{"id":821880,"date":"2024-04-17T14:47:53","date_gmt":"2024-04-17T09:17:53","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=821880"},"modified":"2024-04-17T14:47:53","modified_gmt":"2024-04-17T09:17:53","slug":"ncert-solutions-class-9-science-chapter-3","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-9-science-chapter-3\/","title":{"rendered":"NCERT Solutions Class 9 Science Chapter 3 ‘Atoms and Molecules’ (Free PDF)\u00a0"},"content":{"rendered":"\n

We are providing NCERT Solutions Class 9 Science Chapter 3 ‘Atoms and Molecules’ to help you navigate through your school exams. You can also download a PDF for important questions and answers for quick revision. Let us get started!     <\/p>\n\n\n\n\n\n\n

Download the NCERT Solutions of Class 9 Science Chapter 3 PDF Here!<\/strong> <\/a><\/p>\n\n\n\n

Download Solutions of all the Chapters of Class 9 Science<\/strong>: <\/p>\n\n\n\n

Chapter 1<\/strong><\/td>Chapter 2<\/strong><\/td>Chapter 3<\/td>Chapter 4<\/strong><\/td>Chapter 5<\/strong><\/td>Chapter 6<\/strong><\/td><\/tr>
Chapter 7<\/strong><\/td>Chapter 8<\/strong><\/td>Chapter 9<\/strong><\/td>Chapter 10<\/strong><\/td>Chapter 11<\/strong><\/td>Chapter 12<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

NCERT Solutions Class 9 Science Chapter 3 Atoms and Molecules <\/h2>\n\n\n\n

Here are NCERT Solutions of Class 9 Science Chapter 3 ‘Atoms and Molecules\u2019 to the questions in the exercise section of the lesson.     <\/p>\n\n\n\n

1. A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Mass of the sample compound = 0.24g, <\/p>\n\n\n\n

mass of boron = 0.096g, mass of oxygen = 0.144g<\/p>\n\n\n\n

To calculate the percentage composition of the compound,<\/p>\n\n\n\n

Percentage of boron = mass of boron \/ mass of the compound x 100<\/p>\n\n\n\n

= 0.096g \/ 0.24g x 100  = 40%<\/p>\n\n\n\n

Percentage of oxygen = 100 \u2013 percentage of boron<\/p>\n\n\n\n

= 100 \u2013 40 = 60%<\/p>\n\n\n\n

2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen?  Which law of chemical combination will govern your answer?<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. <\/p>\n\n\n\n

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide. <\/p>\n\n\n\n

C + O2 \u2192 CO2<\/p>\n\n\n\n

As per the given condition, when 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.<\/p>\n\n\n\n

3g + 8g \u219211 g ( from the above reaction)<\/p>\n\n\n\n

The total mass of reactants = mass of carbon + mass of oxygen<\/p>\n\n\n\n

=3g+8g<\/p>\n\n\n\n

=11g<\/p>\n\n\n\n

The total mass of reactants = Total mass of products<\/p>\n\n\n\n

Hence, the law of conservation of mass is proved. <\/p>\n\n\n\n

 It depicts that carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.<\/p>\n\n\n\n

Thus, it further proves the law of constant proportions.<\/p>\n\n\n\n

3 g of carbon must also combine with 8 g of oxygen only.<\/p>\n\n\n\n

This means that (50\u22128)=42g of oxygen will remain unreacted.<\/p>\n\n\n\n

The remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of carbon dioxide will be formed<\/p>\n\n\n\n

The law of constant proportions governs the above answer.<\/p>\n\n\n\n

3. What are polyatomic ions? Give examples.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Ions that contain more than one atom, but behave as a single unit are known as polyatomic ions. <\/p>\n\n\n\n

