Category: CBSE Class 9

Online Education NCERT Solutions for Class 9 Hindi Sparsh Chapter 7  धर्म की आड़

These Solutions are part of Online Education NCERT Solutions for Class 9 Hindi. Here we have given NCERT Solutions for Class 9 Hindi Sparsh Chapter 7 धर्म की आड़.

पाठ्य-पुस्तक के प्रश्न-अभ्यास

मौखिक

निम्नलिखित प्रश्नों के उत्तर एक-दो पंक्तियों में दीजिए-

प्रश्न 1.
आज धर्म के नाम पर क्या-क्या हो रहा है?
उत्तर:
आज धर्म के नाम पर कुछ स्वार्थी लोगों द्वारा उत्पात किया जा रहा है और भोले-भाले लोगों को आपस में लड़ाया जा रहा है।

प्रश्न 2.
धर्म के व्यापार को रोकने के लिए क्या उद्योग होने चाहिए?
उत्तर:
धर्म के व्यापार को रोकने के लिए साहस और दृढ़ता के साथ उसका विरोध होना चाहिए।

प्रश्न 3.
लेखक के अनुसार स्वाधीनता आंदोलन का कौन-सा दिन सबसे बुरा था?
उत्तर:
स्वाधीनता आंदोलन का वह दिन सबसे बुरा था जब स्वाधीनता के काम में मुल्ला, मौलवी, शंकराचार्य जैसे धर्म के आचार्यों को अधिक महत्त्व दिया गया।

प्रश्न 4.
साधारण से साधारण आदमी तक के दिल में क्या बात अच्छी तरह घर कर बैठी है?
उत्तर:
साधारण से साधारण आदमी तक के दिल में यह बात अच्छी तरह घर करके बैठी है कि धर्म और ईमान के नाम पर अपनी जान दे देना उचित है।

प्रश्न 5.
धर्म के स्पष्ट चिह्न क्या हैं?
उत्तर:
शुद्ध आचरण और सदाचार करना धर्म के स्पष्ट चिह्न हैं।

लिखित

(क) निम्नलिखित प्रश्नों के उत्तर (25-30 शब्दों में) लिखिए

प्रश्न 1.
चलते पुरजे लोग धर्म के नाम पर क्या करते हैं?
उत्तर:
चलते पुरज़े लोग धर्म के नाम पर लोगों को बेवकूफ़ बनाकर अपना स्वार्थ सिद्ध करते हैं। वे चाहते हैं कि उनका नेतृत्व कायम रहे। उनका प्रभाव बना रहे।

प्रश्न 2.
चालाक लोग साधारण आदमी की किस अवस्था का लाभ उठाते हैं? [CBSE 2012]
उत्तर:
चालाक लोग साधारण आदमी की धार्मिक भावनाएँ भड़काते हैं। साधारण आदमी धर्माध होकर धर्म के नाम पर मरने-मिटने को तैयार हो जाता है। उसकी इसी स्थिति का लाभ चालाक लोग उठाते हैं।

प्रश्न 3.
आनेवाला समय किस प्रकार के धर्म को नहीं टिकने देगा?
उत्तर:
जो लोग धर्म के प्रति दिखावा मात्र करके लोगों को आपस में लड़वाते हैं, आनेवाला समय उन्हें टिकने नहीं देगा। जन साधारण की समझ में आ गया है कि ऐसे धार्मिक नेता उनकी भावनाओं से खेलते हैं।

प्रश्न 4.
कौन-सा कार्य देश की स्वाधीनता के विरुद्ध समझा जाएगा?
उत्तर:
प्रत्येक व्यक्ति किसी धर्म को मानने और पूजा-उपासना की कोई भी रीति अपनाने को स्वतंत्र है। उसकी इस स्वाधीनता में हस्तक्षेप करने के कार्य को देश की स्वाधीनता के विरुद्ध समझा जाएगा।

प्रश्न 5.
पाश्चात्य देशों में धनी और निर्धन लोगों में क्या अंतर है?
उत्तर:
पाश्चात्य देशों में धनी लोगों के पास पैसा है, ऊँची-ऊँची इमारतें हैं, सुख-सुविधा है। गरीब लोग रोटी के लिए संघर्ष करते हैं और झोंपड़ियों में रहते हैं।

प्रश्न 6.
कौन-से लोग धार्मिक लोगों से अधिक अच्छे हैं?
उत्तर:
जिन लोगों का आचरण अच्छा है, जो दूसरों का कल्याण करते हैं, अपने आचरण से दूसरों को दुख नहीं पहुंचाते हैं तथा जो अपनी स्वार्थ सिद्धि के लिए भोले-भाले लोगों का शोषण नहीं करते हैं, वे धार्मिक लोगों से अधिक अच्छे हैं।

(ख) निम्नलिखित प्रश्नों के उत्तर (50-60 शब्दों में) लिखिए

प्रश्न 1.
धर्म और ईमान के नाम पर किए जाने वाले भीषण व्यापार को कैसे रोका जा सकता है? [CBSE 2012]
उत्तर:
धर्म और ईमान के नाम परं दंगे-फसाद हो रहे हैं। कुछ स्वार्थी आदमी धर्म के नाम पर लोगों को आपस में लड़वाते हैं। अपने निजी स्वार्थों के लिए आम आदमी के प्राण ले लिए जाते हैं। इसको रोकने का उपाय है कि लोगों को उन आदमियों और धर्म की सही शिक्षा के लिए जानकारी दी जाए। लोगों को समझाया जाए कि दंगा करके खून बहाने वालों का कोई धर्म नहीं होता।

प्रश्न 2.
‘बुद्धि को मार’ के संबंध में लेख़क के क्या विचार हैं?” [CBSE 2012]
उत्तर:
बुधि की मार के संबंध में लेखक का विचार है-कुछ चलते-पुरज़े लोगों द्वारा साधारण लोगों के मस्तिष्क में ऐसे विचार भर देना कि वे अपनी बुधि से कुछ भी सोचने-समझने योग्य न रह जाएँ। ऐसे लोगों की धार्मिक भावनाएँ भड़काकर अपने हित साधने योग्य बना लेना ताकि स्वार्थी लोग अपना स्वार्थ आसानी से पूरा कर सकें।

प्रश्न 3.
लेखक की दृष्टि में धर्म की भावना कैसी होनी चाहिए?
उत्तर:
लेखक के अनुसार, धर्म के विषय में मानव स्वतंत्र होना चाहिए। हर व्यक्ति आजाद हो। वह जो धर्म अपनाना चाहे, अपनाए। कोई किसी की स्वतंत्रता में बाधा न खड़ी करे। धर्म का संबंध हमारे मन से, ईमान से, ईश्वर और आत्मा से होना चाहिए। वह मन को शुद्ध करने का मार्ग होना चाहिए, अपने जीवन को ऊँचा उठाने का साधन होना चाहिए, दूसरे को कुचलने का नहीं।

प्रश्न 4.
महात्मा गाँधी के धर्म संबंधी विचारों पर प्रकाश डालिए। [CBSE 2012]
उत्तर:
गांधी जी धर्म को मानने वाले थे। इसके बिना वे एक कदम भी नहीं चलते थे। वे पूजा-पाठ, नमाज पढ़ने जैसी दिखावापूर्ण धार्मिक क्रियाओं को सच्चा धर्म नहीं मानते थे। उनका धर्म पवित्र भावनाओं से भरपूर था। वे धर्म को लोगों के कल्याण का साधन समझते थे। उनका मानना था कि धर्म ऊँचे और उदार तत्वों का हुआ करता है, जिसे अपनाने में किसी को आपत्ति नहीं हो सकती।

प्रश्न 5.
सबके कल्याण हेतु अपने आचरणा को सुधारना क्यों आवश्यक है?
उत्तर:
जब तक हम स्वयं का आचरण ठीक नहीं रखेंगे, दूसरे लोगों को उसकी प्रेरणा नहीं दे सकते। समाज में उदाहरण बनने के लिए हमें स्वयं का आचरण सुधारना होगा। मानव मात्र की भलाई तभी हो सकती है, जब हम निजी स्वार्थ को छोड़कर पूरे समाज की भलाई के बारे में सोचें।

