NCERT Solutions for Class 9 Science Chapter 3 - Atoms And Molecules

Intext Exercise 1

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

Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Ans: Given, 

Mass of sodium carbonate = 5.3 g 

Mass of ethanoic acid = 6 g 

Mass of sodium ethanoate = 8.2 g 

Mass of carbon dioxide = 2.2 g 

Mass of water = 0.9 g 

Now, total mass before the reaction \[ = {\text{ }}\left( {5.3{\text{ }} + {\text{ }}6} \right){\text{ }}g{\text{ }} = {\text{ }}11.3{\text{ }}g\]

And, total mass after the reaction =\[\;\left( {8.2{\text{ }} + {\text{ }}2.2{\text{ }} + {\text{ }}0.9} \right){\text{ }}g{\text{ }} = {\text{ }}11.3{\text{ }}g\]

∴Total mass before the reaction = Total mass after the reaction 

Hence, this is in agreement with the law of conservation of mass.

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?

Ans: In water, H:O (by mass) =$1\;:\;8$

The mass of oxygen gas required to react completely with 1 g of hydrogen gas = 8 g. 

So, 

the mass of oxygen gas required to react completely with 3 g of hydrogen gas = \[(8\; \times \;3)\;g\; = \;24\;g\]

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

Ans: The postulate of Dalton’s atomic theory which is based on the law of conservation of mass is: “Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.”

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

Ans: “The elements consist of atoms having fixed mass and that the number and kind of atoms of each element in a given compound is fixed.” This explains the law of definite proportion.

Intext Exercise 2

1. Define atomic mass unit.

Ans: Mass unit equal to exactly one-twelfth the $\left( {\dfrac{1}{{{{12}^{th}}}}} \right)$mass of one atom of carbon-12 is called one atomic mass unit. It is represented by as ‘a.m.u.’ or ‘u’.

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

Ans: Due to small size of an atom we cannot see them with naked eyes.

Intext Exercise 3

1. Write down the formulae of

(i) Sodium oxide 

Ans:  Sodium Oxide: \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}\]

(ii) Aluminium chloride 

Ans: Aluminium chloride: \[{\text{AlC}}{{\text{l}}_{\text{3}}}\]

(iii) sodium sulphide 

Ans: Sodium sulphide: \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}\]

(iv) Magnesium hydroxide

Ans: Magnesium hydroxide: ${\text{Mg(OH}}{{\text{)}}_{\text{2}}}$

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

i) $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right) \mathrm{_3}$

Ans: Aluminium sulphate

ii) ${\text{CaC}}{{\text{l}}_{\text{2}}}$

Ans: Calcium chloride

iii) ${{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$

Ans: :Potassium sulphate

iv) ${\text{KN}}{{\text{O}}_3}$

Ans: Potassium nitrate

v) \[{\text{CaC}}{{\text{O}}_{\text{3}}}\]

Ans:  : Calcium carbonate

3. What is meant by the term chemical formula?

Ans: The symbolic representation of composition of a compound is known as chemical formula. Chemical formula gives us the idea of number of atoms present.

Example: from the chemical formula ${\text{C}}{{\text{O}}_{\text{2}}}$ of  Carbon Dioxide, we come to know that one carbon atom and two oxygens atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

4. How many atoms are present in a:

i) $\mathrm{H}_{2}$ S molecule

Ans: There are total 3 atoms present in ${{\text{H}}_{\text{2}}}{\text{S}}$ molecule, two hydrogen atoms and one Sulphur atom.

ii) \[{\text{PO}}_4^{3 - }\] ion

Ans:  There are total 5 atoms in \[{\text{PO}}_4^{3 - }\] ion, one phosphorus atom and 4 oxygen atoms.

Intext Exercise 4 

1. Calculate the molecular masses of \[{{\text{H}}_2},{{\text{O}}_2},{\text{C }}{{\text{l}}_2},{\text{C}}{{\text{O}}_2},{\text{C}}{{\text{H}}_4},{{\text{C}}_2}{{\text{H}}_6},{{\text{C}}_2}{{\text{H}}_4},{\text{N}}{{\text{H}}_3},{\text{C}}{{\text{H}}_3}{\text{OH}}.\]

Ans: Molecular mass of ${{\text{H}}_{\text{2}}}$ = $2\; \times $ Atomic mass of H

$ = \;2\; \times \;1\;{\text{u  = }}\;{\text{2 u}}$

Molecular mass of ${{\text{O}}_{\text{2}}}$ = $2\; \times $ Atomic mass of O

$ = \;2\; \times \;16\;{\text{u  = }}\;3{\text{2 u}}$

Molecular mass of ${\text{C }}{{\text{l}}_{\text{2}}}$ = $2\; \times $ Atomic mass of Cl

