## Exercise 5A

#### Question 1

State:

(a) Gay-Lussac's Law of combining volumes.

(b) Avogadro's law

**Answer**

(a) Gay-Lussac's Law of combining volumes — When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."

#### Question 2

(a) What do you mean by stoichiometry?

(b) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.

(c) Differentiate between N_{2} and 2N.

**Answer**

(a) Stoichiometry measures quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry.

(b) Atomicity is the number of atoms in a molecule of an element.

Atomicity of hydrogen is 2, phosphorus is 4 and sulphur is 8.

(c) Difference between N_{2} and 2N

N_{2} | 2N |
---|---|

It means 1 molecule of nitrogen. | It means two atoms of nitrogen. |

It can exist independently. | It cannot exist independently. |

#### Question 3

Explain Why?

(a) "The number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium at the same temperature and pressure."

(b) "When stating the volume of a gas, the pressure and temperature should also be given."

(c) Inflating a balloon seems to violate Boyle's law.

**Answer**

(a) Avogadro's Law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."

Considering equal volumes of hydrogen and helium,

volume of hydrogen gas = volume of helium gas

According to Avogadro's Law:

n molecules of hydrogen = n molecules of helium gas

i.e., nH_{2} = nHe

1 molecule of hydrogen has 2 atoms of hydrogen and 1 molecule of helium has 1 atom of helium

∴ 2H = He

∴ atoms in hydrogen are double the atoms of helium.

(b) Since, the volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard values of temperature and pressure to which gas volumes can be referred.

(c) According to Boyle's law, the volume of a given mass of dry gas is inversely proportional to its pressure at a constant temperature. When we inflate a balloon, the volume of air keeps increasing and at the same time the pressure of air also increases due to which balloon inflates. As pressure and volume of air increase simultaneously, hence this seems to violate Boyle's law.

## Numerical Problems

#### Question 4(a)

Calculate the volume of oxygen at STP required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.

2CO + O_{2} ⟶ 2CO_{2}

**Answer**

From equation:

$\begin{matrix} 2\text{CO} & + & \text{O}_2 & \longrightarrow & 2 \text{CO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

[By Gay Lussac's law]

2 V of CO requires = 1V of O_{2}

∴ 100 litres of CO requires = $\dfrac{1}{2}$ x 100 = 50 litres.

Hence, **required volume of oxygen is 50 litres**.

#### Question 4(b)

200 cm^{3} of hydrogen and 150 cm^{3} of oxygen are mixed and ignited, as per the following reaction,

2H_{2} + O_{2} ⟶ 2H_{2}O

What volume of oxygen remains unreacted?

**Answer**

$\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

2 Vol. of hydrogen reacts with 1 Vol. of oxygen

∴ 200 cm^{3} of hydrogen reacts with = $\dfrac{1}{2}$ x 200 = 100 cm^{3} of oxygen.

Hence, **unreacted oxygen is 150 - 100 = 50cm ^{3}**

#### Question 5

24 cc Marsh gas (CH_{4}) was mixed with 106 cc oxygen and then exploded. On cooling, the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.

**Answer**

This experiment supports Gay-Lussac's law of combining volumes.

Since the unchanged oxygen is 58 cc so, used oxygen 106 - 58 = 48cc

According to Gay-Lussac's law, the volumes of gases reacting should be in a simple ratio.

$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & 2\text{CO}_2 + \text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} \\ 24 \text{ cc} & : & 48 \text{ cc} \end{matrix}$

Hence, **methane and oxygen are in the ratio 1:2 .**

#### Question 6

What volume of oxygen would be required to burn completely 400 ml of acetylene [C_{2}H_{2}]? Also calculate the volume of carbon dioxide formed.

2C_{2}H_{2} + 5O_{2} ⟶ 4CO_{2} + 2H_{2}O (l)

**Answer**

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$

[By Gay Lussac's law]

2 Vol. of C_{2}H_{2} requires 5 Vol. of oxygen

∴ 400 ml C_{2}H_{2} will require $\dfrac{5}{2}$ x 400

= 1000 ml of Oxygen

Hence, **required volume of oxygen = 1000 ml**

Similarly,

2 Vol. of C_{2}H_{2} produces 4 Vol. of Carbon dioxide

∴ 400 ml of C_{2}H_{2} produces $\dfrac{4}{2}$ x 400

= 800 ml of Carbon dioxide

Hence, **carbon dioxide produced = 800 ml**

#### Question 7

112 cm^{3} of H_{2}S (g) is mixed with 120 cm^{3} of Cl_{2} (g) at STP to produce HCl (g) and sulphur (s). Write a balanced equation for this reaction and calculate

(i) the volume of gaseous product formed

(ii) composition of the resulting mixture

**Answer**

$\begin{matrix} \text{H}_2\text{S} & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} & + & \text{S} \\ 1\text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & : & 1\text{ vol.} \\ \end{matrix}$

(i) At STP,

1 mole gas occupies = 22.4 L.

1 mole H_{2}S gas produces = 2 moles HCl gas,

∴ 22.4 L H_{2}S gas produces

= 22.4 × 2

= 44.8 L HCl gas.

Hence, 112 cm^{3} H_{2}S gas will produce

= 112 × 2

= 224 cm^{3} HCl gas.

Hence, **224 cm ^{3} HCl gas is produced.**

(ii) 1 mole H_{2}S gas consumes = 1 mole Cl_{2} gas.

Hence, 22.4 L H_{2}S gas consumes = 22.4 L Cl_{2} gas at STP.

∴ 112 cm^{3} H_{2}S gas consumes = 112 cm^{3} Cl_{2} gas.

120 cm^{3} - 112 cm^{3} = 8 cm^{3} Cl_{2} gas remains unreacted.

Hence, **the composition of the resulting mixture is 224 cm ^{3} HCl gas + 8 cm^{3} Cl_{2} gas.**

#### Question 8

1250 cc of oxygen was burnt with 300 cc of ethane [C_{2}H_{6}]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed:

2C_{2}H_{6} + 7O_{2} ⟶ 4CO_{2} + 6H_{2}O

**Answer**

$\begin{matrix} 2\text{C}_2\text{H}_6 & + & 7\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 6\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$

[By Gay Lussac's law]

2 Vol. of C_{2}H_{6} requires 7 Vol. of oxygen

∴ 300 cc C_{2}H_{6} will require $\dfrac{7}{2}$ x 300

= 1050 cc of Oxygen

Hence, **unused oxygen = 1250 - 1050 = 200 cc**

Similarly,

2 Vol. of C_{2}H_{6} produces 4 Vol. of carbon dioxide

∴ 300 cc C_{2}H_{6} produces $\dfrac{4}{2}$ x 300

= 600 cc of Carbon dioxide

Hence, **carbon dioxide produced = 600 cc.**

#### Question 9

What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C_{2}H_{4}] at 273°C and 380 mm of Hg pressure?

C_{2}H_{4} + 3O_{2} ⟶ 2CO_{2} + 2H_{2}O

**Answer**

$\begin{matrix} \text{C}_2\text{H}_4 & + & 3\text{O}_2 & \longrightarrow & 2\text{CO}_2 + 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 3 \text{ vol.} \\ 11 \text{ lit} & : & 33 \text{ lit} \end{matrix}$

STP | Given Values |
---|---|

P_{1} = 760 mm of Hg | P_{2} = 380 mm of Hg |

V_{1} = x lit | V_{2} = 33 lit |

T_{1} = 273 K | T_{2} = 273 + 273 K = 546 K |

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{760 \times x}{273} = \dfrac{380 \times 33}{546} \\[0.5em] x = \dfrac{380 \times 33 \times 273}{546 \times 760 } \\[0.5em] x = \dfrac{3,423,420}{414,960} \\ \\[0.5em] x = 8.25 \text{ lit}$

Hence, **volume of oxygen required = 8.25 lit.**

#### Question 10

Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at STP.

CH_{4} + 2Cl_{2} ⟶ CH_{2}Cl_{2} + 2HCl

**Answer**

$\begin{matrix} \text{CH}_4 & + & 2\text{Cl}_2 & \longrightarrow & \text{CH}_2\text{Cl}_2 & + & 2\text{HCl} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1 \text{ vol.} & : & 2 \text{ vol.}\\ \end{matrix}$

volume of HCl gas formed = ?

[By Gay Lussac's law]

1 Vol of methane produces = 2 Vol. HCl

∴ 40 ml of methane produces = 80 ml HCl

volume of chlorine gas required = ?

For 1 Vol of methane = 2V of Cl_{2} required

∴ for 40 ml of methane = 40 x 2 = 80 ml of Cl_{2} is required.

Hence, **volume of HCl gas formed = 80 ml and chlorine gas required = 80 ml**

#### Question 11

What volume of propane is burnt for every 500 cm^{3} of air used in the reaction under the same conditions? (assuming oxygen is 1/5th of air)

C_{3}H_{8} + 5O_{2} ⟶ 3CO_{2} + 4H_{2}O

**Answer**

Given, oxygen is 1⁄5^{th} of air = $\dfrac{1}{5}$ of 500 = 100 cm^{3}

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3 \text{ vol.} & \\ \end{matrix}$

[By Gay Lussac's law]

5 Vol. of O_{2} requires 1 Vol. of propane

∴ 100 cm^{3} of O_{2} will require = $\dfrac{1}{5}$ x 100 = 20 cm^{3}

Hence, **propane burnt = 20 cm ^{3} or 20 cc**

#### Question 12

450 cm^{3} of nitrogen monoxide and 200 cm^{3} of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.

2NO + O_{2} ⟶ 2NO_{2}

**Answer**

$\begin{matrix} 2\text{NO} & + & \text{O}_2 & \longrightarrow & 2\text{NO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. of O_{2} reacts with = 2V of NO

200 cm^{3} oxygen will react with

= 200 × 2

= 400 cm^{3} of NO

∴ remaining NO is 450 - 400 = 50 cm^{3}

NO_{2} = ?

1 Vol. of O_{2} produces 2 Vol. of NO_{2}

∴ 200 cm^{3} of oxygen produces = $\dfrac{2}{1}$ x 200 = 400cm^{3}

Hence, **NO _{2} produced = 400 cm^{3} and unused oxygen is 50 cm^{3}, so total mixture = 400 + 50 = 450 cm^{3}**

#### Question 13

If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.

**Answer**

$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. of chlorine reacts with = 1 Vol. of hydrogen

∴ 4 litres of chlorine will react with only 4 litres of hydrogen,

hence, 6 - 4 = 2 litres of hydrogen will remain unreacted.