(ग) निम्नलिखित का आशय स्पष्ट कीजिए

प्रश्न 1.
उबल पड़ने वाले साधारण आदमी को इसमें केवल इतना ही दोष है कि वह कुछ भी नहीं समझता-बूझता, और दूसरे लोग उसे जिधर जोत देते हैं, उधर जुत जाता है।
उत्तर:
कुछ चालू-पुरज़े लोग तथा धर्म के तथाकथिक ठेकेदार साधारण लोगों के दिमाग में यह बात अच्छी तरह बिठा देते हैं कि धर्म और ईमान ही तुम्हारे लिए सब कुछ हैं। इसी से तुम्हारा कल्याण होने वाला है। इसकी रक्षा करते हुए तुम्हें अपनी ज्ञान की परवाह नहीं करनी चाहिए। ये अनपढ़ साधारण भोले लोग धर्म क्या है, यह जाने-समझे बिना तनिक-सा उकसाए जाते ही मरने-कटने के लिए तैयार हो जाते हैं। वे दूसरों के बहकावे में जल्दी आ जाते हैं। इससे उनकी शक्ति और साहस का दुरुपयोग स्वार्थी लोग अपने हित के लिए करते हैं।

प्रश्न 2.
यहाँ है बुद्धि पर परदा डालकर पहले ईश्वर और आत्मा का स्थान अपने लिए लेना, और फिर धर्म, ईमान, ईश्वर और आत्मा के नाम पर अपनी स्वार्थ सिद्धि के लिए लोगों को लड़ाना-भिड़ाना।
उत्तर:
भारत में धार्मिक नेता लोगों की बुद्धि का शोषण करते हैं। पहले वे अपने प्रति अंध श्रद्धा उत्पन्न करते हैं। लोग उन्हें ईश्वर, आत्मा और धर्म का पूज्य प्रतीक मान बैठते हैं। जब लोगों की श्रद्धा उन पर जम जाती है तो वे ईमान, धर्म, ईश्वर या आत्मा का नाम लेकर उन्हें दूसरे धर्म वालों से लड़ाते-भिड़ाते हैं तथा अपने स्वार्थ सिद्ध करते हैं।

प्रश्न 3.
अब तो, आपका पूजा-पाठ न देखा जाएगा, आपकी भलमनसाहत की कसौटी केवल आपका आचरण होगी।
उत्तर:
धर्म और ढोंग में अंतर है। धर्म ईश्वर तक पहुँचाने की कड़ी है। कुछ लोग धर्म का आडंबर करते हुए दो-दो घंटे तक पूजा-पाठ करते हैं, शंख बजाते हैं, नमाज़ पढ़ते हैं। ऐसा करके वे समझते हैं कि वे कुछ भी करने के लिए स्वतंत्र हैं। ऐसे लोग यदि अपना आचरण नहीं सुधारते हैं, तो यह पूर्जा-पाठ सब व्यर्थ हो जाएगा। उनके आचरण-व्यवहार में सज्जनता और दूसरों के कल्याण की भावना निहित होनी चाहिए।

प्रश्न 4.
तुम्हारे मानने ही से मेरा ईश्वरत्व कायम नहीं रहेगा, दया करके, मनुष्यत्व को मानो, पशु बनना छोड़ो और आदमी बनो! [CBSE 2012]
उत्तर:
स्वयं ईश्वर भटके हुए लोगों को कहता है-लोगों को धर्म के नाम पर लड़वाना छोड़ो। अपवित्र काम छोड़ो, खुद की पूजा करवाना छोड़ दो। मानवता को समझो। आदमी बनो और पशु वाला आचरण त्याग दो। आशय यह है कि धर्म के नाम पर कट्टरता त्याग दो और मनुष्य के साथ उदारतापूर्वक सहयोग और स्नेह से व्यवहार करो।

भाषा-अध्ययन

प्रश्न 1.
उदाहरण के अनुसार शब्दों के विपरीतार्थक लिखिए

          सुगम            –      दुर्गम

1. धर्म              –       …………………..
2. ईमान           –       …………………..
3. साधारण       –       …………………..
4. स्वार्थ           –       …………………..
5. दुरुपयोग     –        …………………..
6. नियंत्रित       –       …………………..
7. स्वाधीनता    –       …………………..

उत्तर:

1. धर्म             –        अधर्म
2. ईमान          –         बेईमान
3. साधारण      –        असाधारण
4. स्वार्थ          –        निस्वार्थ
5. दुरुपयोग    –        सदुपयोग
6. नियंत्रित      –        अनियंत्रित
7. स्वाधीनता   –       पराधीनता

प्रश्न 2.
निम्नलिखित उपसर्गों का प्रयोग करके दो-दो शब्द बनाइए-

1. ला,
2. बिला,
3. बे,
4. बद,
5. ना,
6. खुश,
7. हर,
8. गैर

उत्तर:

1. ला – लापता, लाजवाब, लापरवाही।
2. बिला – बिलावजह, बिलानागा।
3. बे – अदब, बेवज़ह, बेवफ़ा, बेशक ।
4. बद – बदनाम, बदसूरत, बदतमीज़ ।
5. ना – नासमझ, नादानी, नामर्द ।
6. खुश – खुशफहमी, खुशगवार।
7. हर – हररोज़, हरदम।
8. गैर – गैरकानूनी, गैरहाज़िर।

प्रश्न 3.
उदाहरण के अनुसार ‘त्व’ प्रत्यय लगाकर पाँच शब्द बनाइए-
उदाहरण : देव + त्व = देवत्व
उत्तर:

1. लघु + त्व = लघुत्व
2. प्रभु + त्व = प्रभुत्व
3. महत् + त्व = महत्त्व
4. नारी + त्व = नारीत्व
5. मनुष्य + त्व = मनुष्यत्व।

प्रश्न 4.
निम्नलिखित उदाहरण को पढ़कर पाठ में आए संयुक्त शब्दों को छाँटकर लिखिए-
उदाहरण : चलते पुरजे
उत्तर:

1. समझता – बूझता,
2. पढ़े – लिखे,
3. इने – गिने,
4. मन – माना,
5. स्वार्थ – सिद्धि,
6. लड़ाना – भिड़ाना,
7. दीन – दीन,
8. नित्य – प्रति,
9. भली – भाँति,
10. दिन – भर,
11. पूजा – पाठ,
12. देश – भर,
13. सुख – दुःख।

प्रश्न 5.
‘भी’ का प्रयोग करते हुए पाँच वाक्य बनाइए-
उदाहरण : आज मुझे बाज़ार होते हुए अस्पताल भी जाना है।
उत्तर:

1. यह भोजन मेरे साथ तुम्हें भी करना है।
2. गाँधीजी के साथ नेहरू भी आए हैं।
3. आज सब्जीमंडी से आम भी लाना।
4. नौकरी के लिए मेहनत ही नहीं, सिफ़ारिश भी करनी पड़ती है।
5. हम मसूरी-नैनीताल ही नहीं, कौसानी भी गए थे।

योग्यता-विस्तार

प्रश्न 1.
‘धर्म एकता का माध्यम है-इस विषय पर कक्षा में परिचर्चा कीजिए।
उत्तर:

1. एक छात्र – धर्म से एकता बढ़ती है। धार्मिक आयोजनों में लोग ऊँच-नीच भूलकर मित्र भाव से भाग लेते हैं।
2. दूसरा छात्र – परंतु कुछ लोग धर्म के नाम पर ही स्वयं को अलग मानते हैं। कोई खुद को सिख कहता है, कोई बौद्ध कहता है, कोई हिंदू तो कोई मुसलमान।।
3. तीसरा छात्र – परंतु वे सब हैं तो आदमी ही। चौथा छात्र-परंतु धर्म का नाम लेते ही वे स्वयं को इनसान नहीं हिंदू, ईसाई या मुसलमान कहने लगते हैं।

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Online Education NCERT Solutions for Class 9 Science Chapter 13 Why Do we Fall Ill

These Solutions are part of Online Education NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 13 Why Do we Fall Ill. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 13 – Why Do We Fall ill solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 13 – Why Do We Fall ill Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

• NCERT Solutions for Class 9 Science
• HOTS Questions for Class 9 Science
• Value Based Questions in Science for Class 9
• NCERT Exemplar Solutions for Class 9 Science
• Previous Year Question Papers for CBSE Class 9 Science

NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
State any two conditions essential for good health.

1. Complete physical fitness free from any disease.
2. Perfect mental (and social) well being.

Answer:
State any two conditions essential for being free of disease.

1. Non-overcrowded living conditions, safe drinking water and clean environment.
2. Balanced diet, personal hygiene, exercise and relaxation.

Question 2.
Are the answers to the above questions same or different ? Why ?
Answer:
Different: Health is a state of perfect physical, mental, and social well being of a person while disease is a state of disease or discomfort of the body or its part.

Question 3.
List any three reasons why you would think that you are sick and ought to see a doctor. If only one of these symptoms were present, would you still go to the doctor ? Why or why not ?
Answer:

1. Cough, cold and fever. The sickness must be due to infection which requires proper diagnosis and treatment which only doctor can do.
2. If only one symptom is present (say cough only or cold only) even then I will prefer to get treatment from the doctor because an untreated infection can spread and cause further damage to the body.