$ = \;2\; \times \;35.5\;{\text{u  = }}\;71{\text{ u}}$

Molecular mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ = Atomic mass of C $ + $ $2\; \times $Atomic mass of  O

$ = \;(12 + 2\; \times \;16)\;{\text{u  = }}\;44{\text{ u}}$

Molecular mass of ${\text{C}}{{\text{H}}_4}$ = Atomic mass of C $ + $ $4\; \times $Atomic mass of H

$ = \;(12 + 4\; \times \;1)\;{\text{u  = }}\;16{\text{ u}}$

Molecular mass of ${{\text{C}}_2}{{\text{H}}_6}$ = $2\; \times $Atomic mass of C $ + $ $6\; \times $Atomic mass of H

$ = \;(2 \times 12 + 6\; \times \;1)\;{\text{u  = }}\;30{\text{ u}}$

Molecular mass of ${{\text{C}}_2}{{\text{H}}_4}$ = $2\; \times $Atomic mass of C $ + $ $4\; \times $Atomic mass of H

$ = \;(2 \times 12 + 6\; \times \;1)\;{\text{u  = }}\;30{\text{ u}}$

Molecular mass of ${\text{N}}{{\text{H}}_3}$ = Atomic mass of N $ + $ $3\; \times $Atomic mass of H

$ = \;(14 + 3\; \times \;1)\;{\text{u  = }}\;17{\text{ u}}$

Molecular mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ = Atomic mass of C $ + $ $3\; \times $Atomic mass of H $ + $ Atomic mass of O 

$ + $ Atomic mass of   $ = \;(12 + 4\; \times \;1\; + \;16)\;{\text{u  = }}\;32{\text{ u}}$

2. Calculate the formula unit masses of \[{\text{ZnO}},\;{\text{N}}{{\text{a}}_2}{\text{O}},\;{{\text{K}}_2}{\text{C}}{{\text{O}}_3}\]given atomic masses of \[{\text{Z}} = 65{\text{u}},{\text{Na}} = 23\;{\text{u}},\;{\text{K}} = 39\;{\text{u}},\;{\text{C}} = 12{\text{u}},\;{\text{and  O}} = 16{\text{u}}{\text{.}}\]

Ans: Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O

\[ = \;(65 + 16)\;{\text{u  =  81}}\;{\text{u}}\]

Formula unit mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}$ =  $2\; \times $ Atomic mass of Na + Atomic mass of O

=  \[(2\; \times \;23\; + \;16)\;{\text{u}}\; = \;62\;{\text{u}}\]

Formula unit mass of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ =  $2\; \times $ Atomic mass of K + Atomic mass of C + $3\; \times $ Atomic mass O=

$=(2 \times 39+12+3 \times 16) \mathrm{u} $

$=138 \mathrm{u}$

Refer to page 42.

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

Ans: Given mass of  One mole of carbon atoms  =  12 g 

Therefore , Mass of \[6.022 \times {10^{23}}\] number of carbon atoms =  12 g 

Mass of 1  atom of carbon will be:   

$ = \dfrac{{12}}{{6.022 \times {{10}^{23}}}}g$

$ = 1.9927 \times {{10}^{ - 23}}{\text{g}}$

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

Ans: Atomic mass of Na = 23 u (Given) 

Then,  gram atomic mass of Na = 23 g 

Now, 23 g of Na contains = \[6.022 \times {10^{23}}\]  number of  Na atoms

Thus, 100 g of Na contains = \[\dfrac{{6.022 \times {{10}^{23}}}}{{23}} \times 100\] number of Na atoms

\[ = 2.6182 \times {10^{24}}{\text{number of}}\;{\text{Na}}\;{\text{atoms}}\]

Atomic mass of Fe = 56 u (Given) 

Then, gram atomic mass of Fe = 56 g 

Now, 56 g of Fe contains = \[6.022 \times {10^{23}}\]  number of  Fe atoms

Thus, 100 g of Fe contains = \[\dfrac{{6.022 \times {{10}^{23}}}}{{56}} \times 100\] number of Fe atoms

\[ = 1.0753 \times {10^{24}}{\text{number of  Fe  atoms}}\]

\[2.6182 \times {10^{24}} > 1.0753 \times {10^{24}}\]

Therefore, 100 grams of sodium contain a greater number of atoms than 100 grams of iron.