Since, vol. of HCl gas formed is twice that of chlorine used,

∴ vol.of HCl formed will be 4 x 2 = 8 litres However HCl dissolves in water.

Hence, **2 litres of hydrogen is the residual gas, as HCl formed dissolves in water.**

#### Question 14

Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.

4NH_{3} + 5O_{2} ⟶ 4NO + 6H_{2}O

If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

**Answer**

$\begin{matrix} 4\text{NH}_3 & + & 5\text{O}_2 & \longrightarrow & 4\text{NO} & + & 6\text{H}_2\text{O} \\ 4 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

9 litres of reactants produces = 4 litres of NO

So, 27 litres of reactants will produces

$=\dfrac{4}{9} \times 27 \\[0.5em] = 12 \text{ litres}$

Hence, **volume of nitrogen monoxide produced = 12 litres**

#### Question 15

A mixture of hydrogen and chlorine occupying 36 cm^{3} was exploded. On shaking it with water, 4 cm^{3} of hydrogen was left behind. Find the composition of the mixture.

**Answer**

According to Gay lussac's law,

$\begin{matrix} \text{H}_2 & + & \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

As, 4 cm^{3} of hydrogen was left behind, hence, 36 - 4 = 32 cm^{3} of mixture of hydrogen and chlorine exploded.

As, 1 Vol. of hydrogen requires 1 Vol. of oxygen

∴ 16 cm^{3} hydrogen requires 16 cm^{3} of oxygen

∴ **Mixture is 20 cm ^{3} (i.e., 16 + 4) of hydrogen and 16 cm^{3} of chlorine.**

#### Question 16

What volume of air (containing 20% O_{2} by volume) will be required to burn completely 10 cm^{3} each of methane and acetylene?

CH_{4} + 2O_{2} ⟶ CO_{2} + 2H_{2}O

2C_{2}H_{2} + 5O_{2} ⟶ 4CO_{2} + 2H_{2}O

**Answer**

$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. CH_{4} requires 2 Vol. of O_{2}

∴ 10 cm^{3} CH_{4} will require 2 x 10

= 20 cm^{3} of O_{2}

Given, air contains 20% O_{2} by volume.

Let $x$ volume of air contain 20 cm^{3} of O_{2}

$\Rightarrow \dfrac{20}{100} \times x = 20 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 20 \\[1em] \Rightarrow x = 100 \text{ cm}^3$

∴ 20 cm^{3} O_{2} is present in **100 cm ^{3} of air.**

Similarly, 2 Vol C_{2}H_{2} requires 5 Vol. of oxygen

∴ 10 cm^{3} C_{2}H_{2} will require $\dfrac{5}{2}$ x 10

= 25 cm^{3} of oxygen

Given, air contains 20% O_{2} by volume

Let $x$ volume of air contain 25 cm^{3} of O_{2}

$\Rightarrow \dfrac{20}{100} \times x = 25 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 25 \\[1em] \Rightarrow x = 125 \text{ cm}^3$

∴ 25 cm^{3} O_{2} is present in **125 cm ^{3} of air.**

**Hence, total volume of air required is 100 + 125 = 225 cm ^{3}**

#### Question 17

LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to the atmosphere.

C_{3}H_{8} + 5O_{2} ⟶ 3CO_{2} + 4H_{2}O

2C_{4}H_{10} + 13O_{2} ⟶ 8CO_{2} + 10H_{2}O

**Answer**

Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litres

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} \\ 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} \\ \end{matrix}$

1 Vol. C_{3}H_{8} produces carbon dioxide = 3 Vol

So, 6 litres C_{3}H_{8} will produce carbon dioxide = 3 x 6 = 18 litres

2 Vol. C_{4}H_{10} produces carbon dioxide = 8 Vol

So, 4 litres C_{4}H_{10} will produce carbon dioxide = $\dfrac{8}{2}$ x 4 = 16 litres

**Hence, 34 (i.e., 18 + 16) litres of CO _{2} is produced.**

#### Question 18

200 cm^{3} of CO_{2} is collected at STP when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in the original mixture.

2C_{2}H_{2} (g) + 5O_{2} (g) ⟶ 4CO_{2} (g)+ 2H_{2}O (g)

**Answer**

According to Gay lussac's law,

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}$

4 Vol of CO_{2} is collected with 2 Vol. of C_{2}H_{2}

So, 200 cm^{3} CO_{2} will be collected with

$=\dfrac{2}{4} \times 200 \\[0.5em] = 100 \text{ cc}$

Similarly, 4 Vol of CO_{2} is produced by 5 Vol of O_{2}

So, 200 cm^{3} CO_{2} will be produced by = $\dfrac{5}{4}$ x 200 = 250 cm^{3}

**Hence, acetylene = 100 cm ^{3} and oxygen = 250 cm^{3}**

#### Question 19

You have collected (a) 2 litres of CO_{2} (b) 3 litres of chlorine (c) 5 litres of hydrogen (d) 4 litres of nitrogen and (e) 1 litres of SO_{2}, under similar conditions of temperature and pressure. Which gas sample will have:

(a) the greatest number of molecules, and

(b) the least number of molecules?

Justify your answers.

**Answer**

According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain same number of molecules. So, under the same conditions of temperature and pressure, if volume of gas is decreased the number of molecules will also decrease.

Hence,

(a) 5 litres of hydrogen contain the greatest number of molecules as it has the highest volume.

(b) 1 litre of SO_{2} contains the least number of molecules since it has the smallest volume.

#### Question 20

The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.

Gas | Volume (in litres) | Number of molecules |
---|---|---|

Chlorine | 10 | |

Nitrogen | 20 | x |

Ammonia | 20 | |

Sulphur dioxide | 5 |

**Answer**

Gas | Volume (in litres) | Number of molecules |
---|---|---|

Chlorine | 10 | x/2 |

Nitrogen | 20 | x |

Ammonia | 20 | x |

Sulphur dioxide | 5 | x/4 |

**Reason** — According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain same number of molecules. If 20 lit of nitrogen contains x molecules then 20 lit of ammonia will also contain x molecules. As volume of chlorine is half that of nitrogen so it will contain half the number of molecules of nitrogen i.e., x/2. Similarly, sulphur dioxide will contain x/4 molecules.

#### Question 21

(i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B?

The gases A and B are under the same conditions of temperature and pressure.

(ii) Name the law on which the above problem is based

**Answer**

(a) Given, 150 cc of gas A contains X molecules. According to Avogadro's law, 150 cc of gas B will also contain X molecules.

So, 75 cc of gas B will contain $\dfrac{x}{2}$ molecules.

(b) The problem is based on Avogadro's law.

## Exercise 5B

#### Question 1

(a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.

(b) What is the value of Avogadro's number ?

(c) What is the value of molar volume of a gas at S.T.P.?

**Answer**

(a) The relative atomic masses of any element is the weighted average of the relative atomic masses of it's natural isotopes. Chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio 3 : 1.

The average relative atomic mass of Cl = $\dfrac{(35 \times 3) + (37 \times 1)}{4}$ = 35.5

(b) 6.022 × 10^{23}

(c) The molar volume of a gas is 22.4 dm^{3} (litre) or 22400 cm^{3} (ml) at S.T.P.

#### Question 2

Define or explain the terms:

(a) Vapour density

(b) Molar volume

(c) Relative atomic mass

(d) Relative molecular mass

(e) Avogadro's number

(f) Gram atom

(g) Mole

**Answer**

(a) **Vapour density** is defined as the ratio between the masses of equal volumes of gas (or vapour) and hydrogen under the same conditions of temperature and pressure.

(b) The **molar volume** of a gas is the volume occupied by one gram-molecular mass or by one mole of the gas at S.T.P. It is equal to 22.4 dm^{3}.

(c) The **Relative atomic mass** of an element is the number of times one atom of the element is heavier than $\dfrac{1}{12}$ times of the mass of an atom of carbon-12.

(d) The **Relative molecular mass** of an element or a compound is the number that represents how many times one molecule of the substance is heavier than $\dfrac{1}{12}$ of the mass of an atom of carbon-12.

(e) **Avogadro's number** is defined as the number of atoms present in 12 g (gram atomic mass) of C-12 isotope, i.e. 6.022 x10^{23} atoms.

(f) The quantity of the element which weighs equal to it's gram atomic mass is called one **gram atom** of that element

(g) A **Mole** is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.

#### Question 3 (a)

(a) What are the main applications of Avogadro's Law?

(b) How does Avogadro's Law explain Gay-Lussac's Law of combining volumes?

**Answer**

(a) The applications of Avogadro's Law are:

- It explains Gay-Lussac's law.
- It predicts atomicity of gases.
- It determines the molecular formula of gases.
- It determines the relation between molecular mass and vapour density.
- It gives the relationship between gram molecular mass and gram molar volume.

(b) According to Avogadro's law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another. This is what Gay Lussac's Law says.

$\begin{matrix} \text{H}_2 \space + \text{Cl}_2 & \longrightarrow & 2\text{HCl} \\ 1 \text{ vol.} \phantom{+} 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \small{\text{(by Gay-Lussac's Law)}} \\ \underset{\text{molecules}}{\text{n}} \phantom{+} \underset{\text{molecules}}{\text{n}} & \longrightarrow & \underset{\text{molecules}}{\text{2n}} & \small{\text{(by Avogadro's Law)}} \\ \end{matrix}$

#### Question 4

Calculate the relative molecular masses of :

(a) Ammonium chloroplatinate [(NH_{4})_{2}PtCl_{6}]

(b) Potassium chlorate [KClO_{3}]

(c) CuSO_{4}.5H_{2}O

(d) (NH_{4})_{2}SO_{4}

(e) CH_{3}COONa

(f) CHCl_{3}

(g) (NH_{4})_{2}Cr_{2}O_{7}

**Answer**

(a) (NH_{4})_{2}PtCl_{6}

= (2N) + (8H) + (Pt) + (6Cl)

= (2 x 14) + (8 x 1) + 195 + (6 x 35.5)

= 28 + 8 + 195 + 213

= **444 a.m.u.**

(b) KClO_{3}

= (K) + (Cl) + (3O)

= 39 + 35.5 + (3 x 16)

= 39 + 35.5 + 48

= **122.5 a.m.u.**

(c) CuSO_{4}.5H_{2}O

= (Cu) + (S) + (4O) + 5(2H + O)

= 63.5 + 32 + (4 x 16) + 5[(2 x 1) + 16]

= 63.5 + 32 + 64 + (5 x 18)

= 63.5 + 32 + 64 + 90

= **249.5 a.m.u.**

(d) (NH_{4})_{2}SO_{4}

= (2N) + (8H) + (S) + (4O)

= (2 x 14) + (8 x 1) + 32 + (4 x 16)

= 28 + 8 + 32 + 64

= **132 a.m.u.**

(e) CH_{3}COONa

= (C) + (3H) + (C) + (2O) + (Na)

= 12 + (3 x 1) + 12 + (2 x 16) + 23

= 12 + 3 + 12 + 32 + 23

= **82 a.m.u.**

(f) CHCl_{3}

= (C) + (H) + (3Cl)

= 12 + 1 + (3 x 35.5)

= 12 + 1 + 106.5

= **119.5 a.m.u.**

(g) (NH_{4})_{2}Cr_{2}O_{7}

= (2N) + (8H) + (2Cr) + (7O)

= (2 x 14) + (8 x 1) + (2 x 51.9) + (7 x 16)

= 28 + 8 + 103.8 + 112

= 251.8 ≈ **252 a.m.u.**

#### Question 5

Find the:

(a) number of molecules in 73 g of HCl,

(b) weight of 0.5 mole of O_{2},

(c) number of molecules in 1.8 g of H_{2}O,

(d) number of moles in 10 g of CaCO_{3},

(e) weight of 0.2 mole of H_{2} gas,

(f) number of molecules in 3.2 g of SO_{2}.

