NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Edited By Shivani Poonia | Updated on Jul 18, 2025 08:46 AM IST

Have you ever wondered how things around us are made? Like the air we breathe, the phone we are currently using to read this article, or the food we eat daily to quench our appetite. The answer to all those questions is atoms and molecules. Everything in our environment, from the tiniest grain of sand to the vast universe, is made up of atoms and molecules. Atoms are the smallest unit of matter, while molecules are made up of a combination of two or more atoms, and they are so small that we cannot see them with the naked eye; to see them, we require a high-end microscope. Even a small drop of water contains countless molecules.

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Download PDF of NCERT Solutions for Class 9 Science Chapter 3
NCERT Solutions for Class 9 Science Chapter 3 (in-text Exercise Questions)
NCERT Solutions Class 9 Science Chapter 3( Exercise Questions with Answers)
NCERT Class 9 Science Chapter 3: Overview
Practice Questions for Class 9 Science Chapter 3
Approach to Solve Questions of Class 9 Science Chapter 3
Topics and Subtopics Covered in the NCERT Textbook
NCERT Solutions for Class 9 Science Chapter 3: Important Formulas and E-book.
NCERT Solutions for Class 9 Science- Chapter Wise
NCERT Solutions for Class 9 Subject Wise
NCERT Books and NCERT Syllabus

The important topics like laws of chemical combinations, chemical formulas, and molecular mass are all discussed in this chapter. The NCERT Solutions for Class 9 Science chapter 3 will offer a very comprehensive and systematic way to present these topics through a series of solved NCERT questions. These NCERT solutions provide clear and detailed explanations that will help you understand the topics in depth. The practice questions are also included in this article to improve your problem-solving skills. We have also added some points that will help you build a good approach for solving the questions.

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Download PDF of NCERT Solutions for Class 9 Science Chapter 3

Students can download the Class 9 science chapter 3 NCERT solutions PDF from the link below and can revise the concepts anytime. You can also learn these topics from the atoms and molecules class 9 notes before attempting the questions.

Download PDF

NCERT Solutions for Class 9 Science Chapter 3 (in-text Exercise Questions)

All the in-text exercise questions with detailed solutions are given below. Atoms and molecules important questions and answers are present in this section.

Exercise-3.1 (Page: 27)

Question.1 In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water

Answer:

Given the reaction

sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water

5.3g 6g 8.2g 2.2g 0.9g

Now,

Total Mass on the Left Hand Side = 5.3g + 6g = 11.3g

Total Mass on the Right Hand Side = 8.2g + 2.2g + 0.9g = 11.3g

As Mass on the LHS is equal to the RHS, the observation is in agreement with the 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 completely with 3 g of hydrogen gas?

Answer:

Given:

Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.

Now,

For every 1 g of Hydrogen, 8 g of oxygen is needed for the reaction to take place.

Therefore, for 3 g of Hydrogen, the mass of oxygen needed = 8 $\times$ 3 = 24 g.

Hence, 24 g of Oxygen is needed to complete a reaction with 3 g of Hydrogen.

Question.3 Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer:

The postulate of Dalton’s atomic theory, which is the result of the law of conservation of mass is

"Atoms can neither be created nor destroyed".

Question.4 Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer:

The postulate of Dalton’s atomic theory can explain the law of definite proportions is

"The relative number and kinds of atoms are equal in given compounds."

Exercise-3.2 (Page: 30)

Question 1. Define the atomic mass unit.

Answer:

An atomic mass unit is a unit of mass used to express the weight of subatomic particles, where one unit is equal to exactly one-twelfth the mass of a carbon-12 atom.

Question 2. Why is it not possible to see an atom with naked eyes?

Answer:

We can't see the atom with the naked eye because they are minuscule in nature. They are measured in nanometres. Also, except for noble gases, all atoms do not exist independently. They exist in the form of any compound.

