NCERT Solutions Class 9 Science Chapter 3

Q.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-

Mass of the given compound = 0.24 g

Mass of boron = 0.096 g

Mass of oxygen = 0.144g

So,

\begin{array}{l}\text{Percentage of boron by weight =}\frac{\text{Mass of boron}}{\text{Mass of the compound}}\text{x 100}\\ \text{=}\frac{\text{0}\text{.096g}}{\text{0}\text{.24g}}\text{x 100}\\ \text{= 40 \%}\end{array}

\begin{array}{l}\text{Percentage of oxygen by weight =}\frac{\text{Mass of oxygen}}{\text{Mass of the compound}}\text{x 100}\\ \text{=}\frac{\text{0}\text{.144g}}{\text{0}\text{.24g}}\text{x 100}\\ \text{= 60 \%}\end{array}

Q.2 When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00g 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?

Ans-

Given that,

\begin{array}{l}{\text{C + O}}_2\to {\text{CO}}_2\\ \text{3g 8g 11g}\end{array}

The given data shows that in this reaction, 3.0 g of carbon and 8.0 g of oxygen react completely to produce 11.0 g of carbon dioxide.

In the second experiment, excess of oxygen (50 g) is present. When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from the total of 50 g of oxygen to produce 11 g of carbon dioxide and the rest of 42 g (50 g – 8 g) of oxygen will be left behind without any change.

The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction i.e., total mass of reactants remains same as the total mass of products.
Here, in terms of mass 3g of carbon and 50 g of oxygen combine to give 11 g of carbon dioxide and 42 g of unused oxygen keeping the total mass of combining chemicals by weight constant before and after the reaction.
Total mass of reactants = 3 + 50= 53 g

Total mass of products = 11 + 42 (un-used oxygen) = 53g
Total mass of reactants = total mass of products

Hence, the law of conservation of mass is proved.

2. Law of constant proportions states that in a chemical compound the elements are always present in definite proportions by mass. Here, irrespective of the excess quantity of oxygen available for burning of same weight of carbon, the amount of carbon dioxide will remain same. This means that (50 – 8) = 42 g of oxygen will remain unreacted.

Q.3 What are polyatomic ions? Give examples.

Ans-

Polyatomic ions are the group of atoms carrying a net charge on them. This charge may be positive or negative.

Examples:

Ammonium ion (NH₄⁺), Nitrate ion (NO₃^–), Carbonate ion (CO₃^2–), Sulphate ion (SO₄^2–)

Q.4 Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

Ans-

(e) Calcium carbonate

(a) Magnesium chloride: MgCl₂

(b) Calcium oxide: CaO

(c) Copper nitrate: Cu(NO₃)₂

(d) Aluminium chloride: AlCl₃

(e) Calcium carbonate: CaCO₃

Q.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-

Q.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

Ans-

(a) Gram atomic mass of carbon = 12 g

Gram atomic mass of hydrogen = 1g

Molar mass of ethyne (C₂H₂) = 2 x 12 + 2 x 1

= 26 g/mol

(b) Gram atomic mass of sulphur = 32 g

Molar mass of sulphur molecule (S₈) = 8 x 32

= 256 g/mol

(c) Gram atomic mass of phosphorus = 31 g

Molar mass of phosphorus molecule (P₄) = 4 x 31

= 124 g/mol

(d) Gram atomic mass of hydrogen = 1g

Gram atomic mass of chlorine = 35.5 g

Molar mass of hydrochloric acid (HCl) = 1 + 35.5

= 36.5 g/mol

(e) Gram atomic mass of hydrogen = 1g

Gram atomic mass of nitrogen = 14 g

Gram atomic mass of oxygen = 16 g

Molar mass of nitric acid (HNO₃) = 1 + 14 + 3 x 16

= 63 g/mol

Q.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) ?