NCERT QUESTIONS:

1. A 0.24 g sample of 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.

Ans:

Given, 

Mass of boron = 0.096 g  

Mass of oxygen = 0.144 g 

Mass of sample = 0.24 g 

The percentage of boron by weight in the compound \[ = \dfrac{{0.096}}{{0.24}} \times 100\% \; = \;40\;\% \]

And, percentage of oxygen by weight in the compound \[ = \dfrac{{0.144}}{{0.24}} \times 100\% \; = \;60\;\% \]

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 combinations will govern your answer?

Ans: Carbon + Oxygen ⎯⎯→ Carbon dioxide 

3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide. 

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen to form11 g of carbon dioxide. 

The remaining (50 –8) = 42 g of oxygen will be left unreacted. 

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

3. What are polyatomic ions? Give examples?

Ans: A polyatomic ion is a group of atoms carrying a charge either positive or negative. 

For example,\[\;{\text{ammonium ion}}\;\left( {{\text{NH}}_4^ + } \right),\;{\text{hydroxide ion}}\;\left( {{\text{O}}{{\text{H}}^ - }} \right),\;{\text{carbonate ion}},\;\left( {{\text{CO}}_3^{2 - }} \right){\text{,}}\;\;{\text{sulphate ion}}\;\left( {{\text{SO}}_4^{2 - }} \right)\]

4. Write the chemical formulae of the following:

(a) Magnesium chloride 

Ans: \[{\text{MgC}}{{\text{l}}_{\text{2}}}\]

(b) Calcium oxide 

Ans: CaO

(c) Copper nitrate 

Ans: \[{\text{Cu}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}\]

(d) Aluminium chloride 

Ans: \[{\text{AlC}}{{\text{l}}_{\text{3}}}\]

(e) Calcium carbonate

Ans: \[{\text{CaC}}{{\text{O}}_{\text{3}}}\]

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

(a) Quick lime 

(b) Hydrogen bromide 

(c) Baking powder 

(d) Potassium sulphate

Ans:

6. Calculate the molar mass of the following substances:

a) Ethyne ${{\text{C}}_{_{\text{2}}}}{{\text{H}}_{\text{2}}}$

Ans: Molar mass of  \[{{\text{C}}_2}{{\text{H}}_2} = 2 \times 12 + 2 \times 1 = 28{\text{g}}/{\text{mol}}\]

b) Sulphur molecule, ${{\text{S}}_{\text{8}}}$

Ans: Molar mass of  \[{{\text{S}}_8} = 8 \times 32 = 256{\text{g}}/{\text{mol}}\]

c) Phosphorus molecule \[{{\text{P}}_{\text{4}}}\] (atomic mass of phosphorus = 31)

Ans: Molar mass of  \[{{\text{P}}_4} = 4 \times 31 = 124{\text{g}}/{\text{mol}}\]

c) Hydrochloric acid, HCl

Ans: Molar mass of  \[{\text{HCl}} = 1 + 35.5 = 36.5{\text{g}}/{\text{mol}}\]

d) Nitric acid, \[{\text{HN}}{{\text{O}}_{\text{3}}}\]

Ans: Molar mass of \[{\text{HN}}{{\text{O}}_3} = 1 + 14 + 3 \times 16 = 63{\text{g}}/{\text{mol}}\]

7. What is the mass of: 

(a) 1 mole of nitrogen atoms? 

Ans: The mass of 1 mole of N- atoms = 14 g

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

Ans:  \[[{\text{Atomic mass of}}\;{\text{Al}} = 27{\text{u}}]\]

The mass of 4 moles of Al-atoms \[ = (4 \times 27){\text{g}}\;{\text{ = }}\;\;{\text{108}}\;{\text{g}}\]

(c) 10 moles of sodium sulphite \[\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}} \right)\]?

Ans:  Atomic mass of Na = 23 u, Atomic mass of S = 32 u,  Atomic mass of O = 16 u

The mass of 10 moles of sodium sulphite \[\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}} \right)\]\[ = 10 \times [2 \times 23 + 32 + 3 \times 16]{\text{g}}\]

\[ = 10 \times 126{\text{g}} = 1260{\text{g}}\]

8. Convert into mole. 

(a) 12 g of oxygen gas 

Ans:   32 g of oxygen gas = 1 mole 

Then, 12 g of oxygen gas $ = \dfrac{{12}}{{32}}\;{\text{moles}}\; = \;1.11\;{\text{moles}}$

(b) 20 g of water 

Ans:   18 g of Water ${{\text{H}}_{\text{2}}}{\text{O}}$ = 1 mole 

Then, 20 g of water ${{\text{H}}_{\text{2}}}{\text{O}}$ $ = \dfrac{{20}}{{18}}\;{\text{moles}}\; = \;0.375\;{\text{moles}}$

(c) 22 g of carbon dioxide

Ans:  44 g of Carbon Dioxide ${\text{C}}{{\text{O}}_{\text{2}}}$ = 1 mole 

Then, 22 g of water Carbon Dioxide  ${\text{C}}{{\text{O}}_{\text{2}}}$ $ = \dfrac{{22}}{{44}}\;{\text{moles}}\; = \;0.5\;{\text{moles}}$

9. What is the mass of: 

(a) 0.2 mole of oxygen atoms? 