Question 4.
In which of the following case do you think the long term effects on your health are likely to be most unpleasant ?
Answer:
(a) If you get jaundice
(b) If you get lice
(c) If you get acne.
Jaundice. It is a severe disease which also takes several days to heal. During this period bile pigments collect at several places in the body and permanently damage some of them.

Question 5.
Why are we normally advised to take bland and nourishing food when we are sick ? (CCE 2011, 2012)
Answer:
It helps in strengthening of the immune system and provides nourishment to body which is being depleted by infectious agent.

Question 6.
What are the different means by which infectious diseases are spread ?
Answer:
By air directly or as droplets, e.g., cold.
By contaminated food and water, e.g., cholera.
By fomites or articles contaminated by the patient, e.g., chicken-pox.

Question 7.
Contagious diseases simply spread by contact.
Answer:
By sexual and blood contact in case of AIDS, syphilis and some other diseases.
By vectors and carriers, e.g., malaria by female Anopheles.

Question 8.
What precautions would you take in your school to reduce incidence of infectious diseases ?
Answer:

1. Airy and well-spaced classrooms.
2. Advising students falling sick not to come to school. If any such student does come, the same should be provided a separate bench.
3. Ensuring safe drinking water.
4. Cleanliness.
5. Protection against flies and mosquitoes.
6. Vaccination against diseases.
7. Regular medical examination of students.

Question 9.
What is immunisation ?
Answer:
Development of immunity or resistance against a pathogen through vaccination is called immunisation.

Question 10.
What are the immunisation programmes available at our nearest health centre in your locality ?
Which of these diseases are two major health problems in your area ?
Answer:

NCERT CHAPTER END EXERCISES

Question 1.
How many times did you fall ill in the last one year ? What were the illnesses ?
(a) Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.
(b) Think of one change you would wish for in your surroundings in order to avoid any of/most of the above illnesses.
Answer:
I suffered last year from cough and common cold (thrice), typhoid (once), malaria (once), loose motions (thrice)
(a) Change in Habits,

1. I should not sit along with those persons who are suffering from cough and cold,
2. I should take care of not eating unprotected food,
3. I should protect myself from mosquito bites.

(b) Change in Surroundings. Sanitary conditions in and around my home should be improved. The drains should be regularly cleaned and kept covered.

Question 2.
A doctor/nurse/health worker is exposed to more sick people than others in the community. Find out how she avoids getting sick herself. (CCE 2012, 2013, 2017)
Answer:
Through

1. vaccination against most of the common diseases,
2. use of gloves,
3. use of different dress or coat while visiting patients, and
4. use of disposable syringes and needles.

Question 3.
Conduct a survey in your neighbourhood to find what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.
Answer:
The three most common diseases in my neighbourhood are malaria, typhoid and diarrhoea. The diseases mostly spread through unhygienic surroundings. The authorities should take care of

1. disposal of garbage,
2. cleaning of drains with occasional spray of insecticides
3. covering of drains and
4. providing clean drinking water.

Question 4.
A baby is not able to tell his/her caretakers that he/she is sick. What would help us to find out
(a) that the baby is sick
(b) what the sickness is ?
Answer:

1. Running fever
2. Cold and cough
3. Excessive crying
4. Loose motions
5. Non-intake of proper diet.

To find the sickness, the baby is taken to a doctor. The doctor is able to diagnose the disease with the help of stethoscope, palpation (feeling with hand), above symptoms and laboratory test.

Question 5.
Under which of the following conditions is a person most likely to fall sick ?
(a) When she is recovering from malaria.
(b) When she has recovered from malaria and is taking care of someone suffering from chicken-pox ?
(c) When she is on a four day fast after recovering from malaria and is taking care of someone suffering from chicken-pox. Why ?
Answer:
(c) Four day fast will make the person weak with reduced immunity. As a result the chances of picking up chicken-pox and falling sick would be more.

Question 6.
Under which of the following conditions are you most likely to fall sick ?
(a) When you are having examinations.
(b) When you have travelled by bus and train for two days.
(c) When your friend is suffering from measles. Why ?
Answer:
(c) Measles is an infectious viral disease of young children which spreads through nasal or throat discharge (droplet method, fomite and other contacts). Visiting such a friend is likely to provide infection.

SELECTION TYPE QUESTIONS

Alternate Response Type Questions
(True/False, Right/Wrong, Yes/No)

Question 1.
Health of an individual depends on the surroundings.
Question 2.
Gainful employment has no relation to individual health.
Question 3.
On the basis of symptoms, physicians look for signs of disease.
Question 4.
High blood pressure is an infectious disease.
Question 5.
Penicillin is effective against bacteria because it inhibits their wall formation.
Question 6.
In open spaces, air borne human diseases spread rapidly.
Question 7.
AIDS spreads through sex, blood to blood contact and from mother to child.
Question 8.
Personal hygiene is basic to prevent infectious diseases.

Matching Type Questions :

Question 9.
Match the contents of the columns I and II (single matching)

Question 10.
Match the contents of columns I, II and III (double matching)

Question 11.
Which type of pathogen(Viral-V, Bacteria-B, Protozoan-P) cause the following discases (key or Check List Items)

Question 12.
Match the stimulus with Appropriate Response.

Fill in the Blanks

Question 13. ………….. medicines are difficult to make as the pathogens have very few biochemical mechanisms of their own.
Question 14. In AIDS patients even small cold can become ……………. .
Question 15. Sleeping sickness is caused by …………. a protozoan.
Question 16. The two important contributory causes of diseases are ……………. and …………… make up.
Question 17. Health is the state of well being physically, ……………. and socially.

Answers:

SOME TYPICAL QUESTIONS

Question 1.
Which parameters to health are difficult to measure ?
Answer:
Mental health and social well being.

Question 2.
Name the biologist who established that pathogen is a disease agent.
Answer:
Robert Koch.

Question 3.
Name the enzyme present in tears which prevents eye infections.
Answer:
Lysozyme.

Question 4.
What kills bacteria in our food in the mouth and stomach ?
Answer:
The enzyme lysozyme in saliva kills bacteria in food in our mouth, and in stomach HC1 of gastric juice kills the bacteria.

Question 5.
If a pregnant mother is suffering from AIDS, would her child get the disease via genes or placenta ?
Answer:
Placenta (Note : AIDS is not a genetic disease).

Question 6.
Why is it difficult to develop vaccines for some diseases ? (CCE 2011)
Answer:
It is difficult to develop vaccines against the diseases caused by viruses. Viruses are very specific to hosts. They have no metabolic machinery of their own. Viruses live and multiply only in the living cells. They cannot be cultured on artificial medium. It is because of these that vaccines are difficult to be prepared in such cases.

Question 7.
While going abroad, why is it essential to get vaccinated against certain diseases ?
Answer:
A person may be carrier of some disease. Such a person may take that particular disease to a foreign country. Therefore, all visitors to a foreign country are vaccinated against the disease which is not prevalent in that country.

Question 8.
Name such a vaccine which saves the life of babies from three diseases.
Answer:
D.P.T. is a vaccine which is three-in-one. Babies should be immunised within the first six weaks of birth D = Diphtheria, P= Pertussis (whooping cough), T = Tetanus.

Question 9.
Who discovered small pox vaccine ?
Answer:
Edward Jenner.

Question 10.
Define the word disease carrier.
Answer:
One who harbours germs of a disease but does not suffer from the disease is termed as carrier.

Question 11.
What are the common symptoms of infection ?
Answer:
In nearly all the infections of the human body, there is a rise in body temperature, an increased rate of heart beat, increased frequency of respiration, dry tongue, poor appetite, concentration of urine and changes in white blood cells circulating the blood.

NCERT Solutions for Class 9 Science Chapter 13 Why Do We Fall ill

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Online Education NCERT Solutions for Class 9 Hindi Sparsh Chapter 15 नए इलाके में … खुशबू रचते हैं हाथ

These Solutions are part of Online Education NCERT Solutions for Class 9 Hindi. Here we have given NCERT Solutions for Class 9 Hindi Sparsh Chapter 15 नए इलाके में … खुशबू रचते हैं हाथ.

