red litmus blue, moist turmeric paper brown and phenolphthalein solution pink.

Solution 22.

(a) Mg₃N₂ + 6H₂O 3Mg(OH)₂ + 2NH₃(b) 2NH₃ + 3CuO 3Cu + 3H₂O + N₂Ammonia acts as a reducing agent. It reduces metallic oxide to give metals, water vapour and nitrogen.
(c) 8NH₃ +3Cl₂ N₂ + 6NH₄Cl
(d) 4 NH₃ +5O₂ 6H₂O + 4NO +Heat
Ostwald process starts with the catalytic oxidation of ammonia to manufacture nitric acid in the presence of catalyst platinum.

Solution 23.

As the ‘A’ turns red litmus blue it is a base. Now the gas ‘A’ combines with ‘B’ in presence of Catalyst to give colourless gas Nitrogen monoxide. It reacts with oxygen to give brown gas which is Nitrogen dioxide.
A= NH₃B= O₂C=NO
D=NO₂

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-23

Solution 24.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-24

Solution 25.

(a) The main refrigerants used are Freon chlorofluorocarbons (CFC). They deplete ozone layer. The chlorofluorocarbons are decomposed by ultraviolet rays to highly reactive chlorine which is produced in the atomic form. selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-25
This causes depletion of ozone layer and chlorine monoxide so formed reacts with atomic oxygen and produces more chlorine free radicals.
ClO + O → Cl + O₂Again this free radical destroys ozone and the process continues thereby giving rise to ozone depletion.
(b) Liquid ammonia can be used as a refrigerant, as an alternative for chlorofluorocarbons.
(c) Advantages of ammonia as refrigerant:

Ammonia is environmentally compatible. It does not deplete ozone layer and does not contribute towards global warming.
It has superior thermodynamic qualities as a result ammonia refrigeration systems use less electricity.

Ammonia has a recognizable odour and so leaks are not likely to escape.

Solution 26.

Disadvantages of ammonia as a refrigerant are as follows:

It is not compatible with copper, so it cannot be used in any system with copper pipes.
It is poisonous in high concentration although it is easily detectable due to its peculiar smell and since it is less dense than air it goes up in the atmosphere not affecting the life too much on earth.

Solution 27.

(a) Explosive: ammonium nitrate
(b) Medicine: ammonium carbonate
(c) Fertilizers: ammonium sulphate
(d) Laboratory reagent: ammonia solution

Solution 28.

(a) Dry air free from carbon dioxide and dry ammonia from Habers process.
(b) The catalyst used in the process is Platinum.
(c) The oxidizing agent used in the process is oxygen.
(d) Ratio of ammonia and air is 1:10.
(e) Quartz is acid resistant and when packed in layers help in dissolving nitrogen dioxide uniformly in water.

Solution 29.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-29

Solution 30.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-30

Solution 1 (2003).

(a) Mg₃N₂ +6H₂O → 3Mg(OH)₂ + 2NH₃(b) Ammonia gas is collected in inverted gas jars by the downward displacement of air.
(c) Ammonia is not collected over water because it is highly soluble in water.
(d) Quicklime is used as a drying agent for ammonia.

Solution 1 (2004).

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-1-2004

Solution 1 (2005).

(a) It is the basic nature of ammonia molecule.
(b) Hydroxyl ion (NH₃+ H₂O → NH₄⁺ + OH^–)
(c) The red litmus paper turns blue in the solution.

Solution 1 (2006).

Pb(NO₃)₂+ NH₄OH →2NH₄NO₃+Pb(OH)₂The chalky white ppt. of lead hydroxide is formed.

Solution 1 (2007).

(a) HCl gas is more dense [V.D.=18.25,V.D. of ammonia =8.5] and it is collected by the upward displacement of air.
(b) NH₃ + HCl → NH₄Cl

Solution 2 (2005).

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-2-2005

Solution 2 (2007).

Balanced equation:

(a) 2NH₃ + 3CuO → 3Cu + 3H₂O + N₂(b) 2NH₃ + 3Cl₂ → N₂ + 6HCl

Solution 2 (2008).