Ans: Mass of 1 mole of oxygen atoms = 16 g 

Then, mass of 0.2 mole of oxygen atoms \[ = {\text{ }}0.2{\text{ }} \times {\text{ }}16g{\text{ }} = {\text{ }}3.2{\text{ }}g\]

(b) 0.5 mole of water molecules?

Ans: Mass of 1 mole of water molecules (${{\text{H}}_{\text{2}}}{\text{O}}$) = 18 g 

Then, mass of 0.5 mole of water molecules (${{\text{H}}_{\text{2}}}{\text{O}}$) \[ = 0.5{\text{ }} \times {\text{ }}18{\text{ }}g{\text{ }} = {\text{ }}9{\text{ }}g\]

10. Calculate the number of molecules of sulphur (${{\text{S}}_{\text{8}}}$) present in 16 g of solid sulphur.

Ans: 1 mole of solid sulphur (${{\text{S}}_{\text{8}}}$) = \[8{\text{ }} \times {\text{ }}32{\text{ }}g{\text{ }} = {\text{ }}256{\text{ }}g\]

i.e., 256 g of solid sulphur contains = \[6.022 \times {10^{23}}\] molecules

Then, 16 g of solid sulphur contains = \[\dfrac{{6.022 \times {{10}^{23}}}}{{256}} \times 16\] molecules

= \[ = 3.76 \times {10^{22}}{text{\;molecules}}\]

11. Calculate the number of aluminium ions present in 0.051 g 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 = 27 u).

Ans: Given, atomic mass of $Al=27 \mathrm{u}$, atomic mass of $\mathrm{O}=16 \mathrm{u}$

1 mole of aluminium oxide $\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)=(2 \times 27+3 \times 16)=102 \mathrm{~g}$

i.e., $102 \mathrm{~g}$ of $\mathrm{Al}_{2} \mathrm{O}_{3}=6.022 \times 10^{23}$ molecules of $\mathrm{Al}_{2} \mathrm{O}_{3}$

Then, $0.051 \mathrm{~g}$ of $\mathrm{Al}_{2} \mathrm{O}_{3}$ contains;

$=\frac{6.022 \times 10^{2}}{102} \times 0.051$ molecules of $\mathrm{Al}_{2} \mathrm{O}_{3}$

$=3.011 \times 10^{20}$ molecules of $\mathrm{Al}_{2} \mathrm{O}_{3}$

The number of aluminium ions $\left(\mathrm{Al}^{3+}\right)$ present in 1 molecule of aluminium oxide is 2 .

Therefore, the number of aluminium ions $\left.(A]^{3}\right)$ present in $3.011 \times 10^{20}$ molecules $(0.051 \mathrm{~g})$ of aluminium oxide $\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)=2 \times 3.011 \times 10^{20}=6.022 \times 10^{20}$

NCERT Solutions for Class 9 Science – Free PDF Download

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The NCERT Solutions for Class 9 Science Chapter 3 on Vedantu can be accessed by students online and can also be downloaded for free. Following these NCERT solutions will help you get an insight into all the concepts covered in this chapter.

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

3.1 Law of Chemical Combination

There are two laws of chemical combination- the law of conservation of mass, and the law of constant proportion. Both these laws are discussed in detail in chapter 3, science class 9.

There is one important question related to the chemical reaction. Is there any change in the total mass of the system when a chemical reaction takes place? According to the law of conservation of mass, mass can neither be created nor be destroyed. The same logic stands true for chemical reactions as well.  Several experiments carried out by Lavoisier standardized this law. This law is discussed in NCERT Solutions for Class 9 Science Chemistry Chapter 3.

The law of chemical combination is based on another law- the law of constant proportion. This law was also the result of experiments conducted by Lavoisier and Joseph L. Proust. According to the law of constant proportion, any compound is made by mixing two or more elements in a constant proportion, and this proportion does not change irrespective of the source of the compounds. For example, in water, the ratio of the mass of Hydrogen to the mass of Oxygen is 1:8. The ratio of the mass is the same for water from different sources. This law is also known as the law of definite proportion. The law of constant proportion is explained thoroughly in Class 9 Science Chapter 3.