पाठ्य-पुस्तक के प्रश्न-अभ्यास

(1) नए इलाके में

प्रश्न 1.
निम्नलिखित प्रश्नों के उत्तर दीजिए-
(क) नए बसते इलाके में कवि रास्ता क्यों भूल जाता है? [CBSE]
(ख) कविता में कौन-कौन से पुराने निशानों का उल्लेख किया गया है?
अथवा
कवि अपने गंतव्य तक पहुँचने के लिए कौन-कौन-सी निशानियाँ ढूँढता है? [CBSE]
(ग) कवि एक घर पीछे या दो घर आगे क्यों चल देता है? [CBSE]
(घ) “वसंत का गया पतझड़’ और ‘बैसाख का गया भादों को लौटा’ से क्या अभिप्राय है?
(ङ) कवि ने इस कविता में समय की कमी की ओर क्यों इशारा किया है? [CBSE]
(च) इस कविता में कवि ने शहरों की किस विडंबना की ओर संकेत किया है?
अथवा
यह कविता किस ओर इशारा करती है, स्पष्ट कीजिए। (CBSE]
उत्तर:
(क) नए इलाके में कवि इसलिए रास्ता भूल जाता है, क्योंकि-

• यहाँ रोज़ नए मकान बनते रहते हैं।
• पुराने मकान ढहाकर नए मकान बनाए जाते हैं।
• नए मकान बनाने के लिए पुराने पेड़ काटने से निशानी नष्ट हो जाती है।
• खाली जमीन पर कोई नया मकान बन जाता है।

(ख) कविता में निम्नलिखित पुराने निशानों का उल्लेख हुआ है-

• पीपल का पेड़
• ढहा घर या खंडहर
• जमीन का खाली टुकड़ा
• बिना रंग वाले लोहे के फाटक वाला इकमंजिला मकान

(ग) कवि एक घर आगे या दो घर पीछे इसलिए चल देता है, क्योंकि नए बस रहे उस इलाके में एक ही दिन में काफ़ी बदलाव आ जाता है। वह अपने घर को पहचान नहीं पाता है कि वह सवेरे किस घर से गया था।

(घ)  ‘वसंत का गया पतझड़’ और ‘बैसाख का गया भादों को लौटा’ से यह अभिप्राय है कि वहाँ एक ही दिन में इतना कुछ नया बन गया है, जितना बनने में पहले नौ-दस महीने या साल भर लगते थे। सुबह का निकला कवि जब शाम को वापस आता है तो एक ही दिन में नौ-दस महीने के बराबर का बदलाव दिखाई देता है।

(ङ) कवि ने कविता में समय की कमी की ओर इसलिए संकेत किया है क्योंकि तेज़ी से आ रहे बदलाव के कारण मनुष्य की व्यस्तता भी बढ़ती जा रही है। इससे उसके पास समय की कमी होती जा रही है।

(च) इस कविता में कवि ने शहरों की उस विडंबना की ओर संकेत किया है, जिसमें शहरों में हो रहे बदलाव, खाली जमीनों में टूटे मकानों की जगह इतने नित नए मकान बनते जा रहे हैं कि सुबह घर से निकले आदमी को शाम के समय अपना मकान खोजना पड़ता है, फिर भी उसे अपना मकान नहीं मिल पाता है।

प्रश्न 2.
व्याख्या कीजिए-
(क) यहाँ स्मृति का भरोसा नहीं
       एक ही दिन में पुरानी पड़ जाती है दुनिया
उत्तर:
नगरों में बसने वाली नई बस्तियाँ इस तरह तेजी से बढ़ती चली जा रही हैं कि आदमी को अपना घर तक ढूँढना कठिन हो गया है। वह कुछ ही दिन बाद अपनी बस्ती में लौटकर आए तो रास्ते तक भूल जाता है। उसकी पुरानी निशानियाँ देखते ही देखते नष्ट हो जाती हैं। इसलिए उसकी पुरानी स्मृतियाँ और निशानियाँ किसी काम नहीं आतीं। दुनिया इतनी तेजी से बदल-बन रही है कि जो निर्माण एक दिन पहले किया जाता है, दूसरे दिन तक पुराना पड़ चुका होता है। उसके बाद नए-नए निर्माण और खड़े हो जाते हैं।

(ख)  समय बहुत कम है तुम्हारे पास।
        आ चला पानी ढहा आ रहा अकास
        शायद पुकार ले कोई पहचाना ऊपर से देखकर
उत्तर:
देखिए व्याख्या क्र. 2..

योग्यता-विस्तार

प्रश्न 1.
पाठ में हिंदी महीनों के कुछ नाम आए हैं। आप सभी हिंदी महीनों के नाम क्रम से लिखिए।
उत्तर:
हिंदी महीनों के नाम-

1. चैत्र,
2. बैसाख,
3. ज्येष्ठ,
4. आषाढ़,
5. श्रावण,
6. भाद्रपक्ष,
7. आश्विन,
8. कार्तिक,
9. मार्गशीर्ष,
10. पौष,
11. माघ,
12. फाल्गुन

(2) खुशबू रचते हैं हाथ

प्रश्न 1.
निम्नलिखित प्रश्नों के उत्तर दीजिए-
(क) “खुशबू रचनेवाले हाथ’ कैसी परिस्थितियों में तथा कहाँ-कहाँ रहते हैं?
(ख) कविता में कितने तरह के हाथों की चर्चा हुई है?
(ग) कवि ने यह क्यों कहा है कि ‘खुशबू रचते हैं हाथ’?
(घ) जहाँ अगरबत्तियाँ बनती हैं, वहाँ का माहौल कैसा होता है?
(ङ) इस कविता को लिखने का मुख्य उद्देश्य क्या है?
उत्तर:
(क) खुशबू रचनेवाले हाथ अत्यंत कठोर परिस्थितियों में गंदी बस्तियों में, गलियों में, कूड़े के ढेर के इर्द-गिर्द तथा नाले के किनारे रहते हैं। वे अस्वच्छ एवं प्रदूषित वातावरण में जीवन बिताते हैं। वे इस दुर्गंधमय वातावरण में रहने को विवश हैं। वे सामाजिक और आर्थिक विषमता के शिकार हैं। दूसरों को खुशबू देने का काम करने । वाले इस प्रकार बदहाली का जीवन बिताते हैं।

(ख) कविता में निम्नलिखित तरह के हाथों की चर्चा हुई है-

1. उभरी नसोंवाले अर्थात् वृद्ध हाथ।
2. घिसे नाखूनोंवाले हाथ श्रमिक वर्ग को प्रतीक है।
3. पीपल के पत्ते जैसे नए-नए हाथ अर्थात् छोटे बच्चों के कोमल हाथ।
4. जूही की डाल जैसे खुशबूदार हाथ अर्थात् नवयुवतियों के सुंदर हाथ।
5. गंदे कटे-पिटे हाथ।
6. जखम से फटे हुए हाथ।

(ग) कवि ने ऐसा इसलिए कहा है क्योंकि इन गरीब मजदूरों के हाथ सुगंधित अगरबत्तियों का निर्माण करते हैं। तथा हमारे जीवन को सुख-सुविधाएँ उपलब्ध कराकर खुशबू से महकाते हैं जिससे ऐसा लगता है कि अत्यंत प्रदूषित वातावरण में रहकर भी इनके हाथ हमारे लिए सुख-सुविधाओं से भरी वस्तुओं का निर्माण करते हैं। जिससे समस्त प्राणियों के जीवन में सुगंध फैल जाती है। ये लोग स्वयं बदहाली का जीवन बिताकर दूसरे लोगों के जीवन में खुशहाली लाते हैं। इन शब्दों द्वारा कवि ने श्रमिकों के श्रम का गुणगान किया है।

(घ) जहाँ अगरबत्तियाँ बनती हैं वहाँ का वातावरण अत्यंत गंदगी भरा होता है। चारों ओर नालियाँ तथा कूड़े-करकट का ढेर जमा होता है। चारों ओर बदबू फैली होती है। ये सुगंधित अगरबत्तियाँ बनाने वाले ऐसे गंदे वातावरण में रहकर भी दूसरों के जीवन में खुशबू बिखेरते हैं पर ऐसे वातावरण में, ऐसी भयावह स्थितियों में रहनी इनकी विवशता है।

(ङ) इस कविता को लिखने का मुख्य उद्देश्य यह है कि हमारे समाज में सुंदरता की रचना करनेवाले गरीब
और उपेक्षित लोगों की ओर हमारा ध्यान आकर्षित करना है ताकि आम लोग इन गरीब मजदूरों के जीवन की वास्तविकता को जान लें और समाज में फैली विषमताओं तथा भेदभावों को मिटाने की कोशिश करें। मजदूरों और कारीगरों की दुर्दशा का चित्रण करना तथा लोगों में उनके उद्धार की चेतना जगाना भी है। कवि अगरबत्तियाँ बनानेवाले कारीगरों का प्रदूषित वातावरण में रहना दिखाकर यह कहना चाहता है कि इनके जीवन स्तर को ऊँचा उठाने के लिए हम सबको मिलकर प्रयास करना चाहिए ताकि इन्हें भी जीवन जीने के लिए। स्वच्छ वातावरण मिल सके।

