NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules

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Updated at 2021-05-07


NCERT solutions for class 9 Science Chapter 3 Atoms And Molecules Exercise Questions

Q1 ) A 0.24g sample of 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.



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

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

\( Percentage \ of \ boron = \) \( mass \ of \ boron \over mass \ of \ the \ compound \) \( × 100 \) \( = \) \( 0.096 \over 0.24 \) × 100 = 40%

Percentage of oxygen = 100 – Percentage of boron

= 100 – 40 = 60%

Q2 ) 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?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

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

Thus carbon and oxygen are combined in the ratio of 3 : 8 to give 11g of carbon dioxide .

Hence for 3 g of carbon and 50 g of oxygen, 8 g of oxygen is used and 11g of carbon is formed while 42 g of oxygen is left unused. This chemical combination is governed by law of definite proportions.

Q3 ) What are polyatomic ions? Give examples.



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

Polyatomic ions are the ions that contain more than one atom but they behave as a single unit. Example: OH \(^- \) , H\(_2 \)PO\( _4^– \)

Q4 ) Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


The chemical formula are:

(a) Magnesium chloride – MgCl\(_2\)

(b) Calcium oxide – CaO

(c) Copper nitrate – Cu(NO3)\(_2\)

(d) Aluminium chloride – AlCl\(_3\)

(e) Calcium carbonate – CaCO\(_3\)

Q5 ) Give the names of the elements present in the following compounds.
(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


The names of the elements present are:

(a) Quick lime – Calcium and Oxygen

(b) Hydrogen bromide – Hydrogen and Bromine

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

(d) Potassium sulphate – Sulphur, Oxygen, Potassium

Q6 )
Calculate the molar mass of the following substances.

(a) Ethyne, C\(_2\)H\(_2\)

(b) Sulphur molecule, S\(_8\)

(c) Phosphorus molecule, P\(_4\) (Atomic mass of phosphorus =31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO\(_3\)



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


We know that molar mass of a substance is the sum of the atomic masses of all the elements.

(a) Molar mass of Ethyne C\(_2\)H\(_2\)
= 2 × Mass of C+2 × Mass of H
= (2 × 12) + (2 × 1) = 24 + 2
= 26g

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

(c) Molar mass of Phosphorus molecule, P4
= 4 × Mass of P = 4 × 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 × Mass of O = 1 + 14 + 3 × 16 = 63g

Q7 )
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 (Na\(_2\)SO\(_3\))?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


(a) Atomic mass of nitrogen atoms = 14u

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

Therefore, 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 \(\Rightarrow \) 4 moles of aluminium atoms = 4 × 27 = 108g

(c) Mass of 1 mole of sodium sulphite Na\(_2\)SO\(_3\)
= Molecular mass of sodium sulphite
= 2 × Mass of Na + Mass of S + 3 × Mass of O
= (2 × 23) + 32 +(3 × 16) = 46 + 32 + 48 = 126g

Therefore, mass of 10 moles of Na\(_2\)SO\(_3\) = 10 × 126 = 1260g

Q8 )
Convert into mole.

(a) 12g of oxygen gas

(b) 20g of water

(c) 22g of carbon dioxide



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


(a) Given mass of oxygen gas = 12g

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

\( Number \ of \ moles \ \)

= \( Given \ mass \over molar \ mass \ of \ oxygen \ gas \) \( = \) \( 12 \over 32 \) \( = 0.375 \ moles \)

(b) Given mass of water = 20g

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

\( Number \ of \ moles \ \)

= \( Given \ mass \over molar \ mass \ of \ water \) \( = \) \( 20 \over 18 \) \( = 1.11 \ moles \)

(c) Given mass of carbon dioxide = 22g

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

\( Number \ of \ moles \)

= \( Given \ mass \over molar \ mass of \ carbon \ dioxide \) \( = \) \( 22 \over 44 \) \( = 0.5 \ moles \)

Q9 ) What is the mass of:
(a) 0.2 mole of oxygen atom?

(b) 0.5 mole of water molecules?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


(a) Mass of 1 mole of oxygen atoms = 16u

\(\therefore \) Mass of 0.2 moles of oxygen atoms = 0.2 × 16 = 3.2u

(b) Mass of 1 mole of water molecules = 18u

\(\therefore \) Mass of 0.5 moles of water molecules = 0.5 × 18 = 9u


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

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

\( Number \ of \ moles \ = \) \( Mass \ given \over molar \ mass \ of \ oxygen \ gas \) \( = \) \( 12 \over 32 \) \( = 0.375 \ moles \)

(b) Mass of water = 20g

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

\( Number \ of \ moles \ = \) \( Mass \ given \over molar \ mass \ of \ water \) \( = \) \( 20 \over 18 \) \( = 1.11 \ moles \)

(c) Mass of carbon dioxide = 22g

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

\( Number \ of \ moles = \) \( Mass \ given \over molar \ mass of \ carbon \ dioxide \) \( = \) \( 22 \over 44 \) \( = 0.5 \ moles \)



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


(a) Mass of 1 mole of oxygen atoms = 16u

∴ Mass of 0.2 moles of oxygen atoms = 0.2 × 16 = 3.2u

(b) Mass of 1 mole of water molecules = 18u

∴ Mass of 0.5 moles of water molecules = 0.5 × 18 = 9u

Q10 ) Calculate the number of molecules of sulphur (S\(_8\)) is present in 16g of solid sulphur.