**Answer**

(a) Number of molecules in 73 g of HCl —

Molecular wt. of any substance contain 6.022 × 10^{23} molecules.

Mass of 1 mole of HCl is 1 + 35.5 = 36.5 g

36.5 g of HCl contains 6.022 × 10^{23} molecules

∴ 73 g of HCl contains $\dfrac{6.022 \times 10^{23} \times 73 }{36.5}$

= **1.2 × 10 ^{24} molecules**

(b) Weight of 0.5 mole of O_{2} —

1 mole of O_{2} weighs = 2O = 2 x 16 = 32 g

∴ 0.5 moles will weigh = $\dfrac{32}{2}$ = **16 g**

(c) Number of molecules in 1.8 g of H_{2}O —

Molecular wt. of any substance contains 6.022 × 10^{23} molecules.

Mass of 1 mole of H_{2}O is (2 x 1) + 16 = 2 + 16 = 18 g

18 g of H_{2}O contains 6.022 × 10^{23} molecules

∴ 1.8 g of H_{2}O contains $\dfrac{6.022 \times 10^{23} \times 1.8 }{18}$

= **6.02 × 10 ^{22} molecules**

(d) Number of moles in 10 g of CaCO_{3} —

Mass of 1 mole of CaCO_{3} = 40 + 12 + 3(16) = 52 + 48 = 100 g

100 g of CaCO_{3} = 1 mole

∴ 10 g of CaCO_{3} = $\dfrac{1 \times 10}{100}$

= **0.1 mole**

(e) Weight of 0.2 mole H_{2} gas —

1 mole of H_{2} weighs = 2 g

∴ 0.2 moles will weigh = $\dfrac{2 \times 0.2}{1}$ = **0.4 g**

(f) No. of molecules in 3.2 g of SO_{2} —

Molecular wt. of any substance contain 6 × 10^{23} molecules.

Mass of 1 mole of SO_{2} is 32 + 2(16) = 32 + 32 = 64 g

64 g of SO_{2} contains 6 × 10^{23} molecules

∴ 3.2 g of SO_{2} contains $\dfrac{6 \times 10^{23} \times 3.2 }{64}$

= **3 x 10 ^{22} molecules.**

#### Question 6

Which of the following would weigh most?

(a) 1 mole of H_{2}O

(b) 1 mole of CO_{2}

(c) 1 mole of NH_{3}

(d) 1 mole of CO

**Answer**

1 mole of CO_{2}

**Reason** —

Weight of H_{2}O = 2 + 16 = 18 g

Weight of CO_{2} = 12 + (2 x 16) = 12 + 32 = 44 g

Weight of NH_{3} = 14 + (3 x 1) = 14 + 3 = 17 g

Weight of CO = 12 + 16 = 28 g

As weight of CO_{2} is maximum, hence 1 mole of CO_{2} will weigh the most.

#### Question 7

Which of the following contains the maximum number of molecules?

(a) 4 g of O_{2}

(b) 4 g of NH_{3}

(c) 4 g of CO_{2}

(d) 4 g of SO_{2}

**Answer**

4 g of NH_{3}

**Reason** —

(a) No. of molecules in 4 g of O_{2}

Molecular wt. of any substance contain 6.022 × 10^{23} molecules.

Mass of 1 mole of O_{2} is 2(16) = 32 g

32 g of O_{2} contains 6.022 × 10^{23} molecules

∴ 4 g of O_{2} contains $\dfrac{6.022 \times 10^{23} \times 4}{32}$

= 7.5 x 10^{22} molecules.

Similarly,

(b) 4 g of NH_{3} [14 + 3 = 17g ] contains $\dfrac{6.022 \times 10^{23} \times 4}{17}$

(c) 4 g of CO_{} [12 + 16 = 28g ] contains $\dfrac{6.022 \times 10^{23} \times 4}{28}$

(d) 4 g of SO_{2} [32 + 32 = 64g ] contains $\dfrac{6.022 \times 10^{23} \times 4}{64}$

∴ 4g of NH_{3} having minimum molecular mass contains maximum molecules.

Note : The fraction with lowest denominator gives the highest value. Hence, by observation we can say that 4 g of NH_{3} has maximum number of molecules.

#### Question 8(a)

Calculate the number of particles in 0.1 mole of any substance.

**Answer**

No. of particles in 1 mole = 6.022 × 10^{23}

∴ No. of particles in 0.1 mole = $\dfrac{6.022 \times 10^{23} \times 0.1}{1}$

= **6.022 × 10 ^{22}**

#### Question 8(b)

Calculate the number of hydrogen atoms in 0.1 mole of H_{2}SO_{4}.

**Answer**

1 mole of H_{2}SO_{4} contains (2 × 6.022 × 10^{23}) hydrogen atoms

∴ 0.1 mole of H_{2}SO_{4} contains = $\dfrac{6.022 \times 10^{23} \times 2 \times 0.1}{1}$

= **1.2 × 10 ^{23}** atoms of hydrogen

#### Question 8(c)

Calculate the number of molecules in one kg of calcium chloride.

**Answer**

Mass of 1 mole of CaCl_{2} = Ca + 2Cl = 40 + (2 x 35.5) = 40 + 71 = 111 g

111 g of CaCl_{2} contains 6.022 × 10^{23} molecules

∴ 1000 g of CaCl_{2} contains $\dfrac{6.022 \times 10^{23} \times 1000}{111}$

= **5.42 × 10 ^{24}** molecules

#### Question 9(a)

How many grams of Al are present in 0.2 mole of it?

**Answer**

1 mole of aluminium has mass = 27 g

0.2 mole of aluminium has mass

= $\dfrac{27}{1}$ x 0.2

= **5.4 g**

#### Question 9(b)

How many grams of HCl are present in 0.1 mole of it?

**Answer**

1 mole of HCl has mass = 1 + 35.5 = 36.5 g

0.1 mole of HCl has mass

= $\dfrac{36.5}{1}$ x 0.1

= **3.65 g**

#### Question 9(c)

How many grams of H_{2}O are present in 0.2 mole of it?

**Answer**

1 mole of H_{2}O has mass = 2(1) + 16 = 2 + 16 = 18 g

0.2 mole of H_{2}O has mass

= $\dfrac{18}{1}$ x 0.2

= **3.6 g**

#### Question 9(d)

How many grams of CO_{2} is present in 0.1 mole of it?

**Answer**

1 mole of CO_{2} has mass = 12 + 2(16) = 12 + 32 = 44 g

0.1 mole of CO_{2} has mass

= $\dfrac{44}{1}$ x 0.1

= **4.4 g**

#### Question 10(a)

The mass of 5.6 litres of a certain gas at S.T.P. is 12 g. What is the relative molecular mass or molar mass of the gas?

**Answer**

5.6 litres of gas at S.T.P. has mass = 12 g

∴ 22.4 litre (molar volume) has mass

= $\dfrac{12}{5.6}$ x 22.4

= **48 g**

#### Question 10(b)

Calculate the volume occupied at S.T.P. by 2 moles of SO_{2}.

**Answer**

1 mole of SO_{2} has volume = 22.4 litres

∴ 2 moles will have = 22.4 × 2 = **44.8 litre**

#### Question 11(a)

Calculate the number of moles of CO_{2} which contain 8.00 g of O_{2}

**Answer**

Oxygen in 1 mole of CO_{2} = 2O = (2 x 16) = 32 g

or we can say, 32 g of oxygen is present in 1 mole of CO_{2}

∴ 8 gm of O_{2} is present in $\dfrac{1}{32}$ x 8

= **0.25 moles**

#### Question 11(b)

Calculate the number of moles of methane in 0.80 g of methane.