Magnesium Nitride

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Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts

February 4, 2024

Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 3 Acids, Bases and Salts. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE Solutions Selina ICSE Solutions

Selina ICSE Solutions for Class 10 Chemistry Chapter 3 Acids, Bases and Salts

Exercise Intext 1

Solution 1.
(a) Acids are defined as compounds which contain one or more hydrogen atoms, and when dissolved in water, they produce hydronium ions (H₃O⁺), the only positively charged ions.
(b) Hydronium ion
(c) H₃O⁺

Solution 2.
H2SO4 + H₂O ⇌ H₃O⁺ + HSO4-
HSO4- + H₂O ⇌ H₃O⁺ + SO4-2

Solution 3.
If water is added to a concentrated acid, the heat generated causes the mixture to splash out and cause severe burns. Thus, water is never added to acid in order to dilute it.

Solution 4.
Basicity: The basicity of an acid is defined as the number of hydronium ions (H₃O+) that can be produced by the ionization of one molecule of that acid in aqueous solution.
The basicity of following compounds are:
Nitric acid: Basicity= 1
Sulphuric acid: Basicity=2
Phosphoric acid: Basicity=3

Solution 5.
(a) Oxyacids: – HNO₃, H₂SO₄(b) Hydracid:- HCl, HBr
(c) Tribasic acid:- H₃PO₄, H₃PO₃(d) Dibasic acid: – H₂SO₄ , H₂CO₃

Solution 6.
(a) The anhydride of following acids are:

Sulphurous acid: SO₂
Nitric acid: N₂O₅
Phosphoric acid: P₂O₅
Carbonic acid : CO₂

(b) Acids present in following are:
Vinegar: Acetic acid
Grapes: Tartaric acid and Malic acid
Lemon: Citric acid
(c)

H⁺ ion turns blue litmus red.
OH^– ion turns red litmus blue.

Solution 7.
Acetic acid is a monobasic acid which on ionization in water produce one hydronium ion per molecule of the acid.

Solution 8.
2NO₂_(g) + H₂O_(l)→ HNO_2(aq) + HNO₃

Solution 9.
The strength of an acid is the extent to which the acid ionizes or dissociates in water. The strength of an acid depends on the degree of ionization and concentration of hydronium ions [H₃O⁺] produced by that acid in aqueous solution.

Solution 10.
(a) Carbonic acid is a dibasic acid with two replaceable hydrogen ions; therefore it forms one acid salt or one normal salt. Hydrochloric acid is a monobasic acid with one replaceable hydrogen ion and so forms only one normal salt.
(b) Strength of an acid is the measure of concentration of hydronium ions it produces in its aqueous solution. Dil. HCl produces high concentration of hydronium ion compared to that of concentrated acetic acid. Thus, dil. HCl is stronger acid than highly concentrated acetic acid.
(c) H₃PO₃ is not a tribasic acid because in oxyacids of phosphorus, hydrogen atoms which are attached to oxygen atoms are replaceable. Hydrogen atoms directly bonded to phosphorus atoms are not replaceable.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 1
(d) The salt produced is insoluble in the solution so the reaction does not proceed. Hence, we do not expect lead carbonate to react with hydrochloric acid.
(e) NO₂ is called double acid anhydride because two acids – nitrous acid and nitric acid – are formed when it reacts with water.
2NO₂ + H₂O → HNO₂ + HNO₃

Solution 11.
Acid rain is a by-product of a variety of human activities which release oxides of sulphur and nitrogen in the atmosphere. Burning of fossil fuels, coal, oil, petrol and diesel produces sulphur dioxide and nitrogen oxide which pollute the air. Polluted air also contains many oxidising agents which produce oxygen because of excessive heat. This oxygen combines with the oxides of sulphur and nitrogen and rain water to form acids.
2SO₂ + O₂ + 2H₂O → 2H₂SO₄4NO₂ + O₂ + 2H₂O → 4HNO₃

Solution 12.
(a) Acids are prepared from non-metals by their oxidation. For example :
Sulphur or phosphorus is oxidized by conc. Nitric acid to form sulphuric acid or phosphoric acid.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 2
(b) Acids are prepared from salt by the displacement reaction. For example :
Nitric acid is prepared by using H₂SO₄and sodium chloride.