Exercise 3.3 and 3.4 (Page: 34)

Question 1. Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Answer:

The formulas of the given compounds are :

(i) sodium oxide :

$\Rightarrow N a_{2} O$

(ii) aluminium chloride:

$\Rightarrow A l C l_{3}$

(iii) sodium sulphide:

$\Rightarrow N a_{2} S$

(iv) magnesium hydroxide:

$\Rightarrow M g (O H)_{2}$

Question 2. Write down the names of compounds represented by the following formulae:

(i) $A l_{2} (S O_{4})_{3}$

(ii) $C a C l_{2}$

(iii) $K_{2} S O_{4}$

(iv) $K N O_{3}$

(v) $C a C O_{3}$

Answer:

The Names of the following compounds are :

(i) $A l_{2} (S O_{4})_{3}$

=Aluminium Sulphate

(ii) $C a C l_{2}$

=Calcium Chloride

(iii) $K_{2} S O_{4}$

= Potassium Sulphate

(iv) $K N O_{3}$

= Potassium Nitrate

(v) $C a C O_{3}$

= Calcium Carbonate.

Question 3. What is meant by the term chemical formula?

Answer:

The chemical formula of a compound is a symbolic representation of its composition. For example, the chemical formula for common salt is NaCl, as it is made up of Sodium (Na) and Chlorine (Cl).

Q 4.(i) How many atoms are present in a

$H_{2} S$ molecule

Answer:

$H_{2} S$ molecule has 2 atoms of Hydrogen and 1 atom of Sulphur.

and hence,

A total of 3 atoms are present in $H_{2} S$ molecule.

Question 4.(ii) How many atoms are present in a

$P O_{4}^{3 -}$ ion?

Answer:

$P O_{4}^{3 -}$ ion has one atom of Phosphorus and 4 atoms of Oxygen.

And hence,

A total of 5 atoms are present in $P O_{4}^{3 -}$ ion.

Exercise-3.5.1-3.5.2 (Page: 35)

Question 1. Calculate the molecular masses of $H_{2}, O_{2}, C l_{2}, C O_{2}, C H_{4}, C_{2} H_{6}, C_{2} H_{4}, N H_{3}, C H_{3} O H$

Answer:

The molecular mass of $H_{2}$ :

= 2 $\times$ Atomic mass of Hydrogen

= 2 $\times$ 1u

=2u.

The molecular mass of $O_{2}$ :

= 2 $\times$ Atomic mass of Oxygen

= 2 $\times$ 16u

= 32u.

The molecular mass of $C l_{2}$ :

= 2 $\times$ Atomic mass of Chlorine

= 2 $\times$ 35.5uu

= 71u.

The molecular mass of $C O_{2}$ :

= Atomic mass of Carbon +2 $\times$ Atomic mass of Oxygen

= 12u + 2 $\times$ 16u

= 44u.

The molecular mass of $C H_{4}$ :

= Atomic mass of Carbon +4 $\times$ Atomic mass of Hydrogen

= 12u + 4 $\times$ 1u

= 16u.

The molecular mass of $C_{2} H_{6}$ :

= 2 $\times$ Atomic mass of Carbon + 6 $\times$ Atomic mass of Hydrogen

= 2 $\times$ 12u+ 6 $\times$ 1u

=24u+ 6u

= 30u.

The molecular mass of $C_{2} H_{4}$ :

= 2 $\times$ Atomic mass of Carbon + 64 $\times$ Atomic mass of Hydrogen

= 2 $\times$ 12u+ 4 $\times$ 1u

=24u+ 4u

= 28u.

The molecular mass of $N H_{3}$ :

= Atomic mass of Nitrogen + 3 $\times$ Atomic mass of Hydrogen

= 14u+ 3 $\times$ 1u

= 17u.

The molecular mass of $C H_{3} O H$ :

= Atomic mass of Carbon +4 $\times$ Atomic mass of Hydrogen + Atomic mass of Oxygen

= 12u+ 4 $\times$ 1u + 16u

= 32u.

Question 2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃ , given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u

Answer:

Given,

the atomic mass of Zn = 65 u,

the atomic mass of Na = 23 u,

the atomic mass of K = 39 u,

the atomic mass of C = 12 u,

and the atomic mass of O = 16 u

Now,

Formula unit mass of ZnO = Atomic mass of Zinc + Atomic mass of O

= 65 u + 16 u

= 81 u.

Formula unit mass of Na₂O = 2 $\times$ Atomic mass of Na+ Atomic mass of O

= 2 $\times$ 23 u + 16 u

= 62 u.

