 # NCERT Solutions for Class 9 Science Chapter 3 - Atoms and Molecules

## NCERT Solutions for Class 9 Science Chapter 3 – CBSE Get Free PDF

NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules is a detailed study material prepared by experts. It provides answers to the questions given in the textbook. NCERT Solutions are very helpful for a better understanding of the concepts and self-analysis.

Questions from all the topics are covered in the solutions. Everything is presented in a way that students can easily understand. It will help them score well in the CBSE examination. The answers to all kinds of long and short questions, MCQs, tricks and tips are provided in the NCERT Solutions for Class 9 Science. Try solving the questions after completing the entire syllabus and overcoming the shortcomings before the CBSE exams arrive.

The chapter Atoms and Molecules form the basis of the upcoming chapters; therefore, it should be dealt with thoroughly. The questions and solutions will help the students clarify all their doubts related to the topic.

The NCERT Solutions for Class 9 carries all the important questions and answers for all the subjects and chapters. The students can refer to these solutions to excel in the CBSE examinations.

## Exercise-3.1 Page: 32

1. In a reaction, 5.3g 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

Solution:

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

5.3g                             6g                 8.2g     2.2g      0.9g

As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of

products.

As per the above reaction, L.H.S. = R.H.S.    i.e., 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g

Hence, the observations are in agreement with the law of conservation of mass.

2. Hydrogen and oxygen combine in a 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?

Solution:

We know hydrogen and water mix in a ratio 1: 8.

For every 1g of hydrogen, it is 8g of oxygen.

Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g

Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas.

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

Solution:

The relative number and types of atoms are constant in a given composition, says Dalton’s atomic theory, which is based on the rule of conservation of mass.

“Atoms cannot be created nor be destroyed in a chemical reaction.”

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

Solution:

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

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

## Exercise-3.2 Page: 35

1. Define the atomic mass unit.

Solution:

An atomic mass unit is a unit of mass used to express the weights of atoms and molecules where one

atomic mass is equal to 1/12th the mass of one carbon-12 atom.

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

Solution:

Firstly, atoms are minuscule in nature, measured in nanometers. Secondly, except for atoms of noble

gases, they do not exist independently. Hence, an atom cannot be visible to the naked eyes.

## Exercise-3.3-3.4 Page: 39

1. Write down the formulae of

(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

Solution:

The following are the formulae:

(i) sodium oxide – Na2O

(ii) aluminium chloride – AlCl3

(iii) sodium sulphide – Na2S

(iv) magnesium hydroxide – Mg (OH)2

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

(i) Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3.

Solution:

Listed below are the names of the compounds for each of the following formulae:

(i) Al2(SO4)3 – Aluminium sulphate

(ii) CaCl2 – Calcium chloride

(iii) K2SO4 – Potassium sulphate

(iv) KNO3 – Potassium nitrate

(v) CaCO3 – Calcium carbonate

3. What is meant by the term chemical formula?

Solution:

Chemical formulas are used to describe the different types of atoms and their numbers in a compound or element. Each element’s atoms are symbolised by one or two letters. A collection of chemical symbols that depicts the elements that make up a compound and their quantities.

For example, the chemical formula of hydrochloric acid is HCl.

4. How many atoms are present in a

(i) H2S molecule and

(ii) PO43- ion?

Solution:

The number of atoms present is as follows:

(i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total.

(ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.

## Exercise-3.5.1-3.5.2 Page: 40

1. Calculate the molecular masses of H2 , O2 , Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Solution:

The following are the molecular masses:

The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u

The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u

The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u

The molecular mass of CO2 – atomic mass of C + 2 x atomic mass of O = 12 + ( 2×16)u = 44u

The molecular mass of CH4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u

The molecular mass of C2H6– 2 x atomic mass of C + 6 x atomic mass of H = (2 x 12) +

(6 x 1)u=24+6=30u

The molecular mass of C2H4– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12) +

(4 x 1)u=24+4=28u

The molecular mass of NH3– atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u

The molecular mass of CH3OH – atomic mass of C + 3x atomic mass of H + atomic mass of O + atomic mass of H = (12 + 3×1+16+1)u=(12+3+17)u = 32u

2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u,

Na = 23 u,  K=39u, C = 12u, and O=16u.