**Answer**

Molar mass of methane (CH_{4}) = C + 4H = 12 + 4 = 16 g

16 g of methane = 1 mole

∴ 0.80 g of methane = $\dfrac{1}{16}$ x 0.80

= **0.05 moles**

#### Question 12

Calculate the weight/mass of :

(a) an atom of oxygen

(b) an atom of hydrogen

(c) a molecule of NH_{3}

(d) 10^{22} atoms of carbon

(e) the molecule of oxygen

(f) 0.25 gram atom of calcium

**Answer**

(a) Number of oxygen atoms in 16 g of atomic oxygen = 6.022 × 10^{23} atoms

∴ mass of 1 atom of oxygen

= $\dfrac{16}{6.022 \times 10^{23}}$

= **2.657 × 10 ^{-23} g**

(b) Number of hydrogen atoms in 1 g of atomic hydrogen = 6.022 × 10^{23} atoms

∴ Mass of 1 atom of hydrogen

= $\dfrac{1}{6.022 \times 10^{23}}$

= **1.666 × 10 ^{-24} g**

(c) Gram molecular mass of NH_{3} = 14 + 3 = 17 g

Number of NH_{3} molecules in 17 g of NH_{3} = 6.022 × 10^{23} molecules

Mass of 6.022 × 10^{23} molecules of NH_{3} = 17g

∴ Mass of 1 molecule of NH_{3} = $\dfrac{17}{6.022 \times 10^{23}}$

= **2.823 × 10 ^{-23} g**

(d) Mass of 6.022 × 10^{23} atoms of atomic carbon = 12 g

∴ Mass of 10^{22} atoms of carbon = $\dfrac{12}{6.022 \times 10^{23}} \times 10^{22}$

= **0.2 g**

(e) Gram molecular mass of oxygen (O_{2}) = 2 x 16 = 32 g

Mass of 6.022 × 10^{23} molecules of O_{2} = 32 g

∴ Mass of 1 molecule of O_{2} = $\dfrac{32}{6.022 \times 10^{23}}$

= **5.314 × 10 ^{-23} g**

(f) Atomic weight of calcium = 40 g

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

Therefore, 0.25 = $\dfrac{\text{Mass of calcium}}{40}$

Mass of calcium = 40 x 0.25 = **10 g**

#### Question 13

Calculate the mass of 0.1 mole of each of the following

(a) CaCO_{3}

(b) Na_{2}SO_{4}.10H_{2}O

(c) CaCl_{2}

(d) Mg

(Ca = 40, Na = 23, Mg =24, S = 32, C = 12, Cl = 35.5, O = 16, H = 1)

**Answer**

(a) Mass of 1 mole of CaCO_{3}

= Ca + C + 3O = 40 + 12 + (3 x 16) = 52 + 48 = 100 g

∴ Mass of 0.1 mole of CaCO_{3} = 0.1 x 100 = **10 g**

(b) Mass of 1 mole of Na_{2}SO_{4}.10H_{2}O

= 2Na + S + 4O + 10(2H + O) = (2 x 23) + 32 + (4 x 16) + 10(2 + 16) = 46 + 32 + 64 + 180 = 322 g

∴ Mass of 0.1 mole of Na_{2}SO_{4}.10H_{2}O = 0.1 x 322 = **32.2 g**

(c) Mass of 1 mole of CaCl_{2}

= Ca + 2Cl = 40 + (2 x 35.5) = 40 + 71 = 111 g

∴ Mass of 0.1 mole of CaCl_{2} = 0.1 x 111 = **11.1 g**

(d) Mass of 1 mole of Mg = 24 g

∴ Mass of 0.1 mole of Mg = 24 x 0.1 = **2.4 g**

#### Question 14(a)

Calculate the number of oxygen atoms in 0.10 mole of Na_{2}CO_{3}.10H_{2}O.

**Answer**

1 molecule of Na_{2}CO_{3}.10H_{2}O contains 13 atoms of oxygen

∴ 6.022 × 10^{23} molecules (ie., 1 mole) has 13 × 6.022 × 10^{23} atoms

∴ 0.1 mole will have atoms = 0.1 × 13 × 6.022 × 10^{23}

= **7.8 × 10 ^{23} atoms**

#### Question 14(b)

Calculate the number of gram atoms in 4.6 gram of sodium

**Answer**

Atomic mass of Na = 23

23 g of sodium = 1 gram atom of sodium

∴ 4.6 gram of sodium = $\dfrac{\text{{4.6}}}{\text{{23}}}$

= **0.2 gram atom of sodium**

#### Question 14(c)

Calculate the number of moles in 12 g of oxygen gas

**Answer**

32 g of oxygen = 1 mole

∴ 12 g of oxygen = $\dfrac{12}{32}$ = $\dfrac{3}{8}$

= **0.375 mole**

#### Question 15

What mass of Ca will contain the same number of atoms as are present in 3.2 g of S?

**Answer**

1 mole of Sulphur weighs 32 g and contains 6.02 x 10^{23} atoms

∴ 3.2 g of Sulphur will contain = $\dfrac{6.02 \times 10^{23}}{32} \times 3.2$

= 6.02 x 10^{22} atoms.

6.02 x 10^{23} atoms of Ca weighs = 40 g

∴ 6.02 x 10^{22} atoms of Ca will weigh = $\dfrac{40}{ 6.02 \times 10^{23}}$ x 6.02 x 10^{22} = **4 g**.

#### Question 16

Calculate the number of atoms in each of the following:

(a) 52 moles of He

(b) 52 amu of He

(c) 52 g of He

**Answer**

(a) No. of atoms = Moles x 6.022 x 10^{23}

= 52 × 6.022 x 10^{23} = **3.131 × 10 ^{25} atoms**

(b) 4 amu = 1 atom of He

∴ 52 amu = $\dfrac{52}{4}$ = **13 atoms of He**

(c) Mass of 1 mole of He is 4 g

4 g of He contains 6.022 × 10^{23} atoms

∴ 52 g of He contains $\dfrac{6.022 \times 10^{23}}{4} \times 52$

= **7.828 × 10 ^{24} atoms**

#### Question 17

Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

**Answer**

Molecular mass of Na_{2}CO_{3} = 2Na + C + 3O = (2 x 23) + 12 + (3 x 16) = 46 + 12 + 48 = 106 g

(i) 106 g of Na_{2}CO_{3} has = 2 × 6.022 × 10^{23} atoms of Na

∴ 5.3 g of Na_{2}CO_{3} will have = $\dfrac{2 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = **6.022 × 10 ^{22} atoms of Na**

(ii) 106 g of Na_{2}CO_{3} has = 6.022 × 10^{23} atoms of carbon

∴ 5.3 g of Na_{2}CO_{3} will have = $\dfrac{6.022 \times 10^{23} \times 5.3 }{106}$ = **3.01 × 10 ^{22} atoms of carbon**

(iii) 106 g of Na_{2}CO_{3} has 3 x 6.022 × 10^{23} atoms of oxygen

∴ 5.3 g of Na_{2}CO_{3} will have = $\dfrac{3 \times 6.022 \times 10^{23} \times 5.3 }{106}$ = **9.03 × 10 ^{22} atoms of oxygen**

#### Question 18(a)

Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH_{2})_{2}]

[O = 16; N = 14; C = 12 ; H = 1 ]

**Answer**

Molar mass of urea [CO(NH_{2})_{2}] = 12 + 16 + 2(14 + (2 x 1))

= 28 + 2(16)

= 28 + 32

= 60 g

Molar mass of nitrogen = 2 x 14 = 28 g

60 g urea has mass of nitrogen = 28 g

∴ 5000 g urea will have mass

= $\dfrac{28 \times 5000 }{60}$

= 2333 g = **2.33 kg**

#### Question 18(b)

Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P.

[S = 32; O = 16]

**Answer**

Molar mass of sulphur dioxide (SO_{2}) = S + 2O = 32 + (2 x 16) = 32 + 32 = 64 g

64 g of sulphur dioxide has volume = 22.4 litre

∴ 320 g of sulphur dioxide will have volume = $\dfrac{22.4\times 320}{64}$

= **112 litres**

#### Question 19

(a) What do you understand by the statement that 'vapour density of carbon dioxide is 22'?

(b) Atomic mass of Chlorine is 35.5. What is it's vapour density?

**Answer**

(a) Vapour density of carbon dioxide is 22 implies that 1 molecule of carbon dioxide is 22 times heavier than 1 molecule of hydrogen.

(b) Vapour density = $\dfrac{\text{Molecular mass}}{2}$

Molecular mass of chlorine Cl_{2} = 2Cl = 2 x 35.5 = 71 g

Substituting in formula;

Vapour density = $\dfrac{71}{2}$ = 35.5

Hence, vapour density of Chlorine atom is **35.5**

#### Question 20

What is the mass of 56 cm^{3} of carbon monoxide at S.T.P.?

(C = 12, O = 16)

**Answer**

22400 cm^{3} of CO has mass = 12 + 16 = 28 g

∴ 56 cm^{3} will have mass = $\dfrac{28}{22400}$ x 56 = **0.07 g**

#### Question 21

Determine the number of molecules in a drop of water which weighs 0.09 g.

**Answer**

Molecular wt. of any substance contain 6.022 × 10^{23} molecules.

Mass of 1 mole of water is 2H + O = 2 + 16 = 18 g

18 g of H_{2}O contains 6.022 × 10^{23} molecules

∴ 0.09 g of H_{2}O contains $\dfrac{6.022 \times 10^{23} \times 0.09 }{18}$

= **3.01 × 10 ^{21} molecules**

#### Question 22

The molecular formula for elemental sulphur is S_{8}. In a sample of 5.12 g of sulphur:

(a) How many moles of sulphur are present?

(b) How many molecules and atoms are present?

**Answer**

(a) Mass of 1 mole of S_{8} = 8S = 8 x 32 = 256 g

∴ Moles in 5.12 g of sulphur = $\dfrac{5.12}{256}$ = **0.02 moles**

(b) 1 mole = 6.022 × 10^{23} molecules

∴ 0.02 moles will have = 0.02 × 6.022 × 10^{23}

= 1.2044 × 10^{22} ≈ **1.2 × 10 ^{22} molecules**

No. of atoms in 1 molecule of S_{8} = 8

∴ No. of atoms in 1.2044 × 10^{22} molecules = 1.2044 x 10^{22} × 8

= **9.635 × 10 ^{22} molecules**

#### Question 23

If phosphorus is considered to contain P_{4} molecules, then calculate the number of moles in 100 g of phosphorus?

**Answer**

Mass of 1 mole of P_{4} = 4P = 4 x 31 = 124 g

124 g of phosphorus (P_{4}) = 1 mole

∴ 100 g of phosphorus (P_{4}) = $\dfrac{1}{124}$ x 100 = **0.806 moles**

#### Question 24

Calculate:

(a) The gram molecular mass of chlorine if 308 cm^{3} of it at S.T.P. weighs 0.979 g

(b) The volume of 4 g of H_{2} at 4 atmospheres.

(c) The mass of oxygen in 2.2 litres of CO_{2} at S.T.P.

**Answer**

(a) The mass of 22.4 L of a gas at S.T.P. is equal to it's gram molecular mass.

308 cm^{3} of chlorine weighs 0.979 g

∴ 22,400 cm^{3} of chlorine will weigh

= $\dfrac{0.979}{308}$ × 22400 = 71.2 g

(b) Molar mass of H_{2} = 2H = 2 x 1 = 2 g

2g H_{2} at 1 atm has volume = 22.4 dm^{3}

∴ 4 g H_{2} at 1 atm will have volume 2 x 22.4 = 44.8 dm^{3}

Now, For 4 g H_{2}

P_{1} = 1 atm, V_{1} = 44.8 dm^{3}

P_{2} = 4 atm, V_{2} = ?