3.2 What is an Atom?

The issue that the scientist faced with the law of constant proportion is to provide a possible explanation for this. Eminent scientist John Dalton tried to provide an explanation and introduced the concept of atoms. An atom is the smallest unit of matter. They are so small that thousands of them are stacked in your hair. The atom of one element is different from that of the other elements. Therefore, each of these atoms is depicted by different symbols.

Dalton used different combinations of straight lines and circles for denoting various atoms. However, the modern-day symbols consider the first or the first two letters of the element's name to denote an atom of the element. For example, Hydrogen is denoted by H and Oxygen by O. In some cases, and the Latin names are also considered. For example, the Latin name of Potassium is Kalium, and it is denoted by K. A list of names of elements is shown in Atom and Molecules Class 9.

Every atom has a definite mass, called the atomic mass unit. The atomic mass is the characteristic of the atom. The concept of atomic mass supports the law of constant proportion and the law of conservation of mass. A detailed analysis of atomic mass is provided in the NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules.

However, atoms cannot exist freely in nature. It interacts with the same or different atoms to form molecules. They can also form ions. These ions or molecules aggregate to form elements of compounds that we can see. These concepts are covered in Ch 3 Science Class 9.

3.3 What is Molecule?

A group of two or more atoms is joined together to form a molecule. A molecule can exist freely in nature. For example, Hydrogen, Oxygen, Calcium molecules can exist freely in nature. Molecules can be created from the same atoms or different atoms. If the molecules are made from the same kind of atoms, they are called molecules of elements. For example, Sodium and Potassium elements. Examples of molecules of elements are shown in Atoms and Molecules Class 9 NCERT Solutions.

Molecules can also be made from dissimilar atoms. When different atoms come together, they form a compound. The molecules made of different atoms are called molecules of compounds. Atoms and Molecules Class 9 NCERT PDF contains more examples of molecules of compounds.

Sometimes, the atoms can be charged. Charged atoms are called ions. An ion can always be charged positively or negatively charged. Positively charged ones are known as cations, while negative ones are known as anions. Sometimes, a molecule can contain more than one ion. Such molecules make a polyatomic ion. The individual charges on each atom determine the net charge of such ions. To learn more about polyatomic ions, go through Class 9 Science Chapter 3.

3.4 Writing Chemical Formulae

The symbolic representation of the compounds is said to be their chemical formulae. The chemical formula considers the net charge of individual ions in the formula. The charge of individual ions determines the valency of the radical. The valency of the individual ions determines how many ions of a particular kind will take part in the interaction.

For the proper representation of the chemical formula, the valency of individual ions must be balanced. The metal names should come before any non-metal name. In the case of polyatomic ions, the individual elements must be kept in a bracket if their overall valency is more than one. Otherwise, there is no need to use a bracket—for example, NaOH. You will learn more about writing chemical formulas when you go through the NCERT Solutions for Class 9 Science Chapter 3.

3.5 Molecular Mass and Mole Concept

The molecular mass of a compound is the sum of all the atomic masses. Therefore, the molecular mass is represented in atomic mass units. Read NCERT solutions for class 9 science chapter atoms and molecules to learn more about molecular mass.

Formula unit mass is calculated similarly to molecular mass. However, the ions’ mass is considered in calculating the formula unit mass of a compound. Learn more about formula mass units in ch 3 science class 9.

Mole concept is another important concept related to atoms and molecules. It defines the total number of molecules taking part in a chemical reaction. You can learn the mole concept in detail in Atoms and Molecules Class 9.

Exercise 3.5 total Solutions: 11 Questions (6 short question and 5 Long questions).

Key Features of NCERT Solutions for Class 9 Science Chapter 3

NCERT Solutions for Class 9 Science Chapter 3 has the following features.

• Chapter 3 Science Class 9, is presented comprehensively.

• All the concepts covered in the Atom and Molecules Class 9 chapter are explained with the help of relevant examples in these NCERT Solutions.

• The practice exercise of NCERT Class 9 Science Chapter 3 is solved as per the CBSE guidelines so that students can score good marks in the examination.

• The pointwise approach of chemistry ch 3 class 9 NCERT solutions help students revise the chapter before the examination.

• The topics covered in ch 3 science class 9 become easily understandable when students go through the atoms and molecules class 9 NCERT solutions.