Solution 13.
(a) SO₂ +H₂O H₂SO₃(b) P₂O₅ +3H₂O 2H₃PO₄(c) CO₂ + H₂O H₂CO₃

Solution 14.
4(a) Citric acid
(b) Carbonic acid
(c) Oxalic acid
(d) Boric acid

Exercise Intext 2

Solution 1.
An alkali is a basic hydroxide which when dissolved in water produces hydroxyl ions (OH^–) as the only negatively charged ions.
(a) Strong alkalis: Sodium hydroxide , Potassium hydroxide
(b) Weak alkalis: Calcium hydroxide , Ammonium hydroxide

Solution 2.
(a) An alkali and a base:

Alkalis are soluble in water whereas bases may be or may not be soluble in water.
All alkalis are bases but all bases are not alkalis.

(b) The chemical nature of an aqueous solution of HCl and an aqueous solution of NH₃

The aqueoussolution of HCl is acidic in nature. It can turn blue litmus to red.
The aqueoussolution of NH₃is basic in nature. It can turn red litmus to blue.

Solution 3.
(a) Hydroxyl ion (OH^–)
(b) H⁺

Solution 4.
(a) Barium oxide
(b) Sodium hydroxide
(c) Manganese oxide
(d) Cupper hydroxide
(e) Carbonic acid
(f) Ferric hydroxide
(g) Copper oxide
(h) Ammonia
(i) Ammonium hydroxide

Solution 5.
The test tube containing distilled water does not affect the red litmus paper.
The test tube containing acidic solution does not change the red litmus paper.
But the test tube containing basic solution turns red litmus paper blue.

Solution 6.
It is because HCl and HNO₃ ionize in aqueous solution whereas ethanol and glucose do not ionize in aqueous solution.

Solution 7.
(a) DryHCl gas does not contain any hydrogen ions in it, so it does not show acidic behaviour. Hence, dry HCl gas does not change the colour of dry litmus paper.
(b) Lead oxide is a metallic oxide which reacts with hydrochloric acid to produce lead chloride and water, but it is excluded from the class of bases, because chlorine is also produced.
PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O
Thus, lead oxide is not a base.
(c)Yes, basic solutions have H⁺ions, but the concentration of OH^– ions is more than the H⁺ ions which makes the solution basic in nature.

Solution 8.
(a) We can obtain a base from another base by double decomposition. The aqueous solution of salts with base precipitates the respective metallic hydroxide.
FeCl₃ +3NaOH Fe(OH)₃ +3NaCl
(b) An alkali from a base

(c) Salt from another salt

Solution 9.
(a) Mg +2HCl MgCl₂ + H₂(b) HCl + NaOH NaCl + H₂O
(c) CaCO₃ +2HCl CaCl₂+H₂O + CO₂(d) CaSO₃ + 2HCl CaCl₂ + H₂O+ SO₂(e) ZnS + 2HCl ZnCl₂ + H₂S

Solution 10.
As we know that alkalis react with oil to form soap. As our skin contains oil so when we touch strong alkalis, a reaction takes place and soapy solution is formed. Hence we should wear gloves.

Solution 11.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 6

Solution 12.
pH represents the strength of acids and alkalis expressed in terms of hydrogen ion concentration. The solution with pH value 10 will give pink colour with phenolphthalein indicator.