Formula unit mass of K₂CO₃= 2 $\times$ Atomic mass of K+ Atomic mass of C + 3 $\times$ Atomic mass of O

= 2 $\times$ 39 u + 12 u + 3 $\times$ 16 u

= 138 u.

NCERT Solutions Class 9 Science Chapter 3( Exercise Questions with Answers)

The atoms and molecules exercise questions are solved in this section. Go through Class 9 science chapter 3 NCERT solutions PDF for quick revision of concepts.

Question 1. A 0.24 g 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:

The total mass of the compound = 0.24 g

Mass of boron in the compound = 0.096 g

Mass of oxygen in the compound = 0.144 g

Now, as we know,

The percentage of an element in the compound :

$= \frac{t o t a l m a s s o f e l e m e n t}{t o t a l m a s s o f c o m p o u n d} \times 100$

So,

The percentage of Boron in the compound by weight :

$= \frac{0.096}{0.24} \times 100$

$= 40 %$

The percentage of Oxygen in the compound by weight :

$= \frac{0.144}{0.24} \times 100$

$= 60 %$

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

Answer:

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 g of carbon dioxide.

So, by the law of definite proportions,

When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used to produce 11.00 g of carbon dioxide. The remaining 42.00 g of oxygen will remain unreacted. The law of constant proportion is held.

Q 3. What are polyatomic ions? Give examples.

Answer:

Polyatomic ions are ions that contain more than one atom. These atoms can be of the same type or of a different type.

Some examples of polyatomic ions are NH₄+ , OH^- , SO₄^2- , and SO₃^2- .

Question 4. Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(d) Aluminium chloride

(e) Calcium carbonate

Answer:

The chemical formula of the given compounds is:

(a) Magnesium chloride

$= M g C l_{2}$

(b) Calcium oxide

$= C a O$

$= C u (N O_{3})_{2}$

(d) Aluminium chloride

$= A l C l_{3}$

(e) Calcium carbonate

$= C a C O_{3}$

Question 5. Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(d) Potassium sulphate.

Answer:

The names of the elements present in the given compounds are :

(a) Quick lime :

$C a O$ = Calcium and Oxygen.

(b) Hydrogen bromide:

$H B r$ = Hydrogen and Bromine.

$N a H C O_{3}$ = Sodium, Hydrogen, Carbon, and Oxygen.

(d) Potassium sulphate:

$K_{2} S O_{4}$ = Potassium, Sulphur, and oxygen.

Question 6.(a) Calculate the molar mass of the following substances.

Ethylene , $C_{2} H_{2}$

Answer:

The molecular mass of ethyne, $C_{2} H_{2}$ = 2 $\times$ Atomic Mass of C + 2 $\times$ Atomic Mass of H

= 2 $\times$ 12u + 2 $\times$ 1u

= 24u + 2u

= 26u

Question 6. (b) Calculate the molar mass of the following substances.

Sulphur molecule, S₈

Answer:

The molecular mass of Sulphur molecule, = 8 $\times$ Atomic Mass of S

= 8 $\times$ 32u

= 256u

Question 6.(c) Calculate the molar mass of the following substances

Baking soda

Answer:

The molecular formula of Baking soda is NaHCO₃

So the molecular mass will be some of the individual atoms involved in the formula.

Molecular Mass of baking soda = Molecular mass (Na + H + C + 3xO)

= 23 + 1 + 12 + 3x16

= 84 g/mol

Question 6.(d) Calculate the molar mass of the following substances.

Hydrochloric acid, HCl

Answer:

The molecular mass of Hydrochloric acid, HCl = 1 $\times$ Atomic Mass of Cl + 1 $\times$ Atomic Mass of H

= 35.5 u + 1u

= 36.5u

Question 6. (e) Calculate the molar mass of the following substances.