प्रश्न 2.
व्याख्या कीजिए-
(क)
(i)  पीपल के पत्ते-से नए-नए हाथ
      जूही की डाल-से खुशबूदार हाथ
उत्तर:
अगरबत्ती बनाने वाले हाथों में कुछ के हाथ पीपल के नए-नए पत्तों के समान कोमल हैं। आशय यह है कि कुछ नन्हे-नन्हे बच्चे भी अगरबत्ती बनाने के काम में लगे हुए हैं। कुछ हाथ ऐसे हैं जिनमें से जूही की डालों जैसी खुशबू आती है। आशय यह है कि कुछ सुंदर युवतियाँ भी अगरबत्तियाँ बनाने में लगी हुई हैं।

(ii) दुनिया की सारी गंदगी के बीच
      दुनिया की सारी खुशबू
      रचते रहते हैं हाथे
उत्तर:
यद्यपि अगरबत्ती बनाने वाले कारीगर दुनिया भर को सुगंधित अगरबत्ती प्रदान करते हैं और वातावरण में सुगंध फैलाते हैं किंतु उन्हें स्वयं दुनिया भर की गंदगी के बीच रहना पड़ता है। उनके चारों ओर गंदगी का ही साम्राज्य रहता है। वे शोषित हैं, पीड़ित हैं।

(ख) कवि ने इस कविता में ‘बहुवचन’ का प्रयोग अधिक किया है? इसका क्या कारण है?
उत्तर:
कविता में ‘हाथ’ के लिए बहुवचन का प्रयोग किया गया है। इसके माध्यम से कवि बताना चाहता है कि यहाँ एक कारीगर या एक मजदूर की बात नहीं की जा रही। यह समस्या सब मज़दूरों की है।

(ग) कवि ने हाथों के लिए कौन-कौन से विशेषणों का प्रयोग किया है?
उत्तर:
कवि ने हाथों के लिए निम्नलिखित विशेषणों का प्रयोग किया है-

उभरी नसोंवाले
घिसे नाखूनोंवाले
पीपल के पत्ते-से नए-नए
जूही की डाल-से खुशबूदार
गंदे कटे-पिटे
ज़ख्म से फटे हुए।

योग्यता-विस्तार

प्रश्न 1.
अगरबत्ती बनाना, माचिस बनाना, मोमबत्ती बनाना, लिफ़ाफ़े बनाना, पापड़ बनाना, मसाले कूटना आदि लघु उद्योगों के विषय में जानकारी एकत्रित कीजिए।
उत्तर:
आस पड़ोस में रहने वाले किसी मज़दूर या कर्मचारी से बात करके जानिए और उनकी फैक्ट्री में जाकर देखिए। संभव हो तो घर में बनाने का प्रयास कीजिए।

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Online Education HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom

These Solutions are part of Online Education HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 4 Structure of the Atom

Question 1.
Both helium (He) and beryllium (Be) have two valence electrons. Whereas He represents a noble gas element, Be does not. Assign reason.
Answer:
The element He (Z = 2) has two electrons present in the only shell i.e., K-shell. Since this shell can have a maximum of two electrons only therefore, He is a noble gas element. The element Be (Z = 4) has electronic configuration as : 2, 2. Although the second shell has also two electrons but it is not complete. It can still accommodate six more electrons. Therefore, the element beryllium does not represent a noble gas element.

More Resources

• HOTS Questions for Class 9 Science
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• Value Based Questions in Science for Class 9
• NCERT Exemplar Solutions for Class 9 Science
• Previous Year Question Papers for CBSE Class 9 Science

Question 2.
Study the data given below and answer the questions which follow :

1. Write the mass number and atomic number of the particles A, B, C and D.
2. Which represent a pair of isotopes ?

Answer:

1. Particle A : Mass number = 7 ; Atomic number = 3
   Particle B : Mass number = 17 ; Atomic number = 9
   Particle C : Mass number =16, Atomic number = 8
   Particle D : Mass number =18, Atomic number = 8
2. Particles C and D represent a pair of isotopes since they have same atomic number.

Question 3.
Which of the two will be chemically more reactive ; element X with atomic number 17 or element Y with atomic number 16 ?
Answer:
The electronic configuration of the two elements are as follows :
X(Z = 16): K (2), L(8), M(6) ;
Y(Z = 17): K(2), L(8), M(7)
The element X needs two electrons in the M shell to have the noble gas configuration of element, Ar (Z = 18). However, the element Y needs only one electron to achieve this. This means that the element Y has a greater urge or desire to take up one electron from an outside atom. It is therefore, more reactive than the element X which needs two electrons.

Question 4.
The number of protons, neutrons and electrons in particles from A to E are given below :

1. Which one is a cation ?
2. Which one is an anion ?
3. Which represent pair of isotopes ?

Answer:

1. B is a monovalent cation (B⁺)
2. E is a monovalent anion (E^–)
3. A and D represent pair of isotopes.

Question 5.
An atom of an element has three electrons in the third shell which is the outermost shell. Write

1. the electronic configuration
2. the atomic number
3. number of protons
4. valency
5. the name of the element
6. its nature whether metal or non-metal. (CBSE 2012)

Answer:
The third shell is M shell. If the atom of the element has three electrons in the third shell, this means that K and L shells are already filled.

1. Electronic configuration : 2, 8, 3.
2. Atomic number = No. of electrons =13
3. Number of protons = No. of electrons =13
4. Valency of the element = 3
5. The element with Z = 13 is aluminium (Al)
6. It is a metal.

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Online Education HOTS Questions for Class 9 Science Chapter 11 Work, Power and Energy

These Solutions are part of Online Education HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 11 Work, Power and Energy

Question 1.
A light and a heavy object have the same momentum. What is the ratio of their kinetic energy ?
(CBSE 2011, 2012)
Answer:

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• NCERT Solutions for Class 9 Science
• Value Based Questions in Science for Class 9
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• Previous Year Question Papers for CBSE Class 9 Science

Question 2.
Can any object have mechanical energy even if its momentum is zero ? Explain.
(NCERT Question Bank, CBSE 2011)
Answer:

p = 0, then mechanical energy of the object = mgh = P.E.
Thus, a stationary object at a height h above the surface of earth has mechanical energy even if its momentum is zero.

Question 3.
Can any object have momentum even if its mechanical energy is zero ? Explain.
(NCERT Question Bank ; CBSE 2011)
Answer:
Mechanical energy = K.E. + P.E.
Since mechanical energy is zero, so both K.E. and P.E are zero. Kinetic energy is zero means velocity of the object is zero. Hence, momentum of the object is also zero.

Question 4.
Compare the momentum of two objects of mass 10 kg and 40 kg respectively but having same kinetic energy.
Answer:

Question 5.
Compare the kinetic energies of two objects of masses 10 kg and 50 kg respectively but having same momentum. (CBSE 2015)
Answer:

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Online Education NCERT Solutions For Class 9 Science Chapter 3 Atoms and Molecules

These Solutions are part of Online Education NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Chemistry) Chapter 3 – Atoms and Molecules solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 3 – Atoms and Molecules Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT IN TEXT PROBLEMS

IN TEXT QUESTIONS

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6.0 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. (CBSE 2012)
Answer:
The chemical reaction leading to products is :
sodium carbonate + ethanoic acid ———–> sodium ethanoate + carbon dioxide + water.
Mass of reactants = (5.3 + 6.0) = 1.3 g
Mass of products = (8.2 + 2.2 + 0.9) = 11.3 g.
The reactants and products have same mass. This means that there was no loss of mass during the reaction. Therefore, the data is in agreement with law of conservation of mass.

Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react with 3 g of hydrogen gas ?
Answer:
According to available data,
Mass of oxygen combining with 1 g of hydrogen = 8 g.

Question 3:
Which postulate of Dalton’s Atomic theory is the basis of law of conservation of mass ? (CBSE 2012)
Answer:
The law of conservation mass is based on the following postulate of Dalton’s Atomic theory.
“Atoms can neither be created nor destroyed during a physical change or a chemical reaction.”

Question 4.
Which postulate of Dalton’s Atomic theory can explain the law of definite proportions ?
Answer:
The law of definite proportions is based on the following postulate of Dalton’s Atomic theory.
“All atoms of a particular element are identical in every respect. This means that they have same mass, same size and also same chemical properties.”