Using formula P_{1}V_{1} = P_{2}V_{2}

$\text{V}_2 = \dfrac{\text{P}_1\text{V}_1}{\text{P}_2} \\[1em] \text{V}_2 = \dfrac{1 \times 44.8}{4} \\[1em] = \bold{11.2} \space \bold{dm^3}$

(c) Molar mass of oxygen in carbon dioxide = 2O = 2 x 16 = 32 g

Mass of oxygen in 22.4 litres of CO_{2} = 32 g

∴ Mass of oxygen in 2.2 litres of CO_{2}

= $\dfrac{32}{22.4}$ x 2.2 = **3.14 g**

#### Question 25

A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^{-12} g, calculate the number of carbon atoms in the signature.

**Answer**

No. of atoms in 12 g C = 6.022 × 10^{23}

∴ no. of carbon atoms in 10^{-12} g

$\dfrac{6.022 \times 10^{23}}{12}$ x 10^{-12}

= **5.019 × 10 ^{10} atoms**

#### Question 26

An unknown gas shows a density of 3 g per litre at 273°C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?

**Answer**

Given:

P = 1140 mm Hg

Density = D = 3 g per L

T = 273 °C = 273 + 273 = 546 K

gram molecular mass = ?

At S.T.P., the volume of one mole of any gas is 22.4 L

Volume of unknown gas at S.T.P. = ?

By Charle’s law.

V_{1} = 1 L

T_{1} = 546 K

T_{2} = 273 K

V_{2} = ?

$\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$

Hence, V_{2} = $\dfrac{1}{546}$ x 273 = **0.5 L**

Volume at standard pressure = ?

Apply Boyle’s law.

P_{1} = 1140 mm Hg

V_{1} = 0.5 L

P_{2} = 760 mm Hg

V_{2} = ?

P_{1} × V_{1} = P_{2} × V_{2}

V_{2} = $\dfrac{1140 \times 0.5}{760}$ = 0.75 L

Now,

22.4 L = 1 mole of any gas at S.T.P.,

then 0.75 L = $\dfrac{0.75}{22.4}$

= 0.0335 moles

The original mass is 3 g

Molecular mass = $\dfrac{\text{Mass of compound}}{\text{Moles of compound}}$

= $\dfrac{3}{0.0335 }$ = 89.55 ≈ 89.6 g per mole

Hence, the gram molecular mass of the unknown gas is **89.6g**

#### Question 27

Cost of Sugar (C_{12}H_{22}O_{11}) is ₹40 per kg; calculate it's cost per mole.

**Answer**

Molar mass of C_{12}H_{22}O_{11} = 12C + 22H + 11O = (12 x 12) + (22 x 1) + (11 x 16) = 144 + 22 + 176 = 342 g

1000 g of sugar costs = ₹40

∴ 342 g of sugar will cost = $\dfrac{40}{1000}$ x 342 = ₹**13.68 per mole**

#### Question 28

Which of the following weighs the least?

(a) 2 g atom of N

(b) 3 x 10^{25} atoms of carbon

(c) 1 mole of sulphur

(d) 7 g of silver

**Answer**

7 g of silver

**Reason** —

(a) Weight of 1 g atom of N = 14 g

∴ weight of 2 g atom of N = **28 g**

(b) 6.022 × 10^{23} atoms of C weigh = 12 g

∴ 3 × 10^{25} atoms will weigh = $\dfrac{12}{6.022 \times 10^{23}}$ × 3 × 10^{25} = **597.7 g**

(c) 1 mole of sulphur weighs = **32 g**

(d) 7 g of silver

The weight computed in all other options is greater than the weight in option (d). Hence, 7 grams of silver weighs the least.

#### Question 29

Four grams of caustic soda contains:

(a) 6.02 x 10^{23} atoms of it

(b) 4 g atom of sodium

(c) 6.02 x 10^{22} molecules

(d) 4 moles of NaOH

**Answer**

6.02 × 10^{22} molecule

**Reason** —

Molar mass of NaOH = Na + O + H = 23 + 16 + 1 = 40 g

40 g of NaOH contains 6.022 × 10^{23} molecules

∴ 4 g of NaOH contains

= $\dfrac{6.022 \times 10^{23}}{40}$ x 4

= **6.02 × 10 ^{22} molecules**

#### Question 30

The number of molecules in 4.25 g of ammonia is:

(a) 1.0 × 10^{23}

(b) 1.5 × 10^{23}

(c) 2.0 × 10^{23}

(d) 3.5 × 10^{23}

**Answer**

1.5 × 10^{23}

**Reason** —

Molar mass of ammonia = N + 3H = 14 + 3 = 17 g

The number of molecules in 17 g of ammonia = 6.022 × 10^{23}

∴ No. of molecules in 4.25 g of ammonia

= $\dfrac{6.022 \times 10^{23}}{17}$ x 4.25

= **1.5 × 10 ^{23}**

#### Question 31

Correct the statements, if required

(a) One mole of chlorine contains 6.023 × 10^{23} atoms of chlorine.

(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour.

(c) Relative atomic mass of an element is the number of times one molecule of an element is heavier than $\dfrac{1}{12}$ the mass of an atom of carbon [C^{12}].

(d) Under the same conditions of the temperature and pressure, equal volumes of all gases contain the same number of atoms.

**Answer**

(a) One mole of chlorine contains ** 6.022** × 10

^{23}atoms of chlorine.

(b) Under similar conditions of temperature and pressure, ** four** volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour.

(c) Relative atomic mass of an element is the number of times one ** atom** of an element is heavier than $\dfrac{1}{12}$ the mass of an atom of carbon [C

^{12}].

(d) Under the same conditions of the temperature and pressure, equal volumes of all gases contain the same number of ** molecules**.

## Exercise 5C

#### Question 1

Give three kinds of information conveyed by the formula H_{2}O.

**Answer**

Information conveyed by formula [H_{2}O] —

- One molecule of water (H
_{2}O) is made of two atoms of Hydrogen and one atom of Oxygen. - As atomic weight of hydrogen is 1 and that of oxygen is 16. Therefore, ratio by weight of hydrogen and oxygen is $\dfrac{2\text{H}}{\text{O}}$ = $\dfrac{2}{16}$ = $\dfrac{1}{8}$
- Molecular weight of H
_{2}O is 2H + O = 2 + 16 = 18g.

#### Question 2

Explain the terms empirical formula and molecular formula.

**Answer**

The **empirical formula** of a compound is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.

The **molecular formula** of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

#### Question 3

Give the empirical formula of:

(a) C_{6}H_{6}

(b) C_{6}H_{18}O_{3}

(c) C_{2}H_{2}

(d) CH_{3}COOH

**Answer**

(a) Molecular formula is C_{6}H_{6}

∴ Ratio of C and H is 6 : 6

Simple ratio is 1 : 1

Hence, empirical formula = **CH**

(b) Molecular formula is C_{6}H_{18}O_{3}

∴ Ratio of C, H and O is 6 : 18 : 3

Simple ratio is 2 : 6 : 1

Hence, empirical formula = **C _{2}H_{6}O**

(c) Molecular formula is C_{2}H_{2}

∴ Ratio of C and H is 2 : 2

Simple ratio is 1 : 1

Hence, empirical formula = **CH**

(d) Molecular formula is CH_{3}COOH i.e. C_{2}H_{4}O_{2}

∴ Ratio of C, H and O is 2 : 4 : 2

Simple ratio is 1 : 2 : 1

Hence, empirical formula = **CH _{2}O**

#### Question 4

Find the percentage of water of crystallisation in CuSO_{4}.5H_{2}O. (At. Mass Cu = 64, H = 1, O = 16, S = 32)

**Answer**

Relative molecular mass of CuSO_{4}.5H_{2}O

= 64 + 32 + (4×16) + [5(2+16)]

= 96 + 64 + 90 = 250

250 g of CuSO_{4}.5H_{2}O contains 90 g of water of crystallisation

∴ 100 g of CuSO_{4}.5H_{2}O contains

= $\dfrac{90}{250}$ x 100 = **36%**

#### Question 5

Calculate the percentage of phosphorus in

(a) Calcium hydrogen phosphate Ca(H_{2}PO_{4})_{2}

(b) Calcium phosphate Ca_{3}(PO_{4})_{2}

**Answer**

(a) Molecular mass of Ca(H_{2}PO_{4})_{2}

= Ca + 2[2H + P + 4O]

= 40 + 2[2(1) + 31 + 4(16)]

= 40 + 2[2 + 31 + 64]

= 40 + 194

= 234

234 g of Ca(H_{2}PO_{4})_{2} contains 62 g of P

∴ 100 g of Ca(H_{2}PO_{4})_{2} contains

= $\dfrac{62}{234}$ x 100 = **26.5%**

(b) Molecular mass of Ca_{3}(PO_{4})_{2}

= 3Ca + 2[P + 4O]

= (3 x 40) + 2[31 + 4(16)]

= 120 + 2[31 + 64]

= 120 + 190

= 310

310 g of Ca_{3}(PO_{4})_{2} contains 62 g of P

∴ 100 g of Ca_{3}(PO_{4})_{2} contains

= $\dfrac{62}{310}$ x 100 = **20 %**

#### Question 6

Calculate the percent composition of Potassium chlorate KClO_{3}.

**Answer**

Molecular mass of KClO_{3}

= K + Cl + 3O

= 39 + 35.5 + (3 x 16)

= 39 + 35.5 + 48

= 122.5 g

% of K = ?

Since, 122.5 g of KClO_{3} contains 39 g of K

∴ 100 g of KClO_{3} contains

= $\dfrac{39}{122.5}$ x 100

= **31.83%**

Similarly, 122.5 g of KClO_{3} contains 35.5 g of Cl

∴ 100 g of KClO_{3} contains

= $\dfrac{35.5}{122.5}$ x 100

= **28.98%**

And, 122.5 g of KClO_{3} contains 48 g of O

∴ 100 g of KClO_{3} contains

= $\dfrac{48}{122.5}$ x 100

= **39.18%**

#### Question 7

Find the empirical formula of the compounds with the following percentage composition:

Pb = 62.5%, N = 8.5%, O = 29.0%

**Answer**

Element | % composition | At. mass | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Pb | 62.5 | 207 | $\dfrac{62.5 }{207}$ = 0.301 | $\dfrac{0.301}{0.301 }$ = 1 |

N | 8.5 | 14 | $\dfrac{8.5}{14}$ = 0.607 | $\dfrac{0.607}{0.301 }$ = 2 |

O | 29 | 16 | $\dfrac{29}{16}$ = 1.81 | $\dfrac{1.81}{0.301 }$ = 6 |

Hence, Simplest ratio of whole numbers = Pb : N : O = 1 : 2 : 6

Hence, **empirical formula is Pb(NO _{3})_{2}**.