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


Molecular mass of Sulphur (S\(_8\))
= 8 × Mass of Sulphur = 8 × 32 = 256g

Given mass of solid sulphur = 16g

\( Number \ of \ moles \) = \( Given \ mass \over Molar \ mass \ of \ Sulphur \) = \( 16 \over 256 \) = 0.0625 moles

Number of molecules

= Number of moles × Avogadro number

\( = 0.0625 × 6.022 × 10^{23} \)

\( = 3.763 × 10^{22} \) molecules

Q11 ) 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)



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :


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

1 mole of aluminium oxide Al2O3

= 2 × Mass of aluminium + 3 × Mass of Oxygen

= (2 × 27) + (3 × 16) = 54 + 48 = 102g

1 mole of aluminium oxide \( = 102g = 6.022 × 10^{23} \) molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has

\( = \) \( 0.051 × 6.022 × 10^{23} \over 102 \)

\( = 3.011 × 10^{20} \) molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions present in 0.051g of aluminium oxide

\( = 2 ×3.011 × 10^{20} \) ions of aluminium

\( = 6.022 × 10^{20} \) ions of aluminium

NCERT solutions for class 9 Science Chapter 3 Atoms And Molecules Intext Exercise

Q1 ) 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 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 \(\rightarrow \) Sodium acetate + carbon dioxide + water



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

Sodium carbonate + Acetic acid \(\rightarrow \) Sodium acetate + Carbon dioxide + Water

(5.3g) + (6g) \(\rightarrow \) (8.2g) + (2.2g) + (0.9g)

\(\Rightarrow \) 11g = 11g

\(\Rightarrow \) LHS = RHS

\(\therefore \) Mass of reactant = Mass of product

which satisfies the law of conservation of mass.

Q2 ) 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?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

Given, ratio of H : O by mass to form water = 1 : 8
\(\Rightarrow \) 1 g of hydrogen required 8 g of oxygen to react completely.

We have to find the mass of oxygen that is required to completely react with 3 g of hydrogen.

So, 3 g of hydrogen will require \( (8 × 3) \)g of oxygen

\(\therefore \) 24 g of oxygen is required to completely react with 3 g of hydrogen.

Q3 ) Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

The postulate of Dalton’s Atomic theory which is a result of the law of conservation of mass is, “Atoms are indivisible particles, which can neither be created nor be destroyed”.

Q4 ) Which postulate of Dalton’s atomic theory can explain the law of definite proportions?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

The postulate of Dalton’s atomic theory that can explain the law of definite proportions is, " The relative number and kinds of atoms in a given compound remain constant."

Q1 ) Define the atomic mass unit?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

One atomic mass is equal to 1/12th the mass of one atom of carbon-12 i.e., 1 amu = \( 1 \over 12 \) th mass of one atom of carbon.

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



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

We cannot see atoms with naked eye because their size is extremely small, i.e., their size is in nanometers such that 1nm = 10-9 m . They can be only seen with the help of an electron microscope which has an extremely high resolution.

Q1 ) Write down the formulae of
(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

The formulae are:

(i) sodium oxide \(\rightarrow \) Na\( _2 \)O

(ii) aluminium chloride \(\rightarrow \) AlCl\( _3 \)

(iii) sodium sulphide \(\rightarrow \) Na\( _2 \)S

(iv) magnesium hydroxide \(\rightarrow \)Mg (OH)\( _2 \)

Q2 ) Write down the names of compounds represented by the following formulae:
(i) Al\( _2 \)(SO\( _4 \))\( _3 \)

(ii) CaCl\( _2 \)

(iii) K\( _2 \)SO\( _4 \)

(iv) KNO\( _3 \)

(v) CaCO\( _3 \)



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

The names of the compounds are:

(i) Al\( _2 \)(SO\( _4 \))\( _3 \) \(\rightarrow \) Aluminium sulphate

(ii) CaCl\( _2 \) \(\rightarrow \) Calcium chloride

(iii) K\( _2 \)SO\( _4 \) \(\rightarrow \) Potassium sulphate

(iv) KNO\( _3 \) \(\rightarrow \) Potassium nitrate

(v) CaCO\( _3 \) \(\rightarrow \) Calcium carbonate

Q3 ) What is meant by the term chemical formula?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

A chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule by using chemical element symbols, numbers etc. Chemical formula of a chemical compound is the symbolic representation of its composition. E.g. chemical formula of water is H\( _2 \)O.