Solution:

Given:

The atomic mass of Zn = 65u

The atomic mass of Na = 23u

The atomic mass of K = 39u

The atomic mass of C = 12u

The atomic mass of O = 16u

The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u

The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u

The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u

## Exercise-3.5.3 Page: 42

1. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?

Solution:

Given: 1 mole of carbon weighs 12g

1 mole of carbon atoms = 6.022 x 1023

The molecular mass of carbon atoms = 12g = an atom of carbon mass

Hence, mass of 1 carbon atom = 12 / 6.022 x 1023  = 1.99 x 10-23g

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given the atomic mass of Na = 23u, Fe = 56 u)?

Solution:

(a) In 100 grams of Na:

m = 100g, Molar mass of Na atom = 23g, N0 = 6.022 x 1023, N = ?

N = (Given mass x N0)/Molar mass

N = (100 x 6.022 x 1023)/ 23

N = 26.18 x 1023 atoms

(b) In 100 grams of Fe:

m = 100 g, Molar mass of Fe atom = 56 g, N0 = 6.022 x 1023, N = ?

N = (Given mass x N0)/ Molar mass

N = (100 x 6.022 x 1023)/ 56

N = 10.75 x 1023 atoms

Therefore, the number of atoms is more in 100 g of Na than in 100 g of Fe.

## Exercise Page: 43

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.

Solution:

Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g

To calculate the percentage composition of the compound,

Percentage of boron = mass of boron / mass of the compound x 100

= 0.096g / 0.24g x 100  = 40%

Percentage of oxygen = 100 – percentage of boron

= 100 – 40 = 60%

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?

Solution:

When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

Given that

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

Find out

We need to find out the mass of carbon dioxide that will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen.

Solution

First, let us write the reaction taking place here.

C + O2 → CO2

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.

3g + 8g →11 g ( from the above reaction)

The total mass of reactants = mass of carbon + mass of oxygen

=3g+8g

=11g

The total mass of reactants = Total mass of products

Therefore, the law of conservation of mass is proved.

Then, it also depicts that carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.

Thus, it further proves the law of constant proportions.

3 g of carbon must also combine with 8 g of oxygen only.

This means that (50−8)=42g of oxygen will remain unreacted.

The remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of carbon dioxide will be formed

The above answer is governed by the law of constant proportions.

3. What are polyatomic ions? Give examples.

Solution:

Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.

Example: CO32-, H2PO4

4. Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Solution:

The following are the chemical formula of the above-mentioned list:

(a) Magnesium chloride – MgCl2

(b) Calcium oxide – CaO

(c) Copper nitrate – Cu(NO3)2

(d) Aluminium chloride – AlCl3

(e) Calcium carbonate – CaCO3

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

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate

Solution:

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

(a) Quick lime – Calcium and oxygen (CaO)

(b) Hydrogen bromide – Hydrogen and bromine (HBr)

(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)

(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)

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) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

Solution:

Listed below is the molar mass of the following substances:

(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g

(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8  x 32 = 256g

(c) Molar mass of  Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g

(e) Molar mass of Nitric acid, HNO3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+

3×16 = 63g

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)?

Solution:

The mass of the above-mentioned list is as follows:

(a) Atomic mass of nitrogen atoms = 14u

Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen atoms

Therefore, the mass of 1 mole of nitrogen atom is 14g.

(b) Atomic mass of aluminium =27u

Mass of 1 mole of aluminium atoms = 27g

1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g

(c) Mass of 1 mole of sodium sulphite Na2SO3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O =  (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g

Therefore, mass of 10 moles of Na2SO3  = 10 x 126 = 1260g

8. Convert into a mole.

(a) 12g of oxygen gas

(b) 20g of water

(c) 22g of carbon dioxide

Solution:

Conversion of the above-mentioned molecules into moles is as follows:

(a) Given: Mass of oxygen gas=12g

Molar mass of oxygen gas = 2 Mass of Oxygen =  2 x 16 = 32g

Number of moles = Mass given / molar mass of oxygen gas = 12/32 =  0.375 moles

(b) Given: Mass of water = 20g

Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g

Number of moles = Mass given / molar mass of water

= 20/18 = 1.11 moles

(c) Given: Mass of carbon dioxide = 22g

Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g

Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Solution:

The mass is as follows:

(a) Mass of 1 mole of oxygen atoms = 16u; hence, it weighs 16g.

Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2g

(b) Mass of 1 mole of water molecules = 18u; hence, it weighs 18g.

Mass of 0.5 moles of water molecules = 0.5 x 18 = 9g

10. Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.

Solution:

To calculate the molecular mass of sulphur,

Molecular mass of Sulphur (S8) = 8xMass of Sulphur = 8×32 = 256g

Mass given = 16g

Number of moles = mass given/ molar mass of sulphur

= 16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur,

Number of molecules = Number of moles x Avogadro number

= 0.0625 x 6.022 x 10²³ molecules

= 3.763 x 1022 molecules

11. Calculate the number of aluminium ions present in 0.051g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)

Solution:

To calculate the number of aluminium ions in 0.051g of aluminium oxide,

1 mole of aluminium oxide = 6.022 x 1023 molecules of aluminium oxide

1 mole of aluminium oxide (Al2O3) = 2 x Mass of aluminium + 3 x Mass of oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 1023 molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 6.022 x 1023 / 102 x 0.051

= 3.011 x 1020 molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions; hence, the number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 1020 molecules of aluminium oxide.

= 6.022 x 1020

## NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules

The chapter holds a weightage of 23 marks in the examinations. The questions are not specific and can be asked from any topic. The students are therefore advised to be thorough with the entire chapter. The chemical formulae and the numerical should be practised well.

The important topics provided in this chapter include

3.1 Laws of Chemical Combination

3.2 What Is an Atom?

3.3 What Is a Molecule?

3.4 Writing Chemical Formulae

3.5 Molecular Mass and Mole Concept

Exercise Solutions 11 Questions (8 numerical, 3 short)

Molecular mass and Mole concept – 8 numerical

Chemical Formula – 2 Questions

What Is an Atom – 1 Question

### NCERT Solutions for Class 9 Science Chapter 3

The smallest unit of matter is an atom. It has the properties of an element. An atom comprises a dense core called a nucleus, surrounded by a series of outer shells. The electrons are present in these shells. The nucleus contains protons and neutrons. Protons have a positive charge, while neutrons are neutral.

Two or more atoms tightly bound together form a molecule. The molecules made up of two atoms are known as diatomic. Oxygen, nitrogen, hydrogen, and iodine are diatomic molecules. Earth’s atmosphere is comprised mainly of diatomic molecules. A molecule is the smallest part of a compound.

### Key Features of NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules

• The answers are provided by Science experts.
• The Solutions have questions from every important topic.
• It gives a thorough understanding of the concepts.

Disclaimer:

Dropped Topics – Mole concept.

## Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 3

Q1

### Mention the topics included in Chapter 3 of NCERT Solutions for Class 9 Science.

The topics included in Chapter 3 of NCERT Solutions for Class 9 Science are
3.1 Laws of Chemical Combination
3.2 What Is an Atom?
3.3 What Is a Molecule?
3.4 Writing Chemical Formulae
3.5 Molecular Mass and Mole Concept
Q2

### How to solve the numericals present in Chapter 3 of NCERT Solutions for Class 9 Science faster?

Students of Class 9 should solve the numericals present in the NCERT textbook to get an idea about the important formulas present in them. If doubts arise while solving the exercise problems, students can refer to the NCERT Solutions for Class 9 Science from BYJU’S, where each and every problem is solved accurately. The solutions contain explanations for each step to help students understand the method of solving problems without any difficulty.
Q3

### What are the key features of NCERT Solutions for Class 9 Science Chapter 3?

The key features of NCERT Solutions for Class 9 Science Chapter 3 are
1. The answers are prepared by highly experienced science experts.
2. The solutions are accurate without any errors, based on the latest CBSE syllabus.
3. Every important topic is explained in simple language to help students score well in the CBSE exams.
4. Students get a thorough understanding of the important concepts using the NCERT Solutions from BYJU’S.

1. Thanks it helped me a lot 😊😊😊😊😊

• Thanks for help

2. It so good for students

3. Thanks for helping us

4. Thanks for help

5. Thanks it helped me a lot❤️

6. It helped me a lot❤️

7. Thanks It helped me a lot 😁

8. It helped me a lot 😁 😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁

9. The best 💫🙂😀

10. This is very helpful for me ☺️
Thnx a lot 🙏🏻

11. Thnk u so much byjus for help

12. This helps me in my upcoming exams
Thanks a lots By jus to provide it