#### Question 8

Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.

**Answer**

Atomic wt. of Fe = 56 and O = 16

Molecular mass of Fe_{2}O_{3} = 2Fe + 3O

=(2 x 56) + (3 x 16)

= 112 + 48

= 160 g

Iron present in 80% of Fe_{2}O_{3} = $\dfrac{112}{160}$ x 80

= 56 g

∴ Mass of iron in 100 g of iron ore = 56 g

Hence, mass of iron present in 10 kg (i.e., 10,000 g) of iron ore = $\dfrac{56}{100}$ x 10000

= 5600 g = **5.6 kg**

#### Question 9

If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula.

**Answer**

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 13

Molecular weight = 2 x V.D. = 2 x 13

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 13}{13} = 2$

∴ Molecular formula = n[E.F.] = 2[CH] = **C _{2}H_{2}**

Similarly,

Empirical formula is CH

Empirical formula weight = 12 + 1 = 13

Vapour density (V.D.) = 39

Molecular weight = 2 x V.D. = 2 x 39

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 39}{13} = 6$

∴ Molecular formula = n[E.F.] = 6[CH] = **C _{6}H_{6}**

#### Question 10

Find the empirical formula of a compound containing 17.64% hydrogen and 82.35% nitrogen.

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Nitrogen | 82.35 | 14 | $\dfrac{82.35 }{14}$ = 5.88 | $\dfrac{5.88}{5.88}$ = 1 |

Hydrogen | 17.64 | 1 | $\dfrac{17.64}{1}$ = 17.64 | $\dfrac{17.64}{5.88}$ = 3 |

Simplest ratio of whole numbers = N : H = 1 : 3

Hence, **empirical formula is NH _{3}**

#### Question 11

On analysis, a substance was found to contain

C = 54.54%, H = 9.09%, O = 36.36%

The vapour density of the substance is 44, calculate;

(a) it's empirical formula, and

(b) it's molecular formula

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Carbon | 54.54 | 12 | $\dfrac{54.54}{12}$ = 4.545 | $\dfrac{ 4.545 }{2.275 }$ = 1.99 = 2 |

Hydrogen | 9.09 | 1 | $\dfrac{9.09 }{1}$ = 9.09 | $\dfrac{9.09 }{ 2.275 }$ = 3.99 = 4 |

Oxygen | 36.36 | 16 | $\dfrac{36.36}{16}$ = 2.275 | $\dfrac{2.275 }{ 2.275 }$ = 1 |

Simplest ratio of whole numbers = C : H : O = 2 : 4 : 1

Hence, **empirical formula is C _{2}H_{4}O**

Empirical formula weight = 2(12) + 4(1) + 16 = 24 + 4 + 16 = 44

V.D. = 44

Molecular weight = 2 x V.D. = 2 x 44 = 88

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{88}{44} = 2$

So, molecular formula = (C_{2}H_{4}O)_{2} = C_{4}H_{8}O_{2}

#### Question 12

An organic compound, whose vapour density is 45, has the following percentage composition

H = 2.22%, O = 71.19% and remaining carbon.

Calculate,

(a) it's empirical formula, and

(b) it's molecular formula

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Hydrogen | 2.22 | 1 | $\dfrac{2.22}{1}$ = 2.22 | $\dfrac{2.22 }{2.21}$ = 1 |

Oxygen | 71.19 | 16 | $\dfrac{71.19}{16}$ = 4.44 | $\dfrac{4.44}{2.21 }$ = 2 |

Carbon | 26.59 | 12 | $\dfrac{26.59}{12}$ = 2.21 | $\dfrac{ 2.21 }{2.21}$ = 1 |

Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1

Hence, **empirical formula is CHO _{2}**

Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45

V.D. = 45

Molecular weight = 2 x V.D. = 2 x 45 = 90

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{90}{45} = 2$

So, molecular formula = 2(CHO_{2}) = **C _{2}H_{2}O_{4}**

#### Question 13

An organic compound contains 4.07% hydrogen, 71.65% chlorine and remaining carbon. Its molar mass is 98.96. Find its,

(a) Empirical formula

(b) Molecular formula

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Hydrogen | 4.07 | 1 | $\dfrac{4.07 }{1}$ = 4.07 | $\dfrac{4.07 }{2.01}$ = 2 |

chlorine | 71.65 | 35.5 | $\dfrac{ 71.65}{35.5}$ = 2.01 | $\dfrac{2.01}{2.01 }$ = 1 |

Carbon | 24.28 | 12 | $\dfrac{24.28 }{12}$ = 2.02 | $\dfrac{ 2.02 }{2.01}$ = 1 |

Simplest ratio of whole numbers = H : Cl : C = 2 : 1 : 1

Hence, **empirical formula is CH _{2}Cl**

Empirical formula weight = 12 + (2 x 1) + 35.5 = 49.5

molar mass = 98.96

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{98.96}{49.5} = 1.99 = 2$

So, molecular formula = 2(CH_{2}Cl) = **C _{2}H_{4}Cl_{2}**

#### Question 14

A hydrocarbon contains 4.8 g of carbon per gram of hydrogen. Calculate

(a) the g atom of each

(b) find the empirical formula

(c) find molecular formula, if it's vapour density is 29.

**Answer**

(a) Given, hydrocarbon contains 4.8 g of carbon per gram of hydrogen

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of carbon = $\dfrac{4.8}{12}$ = **0.4** and

g atom of hydrogen = $\dfrac{1}{1}$ = **1**

(b)

Element | Mass | At. wt. | Gram atoms | Simplest ratio |
---|---|---|---|---|

Hydrogen | 1 | 1 | $\dfrac{1 }{1}$ = 1 | $\dfrac{1 }{0.4 }$ = $\dfrac{5 }{2}$ |

Carbon | 4.8 | 12 | $\dfrac{4.8 }{12}$ = 0.4 | $\dfrac{ 0.4 }{0.4 }$ = 1 |

Simplest ratio of whole numbers = H : C = $\dfrac{5 }{2}$ : 1 = 5 : 2

Hence, **empirical formula is C _{2}H_{5}**

(c) Empirical formula weight = (2 x 12) + (5 x 1) = 24 + 5 = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29 = 58

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{58}{29} = 2$

So, molecular formula = 2(C_{2}H_{5}) = **C _{4}H_{10}**

#### Question 15

0.2 g atom of silicon combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.

**Answer**

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

g atom of silicon = 0.2 = $\dfrac{\text{Mass of silicon}}{28}$

∴ Mass of silicon = **5.6 g** and

Mass of chlorine = **21.3 g**

Element | Mass | At. wt. | gram atoms | Simplest ratio |
---|---|---|---|---|

Silicon | 5.6 | 28 | $\dfrac{5.6 }{28}$ = 0.2 | $\dfrac{0.2 }{0.2}$ = 1 |

Chlorine | 21.3 | 35.5 | $\dfrac{ 21.3}{35.5}$ = 0.6 | $\dfrac{ 0.6 }{0.2}$ = 3 |

Simplest ratio of whole numbers = Si : Cl = 1 : 3

Hence, **empirical formula is SiCl _{3}**

#### Question 16

A gaseous hydrocarbon contains 82.76% of carbon. Given that it's vapour density is 29, find it's molecular formula.

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Carbon | 82.76 | 12 | $\dfrac{82.76}{12}$ = 6.89 | $\dfrac{6.89}{6.89}$ = 1 |

Hydrogen | 17.24 | 1 | $\dfrac{17.24}{1}$ = 17.24 | $\dfrac{17.24}{6.89}$ = $\dfrac{5 }{2}$ |

Simplest ratio of whole numbers = C : H = 1 : $\dfrac{5 }{2}$ = 2 : 5

Hence, **empirical formula is C _{2}H_{5}**

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 29}{29} =2$

∴ Molecular formula = n[E.F.] = 2[C_{2}H_{5}] = **C _{4}H_{10}**

#### Question 17

In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions.

(a) How many gram-atoms of magnesium are equal to 18g?

(b) How many gram-atoms of nitrogen are equal to 7g of nitrogen?

(c) Calculate simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.

**Answer**

(a) Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of magnesium = $\dfrac{18}{24}$ = $\dfrac{3}{4}$

Hence, $\dfrac{3}{4}$ gram atoms of magnesium are equal to 18g of magnesium.

(b) g atom of nitrogen = $\dfrac{7}{14}$ = $\dfrac{1}{2}$

Hence, $\dfrac{1}{2}$ gram atoms of nitrogen are equal to 7g of nitrogen.

(c) simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen

= $\dfrac{\dfrac{3}{4}}{\dfrac{1}{2}}$ = $\dfrac{3}{2}$ = magnesium : nitrogen

So, the formula is **Mg _{3}N_{2}**

#### Question 18

Barium chloride crystals contain 14.8% water of crystallisation. Find the number of molecules of water of crystallisation per molecule.

**Answer**

Barium chloride = BaCl_{2}.xH_{2}O

Molecular weight of BaCl_{2}.xH_{2}O = Ba + 2Cl + x(2H + O)

= 137 + (2 x 35.5) + x(2+16)

= 137 + (2 x 35.5) + x(2+16)

= 137 + 71 + 18x

= (208 + 18 x)

(208 + 18 x) contains 14.8% of water of crystallisation in BaCl2.x H_{2}O

∴ 14.8% of (208 + 18 x) = 18x

$\Big[\dfrac{14.8}{100}\Big]$ x [208 + 18 x] = 18x

[0.148 x 208 ] + [0.148 x 18x] = 18x

30.784 = 18x - [0.148 x 18x]

30.784 = 18x - 2.664x

30.784 = 15.336x

x = $\dfrac{30.784}{15.336}$ = 2

Hence, Barium chloride crystals contain **2 molecules of water of crystallisation** per molecule.

#### Question 19

Urea is a very important nitrogenous fertilizer. It's formula is CON_{2}H_{4}. Calculate the percentage of nitrogen in urea. (C = 12, O = 16, N = 14 and H = 1).