Q4 ) How many atoms are present in a
(i) H\( _2 \)S molecule and

(ii) PO\( _4 \)\(^{3-} \) ion?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

The number of atoms present are:
(i) H\( _2 \)S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence H\( _2 \)S molecule a total of 3 atoms.

(ii) PO\( _4 \)\(^{3-} \) ion has 1 atom of phosphorus and 4 atoms of oxygen hence PO\( _4 \)\(^{3-} \) ion has a total of 5 atoms.

Q1 ) Calculate the molecular masses of H\( _2 \) , O\( _2 \) , Cl\( _2 \), CO\( _2 \), CH\( _4 \), C\( _2 \)H\( _6 \), C\( _2 \)H\( _4 \), NH\( _3 \), CH\( _3 \)OH.



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

We know that molecular mass of a substance is the sum of the atomic masses of all the atoms in the molecule of the substance.

The molecular mass of H\( _2 \)  = atomic mass of H + atomic mass of H = 2 × atoms atomic mass of H = 2 × 1 = 2u

The molecular mass of O\( _2 \)  = 2 × atoms atomic mass of O = 2 × 16 = 32u

The molecular mass of Cl\( _2 \)  = 2 × atoms atomic mass of Cl = 2 × 35.5 = 71u

The molecular mass of CO\( _2 \)   = atomic mass of C + 2 × atomic mass of O = 12 + ( 2 × 16) = 44u

The molecular mass of CH\( _4 \)  = atomic mass of C + 4 × atomic mass of H = 12 + ( 4 × 1) = 16u

The molecular mass of C\( _2 \)H\( _6 \)  = 2 × atomic mass of C + 6 × atomic mass of H = (2 × 12) + (6 × 1) = 24 + 6 = 30u

The molecular mass of C\( _2 \)H\( _4 \)  = 2 × atomic mass of C + 4 × atomic mass of H = (2 × 12) + (4 × 1) = 24 + 4 = 28u

The molecular mass of NH\( _3 \)  = atomic mass of N + 3 × atomic mass of H = 14 + (3 × 1) = 17u

The molecular mass of CH\( _3 \)OH   = atomic mass of C + 3 × atomic mass of H + atomic mass of O + atomic mass of H = 12 + (3 × 1) + 16 + 1 = 12 + 3 + 17 = 32u

Q2 ) Calculate the formula unit masses of ZnO, Na\( _2 \)O, K\( _2 \)CO\( _3 \), given atomic masses of Zn = 65u, Na = 23 u, K=39u, C = 12u, and O=16u.



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

Given, atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, O = 16u

We know that formula unit mass of a substance is the sum of the atomic masses of all the atoms in the formula unit of the substance.

So, the formula unit mass of ZnO  = Atomic mass of Zn + Atomic mass of O = 65 + 16 = 81u

The formula unit mass of Na\( _2 \)O  = 2 × Atomic mass of Na + Atomic mass of O = (2 × 23) +16 = 46 + 16 = 62u

The formula unit mass of K\( _2 \)CO\( _3 \)  = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O = (2 × 39) + 12 + (3 × 16) = 78 + 12 + 48 = 138u

Q1 ) If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

Given, 1 mole of carbon weighs = 12g

1 mole of carbon has \( = 6.022 × 10^{23} \)atoms

i.e., \( 6.022 × 10^{23} \)atom = 12g

So, mass of 1 atom of carbon \( = \) \( 12 \over 6.022 × 10^{23} \) \( = 1.99 × 10^{-23} \)g

Q2 ) Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23u, Fe = 56 u)?



NCERT Solutions for Class 9 Science Chapter 3 Atoms And Molecules


Answer :

Given, Atomic mass of Na=23u

Atomic mass of Fe= 56u

Now,
23g of Na contains \( = 6.022 × 10^{23} \) atoms

1g of Na contains = \( 6.022 × 10^{23} \over 23 \)atoms

100g of Na contains = \( 6.022 × 10^{23} \over 23 \) × 100 \( = 2.6182 × 10^{24} \) atoms

Now, 56g of Fe contains = 6.022 × 1023 atoms

1g of Fe contains = \( 6.022 × 10^{23} \over 56 \)atoms

100g of Fe contains = \( 6.022 × 10^{23} \over 56 \) × 100 \( = 1.075 × 10^{24} \) atoms

Hence 100g of Na has more atoms.



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