Question 5.
Define atomic mass unit.
Answer:
Atomic mass unit may be defined as :
The mass of one-twelfth (1/12) of the mass of one atom of carbon taken as 12 u. It is represented as 1 u (unified mass).

Question 6.
Why is not possible to see an atom with naked eye ?
Answer:
It is not possible to see an atom with naked eye because of its extremely small size. For example, the radius of an atom of hydrogen is of the order of 10^-10 m. Actually an atom is regarded as a microscopic particle. These microscopic particles cannot be seen with naked eye.

Question 7.
Write down the formulae of :
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide.
Answer:

Question 8.
Write the names of the compounds represented by the following formulae :
(i) Al₂(SO₄)₃
(ii) CaCl₂
(Hi) K₂SO₄
(iv) KNO₃
(v) CaCO₃.
Answer:
(i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

Question 9.
What is meant by the term chemical formula ?
Answer:
Molecule represents a group of two or more atoms (same or different) chemically bonded to each other and held tightly by strong attractive forces. Molecules are represented in terms of symbols of constituting atoms and it is known as chemical formula.

Question 10.
How many atoms are present in
(i) H₂S molecule
(ii) PO₄^3- ion ?
Answer:
(i) Three
(ii) Five.

Question 11.
Calculate the molecular masses of :
(i) H₂
(ii) O₂
(iii) Cl₂
(iv) CO₂
(v) CH₄
(vi) C₂H₆
(vii) C₂H₄
(viii) NH₃
(ix) CH₃OH.
Answer:
(i) Hydrogen (H₂)
Molecular mass of H₂ = 2 x Atomic mass of H = (2 x 1 u) = 2 u.
(ii) Oxygen (O₂)
Molecular mass of O₂ = 2 x Atomic mass of O = (2 x 16 u) = 32 u.
(iii) Chlorine (CI₂)
Molecular mass of Cl₂ = 2 x Atomic mass of Cl = (2 x 35’5 u) = 71 u.
(iv) Carbon dioxide (CO₂)
Molecular mass of CO₂ = (1 x Atomic mass of C) + (2 x Atomic mass of O)
= (1 x 12 u) + (2 x 16 u) = 12 u + 32 u = 44 u.
(v) Methane (CH₄)
Molecular mass of CH₄ = ( 1 x Atomic mass of C) + (4 x Atomic mass of H)
= ( 1 x 12 u) + (4 x 1 u) = 16 u.
(vi) Ethane (C₂H₆)
Molecular mass of C₂H₆ = (2 x Atomic mass of C) + (6 x Atomic mass of H)
= (2 x 12 u) + (6 x 1 u) = 30 u.
(vii) Ethylene (C₂H₄)
Molecular mass of C₂H₄ =(2 x Atomic mass of C) + (4 x Atomic mass of H)
= (2 x 12 u) + (4 x 1 u) = 28 u.
(viii) Ammonia (NH₃)
Molecular mass of NH₃ = (1 x Atomic mass ofN) + (3 x Atomic mass of H)
= (1 x 14 u) + ( 3 x 1 u) = 17 u.
(ix) Methyl alcohol (CH₃OH)
Molecular mass of CH₃OH = (1 x Atomic mass of C) + (4 x Atomic mass of H)
+ (1 x Atomic mass of O)
= (1 x 12 u) + (4 x 1 u) + (1 x 16 u) = 32 u.

Question 12.
Calculate the formula unit mass of :
(i) ZnO
(ii) Na₂O
(iii) K₂CO₃.
Given : Atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u and O = 16 u.
Answer:
(i) Formula unit mass of ZnO (Zinc oxide)
= (1 x Atomic mass of Zn) +(1 x Atomic mass of O)
= (1 x 65 u) + (1 x 16 u) = 81 u.
(ii) Formula unit mass of Na₂O (Sodium oxide)
= (2 x Atomic mass of Na) + (1 x Atomic mass of O)
= (2 x 23 u) + (1 x 16 u) = 62 u.
(iii) Formula unit mass of K₂CO₃ (Potassium carbonate).
= (2 x Atomic mass of K) + (1 x Atomic mass of C) + (3 x Atomic mass of O)
= (2 x 39 u) + (1 x 12 u) + (3 x 16 u) = 138 u

Question 13.
Calculate the number of moles of the following :
(i) 52 g of He
(ii) 12.044 x 10²³ atoms of He.
Answer:

Question 14.
Calculate the number of particles in each of the following :
(i) 46 g of sodium atoms
(ii) 8 g of oxygen (O₂)
(iii) 0.1 mole of carbon atoms.
Answer:

Question 15.
If one mole of carbon weighs 12 grams, what is the mass (in gram) of one atom of carbon ?
Answer:

Question 16.
Which has more number of atoms ?
(a) 100 grams of sodium
(b) 100 grams of iron
(Given : atomic mass of Na = 23 u ; Fe = 56 u)
Answer:

NCERT END EXERCISE

Question 1.
0.24 g of sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron
and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:

Question 2.
When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is formed. What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen ? Which law of chemical combination will govern your answer ? (CBSE 2011, 2012)
Answer:
Carbon and oxygen react to form carbon dioxide according to the equation Carbon (C) + Oxygen (O₂) > Carbon dioxide (CO₂)

In the first case :
3.0 g of carbon are burnt in 8.0 g of oxygen to form 11.0 g of CO₂ In the second case :
3.0 g of carbon must also combine with 8.0 g of oxygen only. This means that (50 – 8) = 42 g of oxygen will remain unreacted.
The mass of CO₂ in this case must be also 11 g.
The answer is based on Law of constant proportions.
In the second case, only 8.0 g of oxygen react although 50.0 g are available. This shows that the mass of carbon dioxide (11.0 g) formed depends upon the mass of carbon (3.0 g) or the substance present in smaller amount. In general, the substance (element or compound) present in smaller amount in a reaction limits the participation of the other reactants. It is quite often called limiting reactant. Carbon is the limiting reactant in this case. It limits the participation of oxygen and also the formation of carbon dioxide.

Question 3.
What are polyatomic ions ? Give examples. (CBSE 2015)
Answer:
Polyatomic ions are the group of atoms which carry either positive charge (cations) or negative charge (anions). For example,

1. Carbonate ion (CO₃)^2-
2. Nitrate ion (NO₃)^–
3. Ammonium ion (NH₄)⁺
4. Phosphate ion (PO₄)^3-.

Question 4.
Write the chemical formulae of the following :
Answer:

Question 5.
Give the names of the elements present in the following compounds :
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer:
The names of elements present can be given only if the chemical formula of the compound is known. For example,
(a) Quick lime : It is the commercial name of the compound. Its chemical name is calcium oxide and the
chemical formula is CaO.
Elements present : calcium (Ca) : oxygen (O).
(b) Hydrogen bromide : The chemical formula of the compound is HBr
Elements present : hydrogen (H) ; bromine (Br).
(c) Baking powder : It is the commercial name of the compound. Its chemical name is sodium hydrogen carbonate and the chemical formula is NaHCO₃
Elements present : sodium (Na), hydrogen (H), carbon (C), oxygen (O).
(d) Potassium sulphate : The chemical formula of the compound is K₂SO₄
Elements present : potassium (K), sulphur (S), oxygen (O).

Question 6.
Calculate the molar mass of the following substances :
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Nitric acid, HNO₃
(e) Hydrochloric acid, HCl.
Answer:
(a) Ethyne, C₂H₂
Molar mass of C₂H₂ = (2 x Atomic mass of C) + (2 x Atomic mass of H)
= (2 x 12 u) + (2 x 1 u) = 26 u.
(b) Sulphur molecule, S₈
Molar mass of S₈ = 8 x Atomic mass of S = (8 x 32 u) = 256 u
(c) Phosphorus molecule, P₄
Molar mass of P₄ =4 x Atomic mass of P = (4 x 31 u) = 124 u
(d) Nitric acid, HNO₃
Molar mass of HNO₃ = (1 x Atomic mass of H) + (1 x Atomic mass of N) + ( 3 x Atomic mass of O)
= (1 x 1 u) + (1 x 14 u) + (3 x 16 u) = 63 u.
(e) Hydrochloric acid, HCl.
Molar mass of HCl = (1 x Atomic mass of H) + (1 x Atomic mass of Cl)
= (1 x 1 u) + (1 x 35.5 u) = 36.5 u.

