NCERT solution for class 9 science atoms and molecules ( Chapter 3)

Solution for Exercise Questions

1. 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.

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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%

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?

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Answer :

When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. Carbon and oxygen are combined in the ratio 3:8 to give 11 carbon dioxide . Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon is formed, and 42g of oxygen is unused. This is govern by law of definite proportions.

3. What are polyatomic ions? Give examples.

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Answer :

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

4. Write the chemical formula of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

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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\(_3\)

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

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

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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

6. 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\(_3\)

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Answer :

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

(a)So, 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

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

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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 => 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

8. Convert into mole.

(a) 12g of oxygen gas

(b) 20g of water

(c) 22g of carbon dioxide

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Answer :

(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

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

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

(b) 0.5 mole of water molecules?

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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

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Answer :

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

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Answer :

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

Mass of solid sulpur = 16g

\( Number \ of \ moles = \) \( mass \ given \over molar \ mass \ of \ sulphur \) = \( 16 \over 256 \) = 0.0625 moles

Number of molecules = Number of moles × Avogadro number

\( = 0.0625 × 6.022 x 10²³ = 3.763 × 10^{22} \) 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)

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Answer :

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

1 mole of aluminium oxide (Al(S\(_2\))O(S\(_3\))) = 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 \( = 2 ×3.011 × 10^{20} \) molecules of aluminium oxide

\( = 6.022 × 10^{20} \)

Solution for Intext Question

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 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

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Answer :

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

     (5.3g) + (6g) -> (8.2g) + (2.2g) + (0.9g)

     11g = 11g

=>      LHS = RHS

i.e., Mass of reactant = Mass of product
which satisfies 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?

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Answer :

Given, ratio of H:O by mass to form water = 1:8
i.e., 1g of hydrogen required 8g of oxygen to react completely.

To find, mass of oxygen required to completely react with 3g of hydrogen.

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

∴: 24g of oxygen is required to completely react with 3g of hydrogen.

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

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Answer :

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

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

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Answer :

The postulate of Dalton’s atomic theory that can explain the law of definite proportions is, " The relative number and kinds of atoms are equal in given compounds."

1. Define the atomic mass unit?

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Answer :

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

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

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Answer :

We cannot see atoms with naked eye beacuse their size is very small, i.e., in nanometers( \( 1m = 10^9 nm \) ).

1. Write down the formulae of
(i) sodium oxide

(ii) aluminium chloride

(iii) sodium sulphide

(iv) magnesium hydroxide

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Answer :

The formulae are:

(i) sodium oxide -> Na\( _2 \)O

(ii) aluminium chloride -> AlCl\( _3 \)

(iii) sodium sulphide -> Na\( _2 \)S

(iv) magnesium hydroxide -> Mg (OH)\( _2 \)

2. 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\( _3 \)

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Answer :

The names of the compounds are:

(i) Al\( _2 \)(SO\( _4 \))\( _3 \) –->Aluminium sulphate

(ii) CaCl\( _2 \) -> Calcium chloride

(iii) K\( _2 \)SO\( _4 \) -> Potassium sulphate

(iv) KNO\( _3 \) -> Potassium nitrate

(v) CaCO\( _3 \) -> Calcium carbonate

3. What is meant by the term chemical formula?

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Answer :

Chemical formula of a chemical compound is the symbolic representation of its composition. Eg. chemical formula of water is H\( _2 \)O.

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

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

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Answer :

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

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

1. 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.

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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.

i.e., molecular mass of H\( _2 \) = atomic mass of H + atomic mass of H = 2 × atoms atomic mass of H = 2 × 1 = 2u
Similarly,
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

2. 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.

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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

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

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Answer :

Given, 1 mole of carbon weighs = 12g

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

i.e., \( 6.022 × 10^{23} \)atom = 12g

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

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

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Answer :

Given, Atomic mass of Na=23u, and 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

So, 100g of Na has more atoms.