**Answer**

Molar mass of urea (CON_{2}H_{4}) = 12 + 16 + 28 + 4 = 60 g

Molar mass of nitrogen (N_{2}) = 2 x 14 = 28 g

60 g urea has mass of nitrogen = 28 g

∴ 100 g urea will have mass

= $\dfrac{28 \times 100 }{60}$

= **46.67%**

#### Question 20

Determine the formula of the organic compound if it's molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Oxygen | 51.42 | 16 | $\dfrac{51.42 }{16}$ = 3.21 | $\dfrac{3.21}{3.21}$ = 1 |

Hydrogen | 6.48 | 1 | $\dfrac{6.48}{1}$ = 6.48 | $\dfrac{6.48}{3.21}$ = 2 |

Carbon | 42.1 | 12 | $\dfrac{42.1}{12}$ = 3.50 | $\dfrac{3.50}{3.21}$ = 1 |

Simplest ratio of whole numbers = O : H : C = 1 : 2 : 1

Hence, **empirical formula is CH _{2}O**

Since the compound has 12 atoms of carbon, so the formula is **C _{12}H_{24}O_{12}**.

#### Question 21(a)

A compound with empirical formula AB_{2}, has the vapour density equal to it's empirical formula weight. Find it's molecular formula.

**Answer**

Empirical formula = AB_{2}

Empirical formula weight = V.D.

Molecular weight = 2 x V.D.

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \dfrac{\text{2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 2$

∴ Molecular formula = n[E.F.] = 2[AB_{2}] = **A _{2}B_{4}**

#### Question 21(b)

A compound with empirical formula AB has vapour density three times it's empirical formula weight. Find the molecular formula.

**Answer**

Given,

Empirical formula = AB

V.D. = 3 x Empirical formula weight

Hence, Empirical formula weight = $\dfrac{\text{V.D.}}{3}$

and we know, Molecular weight = 2 x V.D.

Substituting in the formula for n we get,

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] \text{n} = \dfrac{\text{ 2 x V.D.}}{\dfrac{\text{V.D.}}{3}} \\[0.5em] \text{n} = \dfrac{\text{ 3 x 2 x V.D.}}{\text{V.D.}} \\[0.5em] \text{n} = 6$

∴ Molecular formula = n[E.F.] = 6[AB] = **A _{6}B_{6}**

#### Question 21(c)

10.47 g of a compound contains 6.25 g of metal A and rest non-metal B. Calculate the empirical formula of the compound (At. wt of A = 207, B = 35.5)

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Metal A | 6.25 | 207 | $\dfrac{6.25 }{207}$ = 0.03 | $\dfrac{0.03}{0.03}$ = 1 |

Non-metal B | 4.22 | 35.5 | $\dfrac{4.22}{35.5}$ = 0.11 | $\dfrac{0.11}{0.03}$ = 3.96 = 4 |

Simplest ratio of whole numbers = A : B = 1 : 4

Hence, **empirical formula is AB _{4}**

#### Question 22

A hydride of nitrogen contains 87.5% percent by mass of nitrogen. Determine the empirical formula of this compound.

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Nitrogen | 87.5 | 14 | $\dfrac{87.5 }{14}$ = 6.25 | $\dfrac{6.25}{6.25 }$ = 1 |

Hydrogen | 12.5 | 1 | $\dfrac{12.5}{1}$ = 12.5 | $\dfrac{12.5}{6.25 }$ = 2 |

Simplest ratio of whole numbers = N : H = 1 : 2

Hence, **empirical formula is NH _{2}**

#### Question 23

A compound has O = 61.32%, S = 11.15%, H = 4.88% and Zn = 22.65%.The relative molecular mass of the compound is 287 a.m.u. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallisation.

**Answer**

Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|

Zn | 22.65 | 65 | $\dfrac{22.65}{65}$ = 0.3484 | $\dfrac{0.3484}{0.3484}$ = 1 |

S | 11.15 | 32 | $\dfrac{11.15}{32}$ = 0.3484 | $\dfrac{0.3484}{0.3484}$ = 1 |

O | 61.32 | 16 | $\dfrac{61.32 }{16}$ = 3.832 | $\dfrac{3.832}{0.3484}$ = 10.99 = 11 |

H | 4.88 | 1 | $\dfrac{4.88}{1}$ = 4.88 | $\dfrac{4.88 }{0.3484}$ = 14 |

Simplest ratio of whole numbers = Zn : S : O : H = 1 : 1 : 11 : 14

Hence, **empirical formula is ZnSO _{11}H_{14}**

Molecular weight = 287

Empirical formula weight = 65 + 32 + 11(16) + 14(1) = 65 + 32 + 176 + 14 = 287

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{287}{287} = 1$

Molecular formula = n[E.F.] = 1[ZnSO_{11}H_{14}] = ZnSO_{11}H_{14}

Since all the hydrogen in the compound is in combination with oxygen as water of crystallization .

Therefore, 14 atoms of hydrogen and 7 atoms of oxygen = 7H_{2}O and hence, 4 atoms of oxygen remain.

∴ **Molecular formula is ZnSO _{4}.7H_{2}O**.

## Exercise 5D

#### Question 1

Complete the following blanks in the equation as indicated.

CaH_{2} (s) + 2H_{2}O (aq) ⟶ Ca(OH)_{2} (s) + 2H_{2} (g)

(a) Moles: 1 mole + ............... ⟶ ............... + ...............

(b) Grams: 42g + ............... ⟶ ............... + ...............

(c) Molecules: 6.02 x 10^{23} + ............... ⟶ ............... + ...............

**Answer**

(a) Moles: 1 mole + ** 2 mole** ⟶

**+**

*1 mole*

*2 mole*(b) Grams: 42g + **36g ** ⟶ ** 74g** +

*4g*(c) Molecules: 6.02 x 10^{23} + ** 12.04 × 10^{23}** ⟶

**+**

*6.02 x 10*^{23}

*12.04 × 10*^{23}#### Question 2

The reaction between 15 g of marble and nitric acid is given by the following equation:

CaCO_{3} + 2HNO_{3} ⟶ Ca(NO_{3})_{2}+ H_{2}O + CO_{2}

Calculate:

(a) the mass of anhydrous calcium nitrate formed

(b) the volume of carbon dioxide evolved at S.T.P.

**Answer**

(a)

$\begin{matrix} \text{CaCO}_3 & + \space 2\text{HNO}_3 \longrightarrow & \text{Ca(NO}_3)_2 & + \space \text{H}_2\text{O} \space + \text{CO}_2 \\ 40 + 12 + 3(16) & & 40 + 2(14) + 6(16) \\ = 40 + 12 + 48 & & 40 + 28 + 96 \\ = 100 \text{ g} & & 164 \text{ g} \end{matrix}$

100 g of CaCO_{3} produces = 164 g of Ca(NO_{3})_{2}

∴ 15 g CaCO_{3} will produce = $\dfrac{164}{100}$ x 15

= **24.6 g**

Hence, **mass of anhydrous calcium nitrate formed = 24.6 g**

(b) 100 g of CaCO_{3} produces = 22.4 litres of carbon dioxide

∴ 15 g of CaCO_{3} will produce = $\dfrac{22.4}{100}$ x 15

= **3.36 litres** of CO_{2}

#### Question 3

66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate:

(a) Mass of ammonia required.

(b) The volume of the gas used at S.T.P.

(c) The mass of acid required.

**Answer**

(a) $\begin{matrix} 2\text{NH}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{(NH}_4)_2\text{SO}_4 \\ 2[14 + 3(1)] & & 2(1) + 32 + 4(16) & & 2[14 + 4(1)] + 32 + 4(16) \\ = (2 \times 17) & & = 2 + 32 + 64 & & = 36 + 32 + 64 \\ = 34 \text{ g} & & = 98 \text{ g} & & = 132 \text{ g} \\ 2\text{ mole} \end{matrix}$

132 g ammonium sulphate is produced by 34 g of NH_{3}

∴66 g ammonium sulphate is produced by $\dfrac{34}{132}$ x 66 = 17 g of NH_{3}

Hence, **17g** of NH_{3} is required.

(b) 132 g ammonium sulphate uses 2 x 22.4 L of gas

∴ 66 g of ammonium sulphate will use $\dfrac{2 \times 22.4}{132}$ x 66 = **22.4 litres**

(c) For 132 g ammonium sulphate 98 g of acid is required

∴ For 66 g ammonium sulphate $\dfrac{98}{132}$ x 66 = 49 g

Hence, **49g** of acid is required.

#### Question 4

The reaction between red lead and hydrochloric acid is given below:

Pb_{3}O_{4} + 8HCl ⟶ 3PbCl_{2} + 4H_{2}O + Cl_{2}

Calculate

(a) the mass of lead chloride formed by the action of 6.85 g of red lead,

(b) the mass of the chlorine and

(c) the volume of chlorine evolved at S.T.P.

**Answer**

(a)

$\begin{matrix} \text{Pb}_3\text{O}_4 & + &8\text{HCl} & \longrightarrow \\ 3(207) + 4(16) & & 8[1 + 35.5] & \\ = 621 + 64 & & = 8(36.5) & \\ = 685 \text{ g} & & = 292 \text{ g} & \\ & & & \\ 3\text{PbCl}_2 & + & 4\text{H}_2\text{O} & + & \text{Cl}_2 \\ 3[207 + 2(35.5)] &&&& 2(35.5) \\ = 3[207 + 71] &&&& = 71\text{g} \\ = 834 \text{ g} \end{matrix}$

685 g of Pb_{3}O_{4} gives = 834 g of PbCl_{2}

∴ 6.85 g of Pb_{3}O_{4} will give

= $\dfrac{834}{685}$ x 6.85 = **8.34 g**

(b) 685g of Pb_{3}O_{4} gives = 71g of Cl_{2}

∴ 6.85 g of Pb_{3}O_{4} will give

= $\dfrac{71}{685}$ x 6.85 = **0.71 g** of Cl_{2}

(c) 685 g of Pb_{3}O_{4} produces 22.4 L of Cl_{2}

∴ 6.85 g of Pb_{3}O_{4} will produce

$\dfrac{22.4}{685}$ x 6.85 = **0.224 L** of Cl_{2}

#### Question 5

Find the mass of KNO_{3} required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO_{3} is required for the same purpose.