Solution 13.
A = Strongly acidic
B= neutral
C=Weakly alkaline
D= Strongly alkaline
E= Weakly acidic
(a) Solution A (acidic solution) + MgH₂ + Mg salt
(b) SolutionA (acidic solution) + ZnH₂ + Zn salt

Solution 14.
(a) A common acid-base indicator and a universal indicator:
An acid-base indicator like litmus tells us only whether a given substance is an acid or a base. The universal indicator gives an idea as to how acidic or basic a substance is universal indicator gives different colours with solutions of different pH values.
(b) The acidity of bases and basicity of acids
The acidity of bases: The number of hydroxyl ions which can be produced per molecule of the base in aqueous solution.
Basicity of acid: The basicity of an acid is defined as the number of hydronium ions that can be produced by the ionization of one molecule of that acid in aqueous solution.
(c) Acid and alkali:
An acid is that substance which gives H⁺ ions when dissolved in water.
An alkali is that substance which gives OH^– ions when dissolved in water.

Solution 15.
Substances like chocolates and sweets are degraded by bacteria present in our mouth. When the pH falls to 5.5 tooth decay starts. Tooth enamel is the hardest substance in our body and it gets corroded. The saliva produced by salivary glands is slightly alkaline, it helps to increase the pH, to some extent, but toothpaste which contains basic substance is used to neutralize excess acid in the mouth.

Solution 16.
A universal indicator is a mixture of dyes which identify a gradual change of various colours over a wide range of pH, depending on the strength of the acid. When we use a universal indicator, we see solutions of different acids produce different colours. Indeed, solutions of the same acid with different concentration give different colours.
The more acidic solutions turn universal indicator bright red. A less acidic solution will only turn it orange-yellow. Colour differences can also be observed in case of vinegar which is less acidic and battery acid which is more acidic.

Solution 17.
(a)

The pH can be increased by adding a basic solution.
The pH can be increased by adding an acidic solution.

(a) The solution is basic in nature and the pH value will be more than 7.
(b)Less than 7

Solution 18.
(a) Solution P
(b) Solution R
(c)Solution Q

Exercise Intext 3

Solution 1.
(a) A normal salt: Normal salts are the salts formed by the complete replacement of the ionizable hydrogen atoms of an acid by a metallic or an ammonium ion.
(b) An acidic salt: Acid salts are formed by the partial replacement of the ionizable hydrogen atoms of a polybasic acid by a metal or an ammonium ion.
(c) A mixed salt: Mixed salts are those salts that contain more than one basic or acid radical.
Examples:
(a) A Normal salt: Na₂SO₄ , NaCl
(b) An acid salt: NaHSO₄ , Na₂HPO₄(c) A mixed salt: NaKCO₃ , CaOCl₂

Solution 2.
(a) Salt is a compound formed by the partial or total replacement of the ionizable hydrogen atoms of an acid by a metallic ion or an ammonium ion.
(b) An insoluble salt can be prepared by precipitation.
(c) A salt prepared by direct combination is Iron(III) chloride.
Reaction:
2Fe +3Cl₂ 2FeCl₃(d) By neutralizing sodium carbonate or sodium hydroxide with dilute sulphuric acid:
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