Question 7.
What is the mass of :
(a) 1 mole of nitrogen atoms ?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27) ?
(c) 10 moles of sodium sulphite (Na₂SO₃) ?
Answer:
(a) 1 mole of nitrogen atoms
Mass of 1 mole of nitrogen (N) atoms = 14 u
(b) 4 moles of aluminium atoms
Mass of 1 mole of aluminium (Al) atoms = 27 u
Mass of 4 moles of aluminium (Al) atoms = 4 x 27 = 108 u
(c) 10 moles of sodium sulphite (Na₂SO₃)
Molar mass of Na₂SO₃ = 2 x Atomic mass of Na + Atomic mass of S + 3 x Atomic mass of O
= 2 x 23 + 32 + 3 x 16 = 126 u
1 mole of sodium sulphite has mass = 126 u
10 moles of sodium sulphite have mass = 10 x 126 = 1260 u.

Question 8.
Convert into moles
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
Answer:

Question 9.
What is the mass of :
(a) 0.2 mole of oxygen atoms ?
(b) 0.5 mole of water molecules ?
Answer:
(a) 0.2 mole of oxygen atoms
Mass of 1 mole of oxygen (O) atoms = 16 u
Mass of 0.2 mole of oxygen (O) atoms = 0.2 x 16 = 3.2 u
(b) 0.5 mole of water molecules
Mass of 1 mole of water (H₂O) molecules = 2 x 1 + 16 = 18 u
Mass of 0.5 mole of water (H2O) molecules = 0.5 x 18 = 9 u.

Question 10.
Calculate the number of molecules of sulphur (Sg) present in 16 g of solid sulphur.
Answer:

Question 11.
Calculate the number of aluminium ions in 0.051 g of aluminium oxide (Al₂O₃). (CBSE 2012)
Answer:

VERY SHORT ANSWER QUESTIONS

Question 1.
Out of atoms and molecules, which can exist independently ?
Answer:
Molecules can exist independently. However, the atoms of noble gases (He, Ne, Ar, Kr, Xe) can also exist independently.

Question 2.
What does the symbol ‘u’ represent ?
Answer:
The symbol V represents unified mass.

Question 3.
Write the chemical symbols and Latin names of
(i) gold
(ii) mercury ?
Answer:
(i) Au (Aurum)
(ii) Hg (Hydrargyrum).

Question 4.
Are the mass of a molecule of a substance and its molar mass same ?
Answer:
No, they are different.

Question 5.
How are mass, molar mass and number of moles of a substance related to each other ?
Answer:

Question 6.
Avogardo’s number represents how many particles ?
Answer:
Avogadro’s number (Ng) represents 6.022 x 10²³ particles.

Question 7.
Give an example of
(i) a divalent anion
(ii) a trivalent cation
(iii) a monovalent anion.
Answer:
(i) (SO₄)^2-
(ii) Al₃⁺
(iii) Cl^–

Question 8.
Calculate the molar mass of ethyl alcohol (C₂H₅OH).
Answer:
Molar mass of C₂H₅OH= (2 x gram atomic mass of C) + (6 x gram atomic mass of H)
+ (1 x Atomic mass of O)
= (2 x 12 g) + (6 x 1 g) + (1 x 16 g) = 46 g.

Question 9.
If one mole of oxygen atoms weigh 16 grams, calculate the mass of one atom of oxygen (in grams).
(CBSE 2014)
Answer:

Question 10.
What is the valency of calcium in CaCO₃ ?
Answer:
The valency of Ca in CaCO₃ is 2+ (i.e. Ca²⁺).

Question 11.
What happens to an element ‘A’ if its atom gains two electrons ?
Answer:
It changes to a divalent anion (A^2-).

Question 12.
Why is a cation so named ?
Answer:
When electric current is passed through the solution of a salt like sodium chloride (NaCl), the positive ion (Na⁺) migrates towards cathode (negative electrode). It is therefore, called cation. Please remember that
• Positive ion migrating towards cathode on passing electric current is known as cation.
• Negative ion migrating towards anode on passing electric current is known as anion.

Question 13.
An element Z forms an oxide with formula Z₂O₃. What is its valency ?
Answer:
The valency of the element Z in Z₂O₃ is 3+.

Question 14.
The valency of an element A is 4. Write the formula of its oxide.
Answer:
The formula of the oxide is A₂O₄ or AO₂.

Question 15.
An element X has valency 3 while the element Y has valency 2. Write the formula of the compound between X and Y.
Answer:
The formula of the compound between X and Y is X₂Y₃.

Question 16.
Formula of the carbonate of a metal M is M₂CO₃. Write the fomula of its chloride.
Answer:
The valency of the metal (M) in M₂CO₃ is (1+) i.e. metal exists as M+ ion. Therefore, the formula of metal chloride is MCl.

Question 17.
What do you understand from the statement “relative atomic mass of sulphur is 32”. (CBSE 2012)
Answer:
This means that an atom of sulphur is 32 times heavier as compared to 1/12 of the mass of 1 atom of C — 12(1 u).

Question 18.
Calculate the formula unit mass of CaCl₂.
Answer:
Formula unit mass of CaCl₂ (Calcium chloride)
= (1 x Atomic mass of Ca) + (2 x Atomic mass of Cl)
= (1 x 40 u) + (2 x 35.5 u) = 111 u.

Question 19.
Which of the following represents the correct chemical formula ?
(a) NaSO₄
(b) CaPO₄
(c) ZnS
(d) AlSO₄.
Answer:
The formula (c) represents the correct formula. Both the ions are divalent i.e. Zn²⁺ and S^2-. The name of the compound is zinc sulphide.

Question 20.
Sample A contains one gram molecules of oxygen molecules and sample B contains one mole of oxygen molecules. What is the ratio of the number of molecules in both the sample ?
Answer:
One gram molecules and one gram mole contain the same number of molecules (6.022 x 10²³). Therefore, the ratio is 1 : 1.

Question 21.
Gram molecular mass of ammonia (NH₃) is 17 g. Is it correct to regard it as formula unit mass also ?
Answer:
No, it is not correct. Ammonia exists in molecular form and is not an ionic compound made up of cation and anion. Therefore, it cannot have formula unit mass. It has only molecular mass.

Question 22.
Give one example each of polyatomic element and polyatomic ion.
Answer:
Polyatomic element (P₄) ; Polyatomic ion (SO₄^2-).

Question 23.
Name the element which is used as a reference for the atomic masses of the elements.
Answer:
1/12 of the mass of carbon atom taken as 12 u is used as a reference for the atomic masses of the elements.

Question 24.
Four samples of water [H₂O] are collected from different sources. Each sample on analysis was found to contain same percentage of oxygen. Which law of chemical combination is demonstrated by the above observation ?
(CBSE 2014)
Answer:
The law of constant combination is demonstrated by this observation.

SHORT ANSWER QUESTIONS

Question 25.
List the elements present in
(i) quick lime
(ii) sodium hydrogen carbonate.
Answer:
(i) The chemical name of quick lime is calcium oxide. Its chemical formula is CaO. Elements present are Ca and O.
(ii) The chemical formula of sodium hydrogen carbonate is NaHCO₃. The elements present are Na, H, C and O.

Question 26.
Cenvert into moles :
(i) 20 g of water
(ii) 20 g of carbon dioxide
Answer:

Question 27.
(a) How many moles are present in 11.5 g of sodium ?
(b) The mass of an atom of element (X) is 2.0 x 10^-23 g. Calculate its atomic mass.
Answer:

Question 28.
(a) How many particles are represented by 0.25 mole of an element ?
(b) Out of 4 g of methane (CH₄) and llg of CO₂, which has more molecules ?
Answer:

Question 29.
What is the number of molecules present in 1.5 mole of ammonia (NH₃) ?
Answer:

Question 30.
Write the chemical symbols of two elements
(a) Which are formed from the first letter of the elements name.
(b) Whose name has been taken from the names of the elements in Latin.
Answer:
(a) Boron (B), carbon (C)
(b) Argentum (Ag), Kalium (K).

Question 31.
(a) Four samples of carbon dioxide (CO₂) were prepared by using different methods. Each sample on analysis was found to contain 27.27% carbon by mass. Name the law which is in agreement with this observation.
(b) Explain why the number of atoms in one mole of hydrogen gas is double the number of atoms in one mole of helium gas.
Answer:
(a) Carbon dioxide consists of elements carbon and oxygen. Since the percentage of carbon in each sample is fixed, that of oxygen must be also fixed. This is according to law of constant proportions.
(b) Hydrogen gas is diatomic in nature (H₂) while helium gas is monoatomic (He). As a result, the number of atoms in one mole of hydrogen. (2 x N_A) are expected to be double as compared to number of atoms in one mole of helium (N_A).