Nitric acid, HNO₃

Answer:

The molecular mass of Nitric acid, HNO₃ = 1 $\times$ Atomic Mass of N + 1 $\times$ Atomic Mass of H + 3 $\times$ Atomic Mass of O

= 1 $\times$ 14u + 1 $\times$ 1u + 3 $\times$ 16u

= 14u + 1u + 48u

= 63u

NCERT Class 9 Science Chapter 3: Overview

The chapter Atoms and Molecules will provide you with the basic concepts of atoms and molecules, the building blocks of matter. It begins with Dalton’s Atomic Theory, which explains the nature of atoms and their behavior. Students will learn about the laws of chemical combination; Law of Conservation of Mass and the Law of Constant Proportions. The chapter will explain to you what atoms, molecules, and ions are and how they form compounds. It will also feature atomic and molecular masses, chemical formulas and how to calculate the mole and molar mass using the concept of the Avogadro number (6.022 × 10²³). This chapter will help you understand chemical reactions and formulas.

Practice Questions for Class 9 Science Chapter 3

The atoms and molecules important questions and answers are given below for practice. Learn more from class 9 chapter 3 notes.

Q1. State the Law of Conservation of Mass.
Answer:
The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. The total mass of reactants is equal to the total mass of products.

Q2. What is the chemical formula of water and carbon dioxide? Name the elements present in them.
Answer:

Water (H₂O); Elements – Hydrogen and Oxygen

Carbon dioxide (CO₂); Elements – Carbon and Oxygen

Q3. Define atomicity. Give the atomicity of oxygen (O₂) and phosphorus (P₄).
Answer:
Atomicity is the number of atoms present in a molecule of an element.

Oxygen (O₂); Diatomic (Atomicity = 2)

Phosphorus (P₄); Tetratomic (Atomicity = 4)

Q4. Calculate the molecular mass of carbon dioxide (CO₂).
Answer:
Atomic mass of C = 12 u, O = 16 u
Molecular mass of CO₂ = 12 + (16 × 2) = 44 u

Q5. What is meant by a mole? How many particles are there in one mole of a substance?
Answer:
A mole is the amount of a substance that contains 6.022 × 10²³ particles (atoms, molecules, or ions). This number is called Avogadro’s constant.

Approach to Solve Questions of Class 9 Science Chapter 3

To effectively solve questions of the chapter Atoms and Molecules, follow a systematic and structured approach

1. Review the chapter

Understand the structure of the chapter and break the chapter into manageable sections.

2. Understand the concepts

After that, try to learn and memorize key concepts of the chapter, such as Dalton’s Atomic Theory, the Law of Conservation of Mass, features of atomic and molecular masses, the mole concept, and molar mass, etc.

3. Make notes

Write all the definitions on a separate sheet to solve definition-based questions.

Make a formula sheet related to the mole concept and molar masses to effectively solve numerical. You can also learn the topics from aoms and molecules class 9 notes available on our website.

4. Practice more

Solve the NCERT examples and practice with in-text questions. Also, use some common tricks, like unit conversion, memorize symbols, and valency.

Practice as many questions as asked in previous board exams and solve mock tests. Follow the NCERT solutions for class 9 chapter 3 to learn more.

Topics and Subtopics Covered in the NCERT Textbook

All the topics and subtopics covered in the NCERT textbook are listed below. You can learn these concepts from the NCERT solutions for class 9 science chapter 3.

3.1 Laws of Chemical Combination

3.1.1 Law of Conservation of Mass

3.1.2 Law of Constant Proportions.

3.2 What is an Atom?

3.2.1 What Are the Modern-Day Symbols of Atoms of Different Elements?

3.2.2 Atomic Mass

3.3 What is a Molecule?

3.3.1 Molecules of Elements

3.3.2 Molecules of Compounds

3.3.3 What is An Ion?

3.4 Writing Chemical Formulae

3.4.1 Formulae of Simple Compounds

3.5 Molecular Mass

3.5.1 Formula Unit Mass

NCERT Solutions for Class 9 Science Chapter 3: Important Formulas and E-book.

1 nanometer (nm) is equal to 10^-9 meters (m)
Number of moles (n) = Given quantity of substance (N) /Molar mass of substance (No)
Number of moles of atoms = Number of atoms Avogrdro's number [Avogrdro’s number=6.022 × 10²³]
Mass = Molar mass × Number of moles
Formula unit mass = (Atomic mass of cation × Number of cations) + (Atomic mass of anion × Number of anions)

Atoms and molecules exercise questions require the application of the above formulas. Learn more from class 9 chapter 3 notes.