2 NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Solution 3.
(a) Copper sulphate crystals from a mixture of charcoal and black copper oxide:
The carbon in the charcoal reduces the black copper oxide to reddish-brown copper. The lid must not be removed until the crucible is cool or the hot copper will be re-oxidized by air.
Take dilute sulphuric acid in a beaker and heat it on wire gauze. Add cupric oxide in small quantities at a time, with stirring till no more of it dissolves and the excess compound settles to the bottom.
Filter it hot and collect the filtrate in a china dish. Evaporate the filtrate by heating to the point of crystallization and then allow it to cool and collect the crystals of copper sulphate pentahydrate.
Reaction:
CuO + H₂SO₄ CuSO₄ + H₂O
CuSO₄ + 5H₂O CuSO_4. 5H₂O
(b) Zinc sulphate crystals from Zinc dust:
Take dilute sulphuric acid in a beaker and heat it on wire gauze. Add some granulated zinc pieces with constant stirring. Add till the Zinc settles at the base of the beaker. Effervescences take place because of the liberation of hydrogen gas. When effervescence stops, it indicates that all the acid has been used up. The excess of zinc is filtered off. Collect the solution in a china dish and evaporate the solution to get crystals. Filter, wash them with water and dry them between the folds of paper. The white needle crystals are of hydrated Zinc sulphate.
Reaction:
Zn + H₂SO₄ ZnSO₄ + H₂ZnSO₄ +7 H₂O ZnSO_4. 7 H₂O
(c) Lead sulphate from metallic lead:
Metallic lead is converted to lead oxide by oxidation. Then lead sulphate is prepared from insoluble lead oxide, by first converting it into soluble lead nitrate. Then the lead nitrate solution is treated with sulphuric acid to obtain white ppt. of Lead sulphate.
Reaction:
PbO +2HNO₃ Pb(NO₃)₂ + H₂O
Pb(NO₃)₂ + H₂SO₄ PbSO₄ + 2HNO₃(d)Sodium hydrogen carbonate crystals:
Dissolve 5 grams of anhydrous sodium carbonate in about 25 ml of distilled water in a flask. Cool the solution by keeping the flask in a freezing mixture. Pass carbon dioxide gas in the solution. Crystals of sodium bicarbonate will precipitate out after some time. Filter the crystals and dry it in folds of filter paper.
Reaction:
Na₂CO₃ + CO₂ + H₂O 2NaHCO₃

Solution 4.

Anhydrous ferric chloride: -A (Direct combination of two elements)
2Fe + 3Cl₂ 2FeCl₃
Lead chloride: -E (Reaction of two solutions of salts to form a precipitate)
Pb(NO₃)₂ +2HCl PbCl₂ +2HNO₃
Sodium sulphate: – D( Titration of dilute acid with a solution of soluble base)
2NaOH + H₂SO₄ Na₂SO₄+2H₂O
Coppersulphate:– C (reaction of dilute acid with an insoluble base)
Cu(OH)₂ +H₂SO₄ CuSO₄ + 2H₂O

Solution 5.
(a) Lead chloride
(b) Silver chloride
(c) Barium sulphate and lead sulphate
(d) Basic lead chloride
(e) Sodium hydrogen sulphate
(f) Sodium potassium carbonate
(g) Sodium argentocyanide
(h) Potash alum
(i) Potassium bromide and potassium chloride
(j) Calcium sulphate

Solution 6.
An acid is a compound which when dissolved in water forms hydronium ions as the only positively charged ions. A base is a compound which is soluble in water and contains hydroxide ions. A base reacts with an acid to form a salt and water only. This type of reaction is known as neutralisation.

Solution 7.
(a) Blue litmus will turn into red which will indicate the solution to be acidic.
(b) No change will be observed.
(c) Red litmus will turn into blue will indicate the solution to be basic.

Solution 8.
(a) Since sodium hydroxide and sulphuric acid are both soluble, an excess of either of them cannot be removed by filtration. Therefore it is necessary to find out on small scale, the ratio of solutions of the two reactants.
(b) As iron chloride is highly deliquescent, so it is kept dry with the help of fused calcium chloride.
(c) On heating the hydrate, HCl acid is released and basic salt (FeOCl) or ferric oxide remains. Hence, anhydrous ferric chloride
cannot be prepared by heating the hydrate.

Solution 9.
Zinc Sulphate – Displacement

Ferrous sulphide – synthesis

Barium sulphate – Precipitation

Ferric Sulphate- Oxidation

Sodium sulphate – Neutralisation

Solution 10.
(a) pH of pure water is 7 at 25^oC. No, the pH does not change when common salt is added.
(b) Acids: H₂SO₄ and HNO₃Bases: Ammonium hydroxide and sodium hydroxide.
Salts: Barium chloride and sodium chloride.

Solution 11.
Neutralization is the process by which H⁺ ions of an acid react completely with the [OH]^– ions of a base to give salt and water only.
(a)

(b) Neutralization is simply a reaction between H⁺ ions given by strong acid and OH^– ions given by strong base. In case of all
strong acids and strong bases, the number of H⁺ and OH^– ions produced by one mole of a strong acid or strong base is always same. Hence the heat of neutralization of a strong acid with strong base is always same.