Ans-

a. Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen expressed in grams = 14 g
b. Mass of 4 moles of aluminium atoms = 4 x Atomic mass of aluminium expressed in grams = 4 x 27 = 108 g
c.Molar mass of sodium sulphite (Na2SO3) = (2 x 23 + 32 + 3 x 16) g/mol

\begin{array}{l}\\ \text{= (46 + 32 + 48) g/mol}\\ \text{= 126 g/mol}\\ \text{Mass of 10 moles of sodium sulphite = 10 mol x Molar mass of sodium sulphite}\\ \text{= 10 mol x 126 g/mol}\\ \text{= 1260 g}\end{array}

Q.8 Convert into mole:

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide

Ans-

(a)

\begin{array}{l}\text{Given mass of oxygen gas = 12 g}\\ \text{Molar mass of oxygen gas (O2) = 2 x 16 g/mol}\\ \text{= 32 g/mol}\\ \text{The number of moles (n) =}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{=}\frac{\text{12 g}}{\text{32 g/mol}}\text{= 0}\text{.375 mol}\end{array}

Hence, 12 g of oxygen gas = 0.375 mol of oxygen gas

(b)

\begin{array}{l}\text{Given mass of water = 20 g}\\ \text{Molar mass of water (H2O) = (2 x 1 + 1 x 16) g/mol}\\ \text{= 18 g/mol}\\ \text{The number of moles (n) =}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{=}\frac{\text{20 g}}{\text{18 g/mol}}\text{= 1}\text{.11 mol}\end{array}

Hence, 20 g of water = 1.11 mol of water

\begin{array}{l}\text{(c)}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{Given mass of carbon dioxide = 22 g}\\ \text{Molar mass of carbon dioxide (CO2) = (12 + 2 x 16) g/mol}\\ \text{= 44 g/mol}\\ \text{The number of moles (n) =}\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{=}\frac{\text{22 g}}{\text{44 g/mol}}\text{= 0}\text{.5 mol}\end{array}

Hence, 20 g of carbon dioxide = 0.5 mol of carbon dioxide

Q.9 What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans-

\begin{array}{l}\left(\text{a}\right)\\ \text{Mass of 0}\text{.2 mol atoms of oxygen = 0}\text{.2 mol x Molar mass of oxygen atom}\\ \text{= 0}\text{.2 mol x 16 g/mol}\\ \text{= 3}\text{.2 g}\\ \left(\text{b}\right)\\ \text{Molar mass of water}\left({\text{H}}_2\text{O}\right)\text{= (2 x 1 + 16) g/mol}\\ \text{= 18 g/mol}\\ \text{Mass of 0}\text{.5 mol molecules of water = 0}\text{.5 mol x Molar mass of water}\\ \text{= 0}\text{.5 mol x 18 g/mol}\\ \text{= 9 g}\end{array}

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

Ans-

Mass of solid sulphur = 16 g

Molar mass of sulphur molecule (S₈) = 8 × 32g/mol = 256 g/mol

Avogadro’s number = 6.022 × 10²³

\begin{array}{l}{\text{Number of moles of S}}_8\text{molecules =}\frac{\text{Given mass of sulphur}}{{\text{Molar mass of S}}_8}\\ \text{=}\frac{\text{16 g}}{\text{256 g/mol}}\text{=}\frac{\text{16}}{256}\text{mol}\\ \text{No}{\text{. of S}}_8\text{molecules in the sample = No}{\text{. of moles of S}}_8\text{x Avogadro’s number}\\ \text{=}\frac{16}{256}\text{x 6}{\text{.022 x 10}}^{\text{23}}\\ \text{= 3}{\text{.76 x 10}}^{\text{22}}\text{molecules}\end{array}

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

Ans-

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

\begin{array}{l}\text{Mass of aluminium oxide = 0}\text{.051 g}\\ \text{Molar mass of aluminium oxide}\left({\text{Al}}_2{\text{O}}_3\right)=\left(2\times 27+3\times 16\right)\text{g/mol}\\ \text{= 102 g/mol}\\ \text{Number of moles of aluminium oxide molecules =}\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{0.051\text{g}}{102\text{g/mol}}\\ =5.0\times {10}^{-4}\text{mol}\end{array}

From the stoichiometry,

Al₂O₃ ≡ 2 Al³⁺
∴ 1 mol of Al₂O₃ contains = 2 mol of Al³⁺ ions
∴ 5.0 × 10^–4 mol Al₂O₃ contains = 2 × 5.0 × 10^–4 mol of Al³⁺ ions = 10^–3 mol of Al³⁺ ions

So,
Number of Al³⁺ ions present in 10^–3 mol
= 10^–3 mol × 6.022 × 10²³ ions mol^–1
= 6.022 × 10²⁰ ions

Hence, 0.051 g of aluminium oxide (Al₂O₃) contains 6.022 × 10²⁰ Al³⁺ ions.