Example: CO32-, H2PO4\u2013<\/p>\n\n\n\n

4. Write the chemical formula of the following.<\/strong><\/p>\n\n\n\n

(a) Magnesium chloride<\/strong><\/p>\n\n\n\n

(b) Calcium oxide<\/strong><\/p>\n\n\n\n

(c) Copper nitrate<\/strong><\/p>\n\n\n\n

(d) Aluminium chloride<\/strong><\/p>\n\n\n\n

(e) Calcium carbonate<\/strong><\/p>\n\n\n\n

Solution<\/strong>: <\/p>\n\n\n\n

(a) Magnesium chloride \u2013 MgCl2<\/p>\n\n\n\n

(b) Calcium oxide \u2013 CaO<\/p>\n\n\n\n

(c) Copper nitrate \u2013 Cu(NO3)2<\/p>\n\n\n\n

(d) Aluminium chloride \u2013 AlCl3<\/p>\n\n\n\n

(e) Calcium carbonate \u2013 CaCO3<\/p>\n\n\n\n

5. Give the names of the elements present in the following compounds.<\/strong><\/p>\n\n\n\n

(a) Quick lime<\/strong><\/p>\n\n\n\n

(b) Hydrogen bromide<\/strong><\/p>\n\n\n\n

(c) Baking powder<\/strong><\/p>\n\n\n\n

(d) Potassium sulphate<\/strong><\/p>\n\n\n\n

Solution<\/strong>: <\/p>\n\n\n\n

(a) Quick lime \u2013 Calcium and oxygen (CaO) <\/p>\n\n\n\n

(b) Hydrogen bromide \u2013 Hydrogen and bromine (HBr) <\/p>\n\n\n\n

(c) Baking powder \u2013 Sodium, Carbon, Hydrogen, Oxygen (NaHCO3) <\/p>\n\n\n\n

(d) Potassium sulphate \u2013 Sulphur, Oxygen, Potassium (K2SO4) <\/p>\n\n\n\n

6. Calculate the molar mass of the following substances.<\/strong><\/p>\n\n\n\n

(a) Ethyne, C2H2<\/strong><\/p>\n\n\n\n

(b) Sulphur molecule, S8<\/strong><\/p>\n\n\n\n

(c) Phosphorus molecule, P4 (Atomic mass of phosphorus =31)<\/strong><\/p>\n\n\n\n

(d) Hydrochloric acid, HCl<\/strong><\/p>\n\n\n\n

(e) Nitric acid, HNO3<\/strong><\/p>\n\n\n\n

Solution<\/strong>: <\/p>\n\n\n\n

(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2\u00d712)+(2\u00d71)=24+2=26g<\/p>\n\n\n\n

(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8  x 32 = 256g<\/p>\n\n\n\n

(c) Molar mass of  Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g<\/p>\n\n\n\n

(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g<\/p>\n\n\n\n

(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+3\u00d716 = 63g<\/p>\n\n\n\n

Download the NCERT Solutions of Class 9 Science Chapter 3 PDF Here! <\/strong><\/a><\/p>\n\n\n\n

Download Solutions of all the Chapters of Class 9 Science<\/strong>: <\/p>\n\n\n\n

Chapter 1<\/strong><\/td>Chapter 2<\/strong><\/td>Chapter 3<\/td>Chapter 4<\/strong><\/td>Chapter 5<\/strong><\/td>Chapter 6<\/strong><\/td><\/tr>
Chapter 7<\/strong><\/td>Chapter 8<\/strong><\/td>Chapter 9<\/strong><\/td>Chapter 10<\/strong><\/td>Chapter 11<\/strong><\/td>Chapter 12<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Related Reads<\/strong>: <\/p>\n\n\n\n

Chemistry in Everyday Life<\/strong><\/td>All Branches of Chemistry with Examples, PDF, PPT<\/strong><\/a> <\/strong><\/td><\/tr>
What is the Importance and Scope of Chemistry?<\/strong> <\/strong><\/td>Basic Chemistry Concepts for Competitive Exams<\/strong> <\/strong><\/td><\/tr>
BSc Chemistry Honours: Syllabus, Eligibility, Career Scope<\/strong> <\/strong><\/td>Career in Chemistry: Top Courses, Universities & Scope<\/strong> <\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Explore Notes of All subjects of CBSE Class 9:<\/strong><\/p>\n\n\n\n

CBSE Notes Class 9 English<\/strong><\/td>CBSE Notes Class 9 History <\/strong><\/td>CBSE Notes Class 9 Geography<\/strong><\/td><\/tr>
CBSE Notes Class 9 Civics<\/strong><\/td>CBSE Notes Class 9 Mathematics<\/strong><\/td>CBSE Notes Class 9 Science<\/strong><\/a> <\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

FAQs<\/h2>\n\n\n\n
What is an atom Class 9 short answer Chapter 3?\u00a0<\/strong>

The smallest unit of matter or any element that exhibits its properties is known as an atom.<\/p> <\/div>

What is valency Class 9?\u00a0<\/strong>

The ability of an atom or the combining power of an atom to for bonds with other atoms is called valency.\u00a0<\/p> <\/div>

What is the full form of amu?\u00a0<\/strong>

The full form of amu is \u201cAtomic Mass Unit\u201d.\u00a0<\/p> <\/div> <\/div>\n\n\n\n

Follow Leverage Edu for complete study material on CBSE Notes of Class 9 Science<\/strong><\/a>. <\/p>\n","protected":false},"excerpt":{"rendered":"We are providing NCERT Solutions Class 9 Science Chapter 3 ‘Atoms and Molecules’ to help you navigate through…\n","protected":false},"author":109,"featured_media":821882,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-821880","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"\nNCERT Solutions Class 9 Science Chapter 3 'Atoms and Molecules' (Free PDF)\u00a0 | Leverage Edu Discover<\/title>\n<meta name=\"description\" content=\"Get important questions and answers in NCERT Solutions Class 9 Science Chapter 3 (Atoms and Molecules). 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