KNO_{3} + H_{2}SO_{4} ⟶ KHSO_{4} + HNO_{3}

NaNO_{3} + H_{2}SO_{4} ⟶ NaHSO_{4} + HNO_{3}

**Answer**

$\begin{matrix} \text{KNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{KHSO}_4 & + \text{HNO}_3 \\ 39 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 39 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 101 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$

63 g of HNO_{3} is formed by 101 g of KNO_{3}

∴ 126000 g of HNO_{3} is formed by $\dfrac{101}{63}$ x 126000

= **202000 g = 202 kg**

Similarly,

$\begin{matrix} \text{NaNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{NaHSO}_4 & + \text{HNO}_3 \\ 23 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 23 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 85 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$

63 g of HNO_{3} is formed by 85 g of NaNO_{3}

∴ 126000 g of HNO_{3} is formed by $\dfrac{85}{63}$ x 126000

= **170000 g = 170 kg**

So, a smaller mass of NaNO_{3} is required.

#### Question 6

Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27°C and normal pressure.

CaCO_{3} + 2HCl ⟶ CaCl_{2} + H_{2}O + CO_{2}

Calculate:

(a) The mass of salt required.

(b) The mass of the acid required

**Answer**

(a) Given,

$\begin{matrix} \text{CaCO}_3 & + &2\text{HCl} & \longrightarrow & \text{CaCl}_2 & + \text{H}_2\text{O} + &\text{CO}_2 \\ 40 + 12 + 3(16) & & 2[1 + 35.5] & && & 1\text{ mole} \\ = 40 + 12 + 48 && = 73 \text{ g} \\ = 100 \text{ g} \\ \end{matrix}$

First convert the volume of carbon dioxide to STP:

V_{1} = 2 L

T_{1} = 27 + 273 K = 300 K

T_{2} = 273 K

V_{2} = ?

Using formula:

$\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$

Substituting in the formula,

$\dfrac{2}{300}$ = $\dfrac{\text{V}_2}{273}$

V_{2} = $\dfrac{2}{300}$ x 273 = 1.82 L

As, 22.4 L of carbon dioxide is obtained using 100 g CaCO_{3}

∴ 1.82 L of carbon dioxide is obtained from $\dfrac{100}{22.4}$ x 1.82

= **8.125 g of CaCO _{3}**

(b) Similarly, 22.4 L of carbon dioxide is obtained using 73 g of acid

∴ 1.82 L of carbon dioxide is obtained from $\dfrac{73}{22.4}$ x 1.82

= **5.93 g of acid**

#### Question 7

Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water

**Answer**

$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow 2\text{H}_2 \space + & \text{O}_2 \\ 2(2 + 16) & & 2(16) \\ 36 \text{g} & & 32 \text{g} \\ \end{matrix}$

36 g of water produces 32 g of O_{2}

∴ 18 g of water will produced

= $\dfrac{32}{36}$ x 18 = **16 g** of O_{2}

$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow 2\text{H}_2 \space + & \text{O}_2 \\ 2\text{ mole} & & 1\text{ mole} \\ \end{matrix}$

2 moles of water produces 1 mole of oxygen

∴ 1 mole of water will produce $\dfrac{1}{2} \times 1$ = 0.5 moles of O_{2}

1 mole of O_{2} occupies 22.4 L volume

∴ 0.5 moles will occupy = 22.4 × 0.5

= **11.2 L**

#### Question 8

1.56 g of sodium peroxide reacts with water according to the following equation:

2Na_{2}O_{2} + 2H_{2}O ⟶ 4NaOH + O_{2}

Calculate:

(a) mass of sodium hydroxide formed,

(b) volume of oxygen liberated at S.T.P.

(c) mass of oxygen liberated.

**Answer**

$\begin{matrix} 2\text{Na}_2\text{O}_2 & + \space 2\text{H}_2\text{O} \longrightarrow & 4\text{NaOH}& & + & \text{O}_2 \\ 2[2(23) + 2(16)] & & 4(23 + 16 + 1) & & & 1 \text{mole} \\ 156 \text{g} & & 160\text{g}& & &32 \text{g} \\ \end{matrix}$

(a) 156 g of sodium peroxide produces 160 g of sodium hydroxide

∴ 1.56 g of sodium peroxide will produce $\dfrac{160}{156}$ x 1.56

= **1.6 g of sodium hydroxide**

(b) 156 g of sodium peroxide produces 22.4 L of oxygen

∴ 1.56 g of sodium peroxide will produce $\dfrac{22.4}{156}$ x 1.56

= 0.224 L

Converting L to cm^{3}

As 1 L = 1000 cm^{3}

So, 0.224 L = **224 cm ^{3}**

(c) 156 g of sodium peroxide produces 32 g of oxygen

∴ 1.56 g of sodium peroxide will produce $\dfrac{32}{156}$ x 1.56 = **0.32 g**

#### Question 9

(a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH_{4}Cl by the reaction:

2NH_{4}Cl + Ca(OH)_{2} ⟶ CaCl_{2} +2H_{2}O + 2NH_{3}

(b) What will be the volume of ammonia when measured at S.T.P?

**Answer**

$\begin{matrix} 2\text{NH}_4\text{Cl} & + \text{ Ca(OH)}_2 \longrightarrow \text{CaCl}_2 \space + \space 2\text{H}_2\text{O} \space + & 2\text{NH}_3 \\ 2[14 + 4(1) + 35.5] & & 2[14 + 3(1)] \\ 107 \text{g} & &34\text{g} \\ & &2\text{ mole} \\ \end{matrix}$

(a) 107 g NH_{4}Cl gives 34 g of NH_{3}

∴ 21.4 g NH_{4}Cl will give $\dfrac{34}{107}$ x 21.4

= **6.8 g of NH _{3}**

(b) Volume of ammonia produced by 107 g NH_{4}Cl = 2 x 22.4 L

∴ Volume of ammonia produced by 21.4 g NH_{4}Cl = $\dfrac{2 \times 22.4}{107}$ x 21.4

= **8.96 L**

#### Question 10

Aluminium carbide reacts with water according to the following equation.

Al_{4}C_{3} + 12H_{2}O ⟶ 3CH_{4} + 4Al(OH)_{3}

(a) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?

(b) What volume of methane is obtained from 12g of aluminium carbide?

**Answer**

$\begin{matrix} \text{Al}_4\text{C}_3 & + & 12\text{H}_2\text{O} & \longrightarrow & 4\text{Al(OH)}_3 & + & 3\text{CH}_4 \\ 4(27) + 3(12) & & & & 4(78) & & 3(22.4) \\ = 144 \text{ g} & & & & = 312 \text{ g} & & = 67.2 \text{ lit.} \\ \end{matrix}$

144 g of aluminium carbide forms 312 g of aluminium hydroxide.

∴ 12 g of aluminium carbide will form $\dfrac{312}{144}$ x 12 = 26 g of aluminium hydroxide

Hence, **26 g of aluminium hydroxide is formed.**

(ii) 144 g of aluminium carbide forms 67.2 lit of methane.

∴ 12 g of aluminium carbide will form $\dfrac{67.2}{144}$ x 12 = 5.6 lit.

Hence, **vol. of methane obtained = 5.6 L**

#### Question 11

MnO_{2} + 4HCl ⟶ MnCl_{2} + 2H_{2}O + Cl_{2}

0.02 moles of pure MnO_{2} is heated strongly with conc. HCl. Calculate:

(a) mass of MnO_{2} used

(b) moles of salt formed,

(c) mass of salt formed,

(d) moles of chlorine gas formed,

(e) mass of chlorine gas formed,

(f) volume of chlorine gas formed at S.T.P.,

(g) moles of acid required,

(h) Mass of acid required.

**Answer**

$\begin{matrix} \text{MnO}_2 & + & 4\text{HCl}& \longrightarrow & \text{MnCl}_2 & + & 2\text{H}_2\text{O} & + & \text{Cl}_2 \\ 1 \text{ mole} && 4 \text{ mole} && 1\text{ mole}&&&& 1\text{ mole} \\ 55 +2(16) && 4[1 + 35.5] & & 55 + 2(35.5) & & &&2(35.5) \\ = 87 \text{ g} & & = 146 \text{ g} & & = 126 \text{ g} & & & & = 71\text{ g} \\ \end{matrix}$

(a) 1 mole of MnO_{2} weighs 87 g

∴ 0.02 mole will weigh $\dfrac{87}{1}$ x 0.02 = **1.74 g**

(b) 1 mole MnO_{2} gives 1 mole of MnCl_{2}

∴ 0.02 mole MnO_{2} will give **0.02** mole of MnCl_{2}

(c) As, 1 mole MnCl_{2} weighs 126 g

∴ 0.02 mole MnCl_{2} will weigh $\dfrac{126}{1}$ x 0.02 = **2.52 g**

(d) 1 mole MnO_{2} gives 1 mole of Cl_{2}

∴ 0.02 mole MnO_{2}will form **0.02 moles** of Cl_{2}

(e) 1 mole of Cl_{2} weighs 71 g

∴ 0.02 mole will weigh $\dfrac{71}{1}$ x 0.02 = **1.42 g**

(f) 1 mole of chlorine gas has volume 22.4 dm^{3}

∴ 0.02 mole will have volume $\dfrac{22.4}{1}$ x 0.02 = **0.448 dm ^{3}**

(g) 1 mole MnO_{2} requires 4 moles of HCl

∴ 0.02 mole MnO_{2} will require $\dfrac{4}{1}$ x 0.02 = **0.08 mole**

(e) Mass of 1 mole of HCl = 36.5 g

∴ Mass of 0.08 mole = 0.08 × 36.5 = **2.92 g**

#### Question 12

Nitrogen and hydrogen react to form ammonia.

N_{2} (g) + 3H_{2} (g) ⟶ 2NH_{3} (g)

If 1000 g of H_{2} react with 2000 g of N_{2}:

(a) Will any of the two reactants remain unreacted? If yes, which one and what will be it's mass?

(b) Calculate the mass of ammonia (NH_{3}) that will be formed?

**Answer**

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2& \longrightarrow & 2\text{NH}_3 \\ 2(14) && 6(1) & & 2[14 + 3(1)] \\ 28 \text{g} & & 6 \text{g} & & 34 \text{ g} \\ \end{matrix}$

(a) 28 g of nitrogen requires 6 g of hydrogen

∴ 2000 g of nitrogen requires $\dfrac{6}{28}$ x 2000

= 428.57 g of hydrogen.

So mass of hydrogen left unreacted = 1000 - 428.57 = 571.42 g

**571.42 g** of **hydrogen** is left unreacted.

(b) 28 g of nitrogen forms 34 g NH_{3}

∴ 2000 g of nitrogen forms $\dfrac{34}{28}$ x 2000

= **2428.57 g of NH _{3}**