Question 32.
10²² atoms of an element ‘X’ are found to have a mass of 930 mg. Calculate the molar mass of the element ‘X’?
Answer:

Question 33.
(a) If the valency of carbon is 4 and that of sulphur is 2, write the formula of the compound formed between carbon and sulphur atoms. Also name the compound.
(b) What is wrong with the statement T mole of hydrogen ?
Answer:
(a) The formula of compound can be written by exchanging the valencies (cross-over). Therefore, the expected formula is C₂S₄ or CS₂. The compound is called carbon disulphide.
(b) The statement is not correct. We must always write whether hydrogen is in atomic form or molecular form. The correct statement is :

Question 34.
T mole of hydrogen atoms or one mole of hydrogen molecules.
Identify the cations and anions in the following compounds :
(a) CH₃COONa
(b) NH₃
(c) NH₄Cl
(d) SrCl₂.
Answer:
(a) CH₃COO^–, Na⁺
(b) It is a molecular compound
(c) NH₄⁺ ,Cl^– (d) Sr²⁺, 2Cl^–

Question 35.
Classify the following based on atomicity
(a) O₃
(b) P₄
(c) S₈. (CBSE 2012)
Answer:
Atomicity is the number of atoms present in one molecule of an element. Based upon this, the atomicity of different molecules may be expressed as :
(a) 3
(b) 4
(c) 8.

Question 36.
Give the formulae of the compounds that will be formed from the following sets of elements.
(a) Calcium and fluorine
(b) Magnesium and oxygen
(c) Sodium and sulphur
(d) Carbon and chlorine
(e) Carbon and sulphur
(f) Nitrogen and hydrogen.
Answer:
(a) CaF₂
(b) MgO
(c) Na₂S
(d) CCl₄
(e) CS₂
(f) NH₃.

Question 37.
Which of the following symbols of elements are incorrect. Write correct symbols
(a) Iron (Ir)
(b) Gold (Au)
(c) Manganese (M)
(d) Potassium (Po)
(e) Zinc (ZN)
(f) Calcium (Ca).
Answer:
(a) Iron (Fe)
(c) Manganese (Mn)
(d) Potassium (K)
(e) Zinc (Zn).

Question 38.
Verify by calculation that :
4 moles of CO₂ and 6 moles of H₂O do not have same mass in grams.
Answer:
Molar mass of CO₂ = 12 + 2 x 16 = 44 g
1 mole of CO₂ has mass = 44 g
4 moles of CO₂
have mass = 44 x 4 = 176 g
Molar mass of H₂O = 2 x 1 + 16 =18 g
1 mole of H₂O has mass = 18 g
6 moles of H₂O have mass = 6 x 18 = 108 g.
This shows that these have different masses in grams.

Question 39.
A sample of vitamin C contains 2.48 x 10²⁵ oxygen atoms. How many moles of oxygen atoms are present in the sample ?
Answer:

Question 40.
Calculate the total number of ions in 0.585 g of sodium chloride.
Answer:

Question 41.
A flask contains 4.4 g of CO₂ gas. Calculate
(a) How many moles of CO₂ gas does it contain ?
(b) How many molecules of CO₂ gas are present in the sample.
(c) How many atoms of oxygen are present in the given sample.
(Atomic mass of C = 12 u. O = 16 u)
Answer:

Question 42.
Determine the molecular mass of :
(i) NH₄OH
(ii) K₂CO₃
(iii) CH₃COOH Given atomic masses :
H = 1.0 u ; O = 16.0 u ; C = 12.0 ; K = 39.0 u ; N = 14.0 u
Answer:
(i) Molecular mass of NH₄OH
= Atomic mass of N + 5 x (Atomic mass of H) + Atomic mass of O
= 14 + 5 x 1 + 16 = 14 + 5 + 16 = 35 u
(ii) Molecular mass of K₂CO₃
= 2 x (Atomic mass of K) + Atomic mass of C + 3 x (Atomic mass of O)
= 2 x 39 + 12 + 3 x 16 = 138 u
(iii) Molecular mass of CH₃COOH
= 2 x (Atomic mass of C) + 4 x (Atomic mass of H) + 2 x (Atomic mass of O)
= 2 x 12 + 4 x 1+2 x 16 = 60 u

Question 43.
(a) What is Avogadro Constant ?
(b) Calculate the number of particles present in 56 g of N₂ molecule.
Answer:
(a) Avogadro’s number is the number of particles (atoms, ions, molecules etc.) present in one mole of any substance. It is denoted either as N_A or as N_Q. The number is also called Avogadro’s constant because its value is fixed (6.022 x 10²³) irrespective of the nature of the particles.
(b) Molar mass of N₂ = 14 x 2 = 28 g
28 g of N₂ have molecules or particles = N_A = 6.022 x 10²³
56 g of N₂ have molecules or particles = N_A = 6.022 x 10²³ x 2 = 1.204 x 10²⁴

Question 44.
(a) An element ‘X’ exhibits variable valencies 3 and 5. Write the formulae of the chlorides of the element.
(b) What is the ratio by mass of the elements present in the chemical formula of magnesium oxide.
Answer:
(a) The formulae of the chlorides of the element ‘X’ = XCl₅ and XCl₃.
(b) The chemical formula of magnesium oxide is MgO. The elements are present in the ratio of 24 : 16 or 3 : 2.

Question 45.
Give the example of trivalent cation and monovalent anion. Write the formula of the compound formed by their combination. .
Answer:
Trivalent cation : Al³⁺ ; Monovalent anion : Cl^–
Formula of the compound : AlCl₃.

Question 46.
(a) State the law of conservation of mass
(b) What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate if the law of conservation of mass is true ?
Answer:
(a) The total mass of the products in a physical change or a chemical reaction is equal to the total mass of the reactants that have combined. The law may also be stated in another form. The mass can neither he created nor destroyed in a physical change or a chemical reaction.
(b) The chemical reaction leading to products is :
Mass of reactants = x g + 5.85 g Mass of products = 14.35 g + 8.5 g According to law of conservation of mass
Mass of reactants = Mass of products . (x + 5.85) g = (14.35 + 8.5) g
x = (22.85 – 5.85) g = 17.0 g

LONG ANSWER QUESTIONS

Question 47.
(a) The mass of one molecule of a substance is 4.65 x 10^-23 g. What is its molecular mass ? What could this substance be ?
(b) Which have more molecules? 10 g of sulphur dioxide (SO₂) or 10 g of oxygen (O₂) ?
Answer:

Question 48.
Which has more atoms ?
(a) 10 g of nitrogen (N₂)
(b) 10 g of ammonia (NH₃)
Answer:

Question 49.
(a) Explain with the help of a labelled diagram an activity for the verification of law of conservation of mass.
(b) Find the number of atoms present in 100 g of sodium and 100 g of iron. (Given that Na = 23 u ; Fe = 56 u ;
Answer:
(a) The total mass of the products in a physical change or a chemical reaction is equal to the total mass of the reactants that have combined. The law may also be stated in another form. The mass can neither he created nor destroyed in a physical change or a chemical reaction.
In other words, the mass remains unchanged or conserved in a chemical reaction. The law is also known as the Law of Indestructibility of Matter.
Explanation : Let us try to analyse as to what happens in a chemical reaction. To understand the same, let us consider a chemical reaction between barium chloride and sodium sulphate. When the solutions of these reactants prepared separately in water are mixed, the following chemical reaction takes place :
Barium chloride + Sodium sulphate———– > Barium sulphate + Sodium chloride (white precipitate)
If we look at this reaction, we find that the exchange of constituents has taken place between the reactants, i.e. chloride part of barium chloride has been exchanged by sulphate part of sodium sulphate. As such, no loss or gain in mass is expected. In other words, the mass remains conserved or does not change.

Question 50.
(a) Give one point of difference between an atom and an ion.
(b) Give one example each of a polyatomic cation and a polyatomic anion.
(c) Identify the correct chemical name of FeSO₄ from the given names Ferrous sulphate, Ferrous sulphide, Ferrous sulphite.
(d) Write the chemical formula for the chloride of magnesium.
Answer:
(a) An atom is always neutral in the sense that it does not carry any charge. An ion carries either positive charge (cation) or negative charge (anion).
(b) Ammonium (NH₄)⁺ ion is a polyatomic cation while sulphate (SO₄)^2- ion is a polyatomic anion.
(c) Correct chemical name of FeSO₄ is ferrous sulphite.
(d) Chemical formula for the chloride of magnesium is MgCl₂.

Question 51.
(a) What do the following observations stand for ?
(i) 2O
(ii) 3O₂
(b) Which amongst the following has more number of atoms and how much ?
(i) 11.5 g of sodium
(ii) 15.0 g of calcium
Answer:
(a) (i) 2O represent two atoms of oxygen (or the gram atoms of oxygen
(ii) 3O₂ represent three molecules of oxygen or three gram moles of oxygen.

NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

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