To download the E-Book click on the link given below

NCERT Class 9 Science: Important Formulas For Class 9 Science - Chapterwise

NCERT Solutions for Class 9 Science- Chapter Wise

The links below will give you access to the chapter-wise class 9 NCERT solutions.

NCERT Solutions for Class 9 Chapter 1 Matter in Our Surroundings

NCERT Solutions for Class 9 Chapter 2 Is Matter Around Us Pure

NCERT Solutions for Class 9 Chapter 3 Atoms and Molecules

NCERT Solutions for Class 9 Chapter 4 Structure of The Atom

NCERT Solutions for Class 9 Chapter 5 The Fundamental Unit of Life

NCERT Solutions for Class 9 Chapter 6 Tissues

NCERT Solutions for Class 9 Chapter 7 Motion

NCERT Solutions for Class 9 Chapter 8 Force and Laws of Motion

NCERT Solutions for Class 9 Chapter 9 Gravitation

NCERT Solutions for Class 9 Chapter 10 Work and Energy

NCERT Solutions for Class 9 Chapter 11 Sound

NCERT Solutions for Class 9 Chapter 12 Improvement in Food Resources

NCERT Solutions for Class 9 Subject Wise

Class 9 NCERT Subject-wise solutions are given below. Also, follow the chapter-wise class 9 NCERT solutions for better learning.

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Science

NCERT Books and NCERT Syllabus

Follow the links below to get the syllabus and prescribed books. Learn more from the atoms and molecules class 9 science NCERT notes.

NCERT Books Class 9 Science

NCERT Syllabus Class 9 Science

NCERT Books Class 9

NCERT Syllabus Class 9

Frequently Asked Questions (FAQs)

1. What are the Laws of Chemical Combination according to class 9 NCERT solutions?

There are two main laws-

Law of Conservation of Mass- It states that mass can neither be created nor destroyed in a chemical reaction. The total mass of the reactants must equal the total mass of the products.
Law of Constant Proportions (or Law of Definite Proportions)- It states that in a pure chemical compound, elements are always present in fixed proportions by mass, regardless of the source or method of preparation. (e.g., water (H₂O) always contains hydrogen and oxygen in an 8:1 ratio by mass).

2. What are the three basic parts of an atom and what are their charges?

The three basic parts of an atom are:

Protons(Located in the nucleus): Positively charged particles.
Neutrons(Located in the nucleus): Neutral particles (no charge)
Electrons(orbit the nucleus): Negatively charged particles.

3. What are Ions as discussed in class 9 NCERT solutions? How are they formed?

Ions are charged particles (atoms or groups of atoms). They are formed when an atom gains or loses electrons.

Cation- A positively charged ion formed when an atom loses one or more electrons (e.g., Na⁺, Mg²⁺).
Anion- A negatively charged ion formed when an atom gains one or more electrons (e.g., Cl⁻, O²⁻).

4. What are chemical bonds and why do atoms form them?

Chemical bonds are attractive forces that hold atoms together in molecules and compounds. Atoms form chemical bonds to achieve a more stable electron configuration.

5. What is 'Atomic Mass' and 'Molecular Mass'?

Atomic Mass: The relative mass of an atom of an element compared to 1/12th the mass of a carbon-12 atom. It is expressed in atomic mass units (amu) or unified mass (u). For example, the atomic mass of hydrogen is 1 u.

Molecular Mass: The sum of the atomic masses of all the atoms present in a molecule. It is also expressed in unified mass (u). For example, the molecular mass of water (H₂O) = (2 × Atomic mass of H) + (1 × Atomic mass of O) = (2 × 1 u) + (1 × 16 u) = 18 u.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Download PDF of NCERT Solutions for Class 9 Science Chapter 3

NCERT Solutions for Class 9 Science Chapter 3 (in-text Exercise Questions)

NCERT Solutions Class 9 Science Chapter 3( Exercise Questions with Answers)

NCERT Class 9 Science Chapter 3: Overview

Practice Questions for Class 9 Science Chapter 3

Approach to Solve Questions of Class 9 Science Chapter 3

Topics and Subtopics Covered in the NCERT Textbook

NCERT Solutions for Class 9 Science Chapter 3: Important Formulas and E-book.

NCERT Solutions for Class 9 Science- Chapter Wise

NCERT Solutions for Class 9 Subject Wise

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

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