Solution 12.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 12

Solution 13.
(a) Iron (III) Chloride: Iron chloride is formed by direct combination of elements.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 13

Solution 14.
(a) By neutralisation:
NaOH + HCl → NaCl + H₂O
(b) By precipitation:
Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃(c)CuCO₃ + H₂SO₄→ CuSO₄ + H₂O + CO₂(d) Simple displacement:
Zn + H₂SO₄→ ZnSO₄ + H₂

Solution 15.
(a) Na₂CO₃ + H₂SO₄ (dil) → Na₂SO₄ + H₂O + CO₂(b) CuCO₃ + H₂SO₄ (dil) → CuSO₄ + H₂O + CO₂(c) Fe + H₂SO₄ (dil) → FeSO₄ + H₂(d) Zn + H₂SO₄ (dil) → ZnSO₄ + H₂ZnSO₄ + Na₂CO₃ → ZnCO₃ + Na₂SO₄

Solution 16.
(a) NaHSO₄(b) AgCl
(c) CuSO₄.5H₂O
(d) CuCO₃(d) Pb(NO₃)₂

Solution 17.
(a) acid salt
(b) NaOH+ HCl → NaCl + H₂O

Solution 18.
(a) Alkali
(b) Precipitate
(c) Weak acid

Solution 19.

Copper (II) chloride – B
Iron (II) chloride – A
Iron (III) chloride – C
Lead (II) chloride – E
Sodium chloride – D

Solution 20.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 14

Exercise Intext 4

Solution 1.
It is the amount of water molecules which enter into loose chemical combination with one molecule of the substance on crystallisation from its aqueous solution.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 15

Solution 2.
(a) Crystalline hydrated salts which on exposure to the atmosphere lose their water of crystallisation partly or completely and change into a powder. This phenomenon is called efflorescent and the salts are called efflorescent.
Examples: CuSO₄.5H₂O, MgSO₄.7H₂O, Na₂CO₃.10H₂O
(b) Water-soluble salts which on exposure to the atmosphere absorb moisture from the atmosphere and dissolve in the same and change into a solution. The phenomenon is called deliquescence and the salts are called deliquescent.
Examples: CaCl₂, MgCl₂, ZnCl₂

Solution 3.
Selina Concise Chemistry Class 10 ICSE Solutions Acids, Bases and Salts img 16

Solution 4.
Conc. sulphuric acid is hygroscopic in nature and can remove moisture from other substances; therefore, it is used as a drying agent. It is also used as a dehydrating agent because it has a strong affinity for water and thus absorbs water quickly from compounds.

Solution 5.

blue
red
hydrogen gas
basic, alkaline
graphite

Solution 6.
(a) Sodium hydrogensulphate [NaHSO₄] is an acid salt and is formed by the partial replacement of the replaceable hydrogen ion in a dibasic acid [H₂SO₄]. The [H] atom in NaHSO₄ makes it behave like an acid.
So, on dissolving in water, it gives hydrogen ions.
(b) Desiccating agentsare used to absorb moisture. Anhydrous calcium chloride (CaCl₂) has the capacity of absorbing moisture as it is hygroscopic in nature. So, it is used in a desiccator.

Solution 7.
(a) Increase
(b) Increase
(c) Decrease
(d) Increase
(e) Increase

Solution 8.
(a) Table salt turns moist and ultimately forms a solution on exposure to air especially during the rainy season. Although pure sodium chloride is not deliquescent, the commercial version of the salt contains impurities (such as magnesium chloride) which are deliquescent substances.
(b) The impurity can be removed by passing a current of dry hydrogen chloride gas through a saturated solution of the affected salt. Pure sodium chloride is produced as a precipitate which can be recovered by filtering and washing first with a little water and finally with alcohol.
(c) Conc. sulphuric acid
(d) Common salt and sugar

Solution 9.
(a) Water of crystallization
(b) White
(c) By heating with any dehydrating agent
(d) Anhydrous calcium chloride

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