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Mole Concept & Stoichiometry Miscellaneous Exercises Solutions - Chemistry Class 10 ICSE - Mole Concept & Stoichiometry Miscellaneous Exercises | Chemistry Class 10 ICSE | Bright Tutorials

Mole Concept & Stoichiometry Miscellaneous Exercises

Solutions for Chemistry, Class 10, ICSE

Miscellaneous Exercises Long Answer Type

4 questions

Question 1

Give three kinds of information conveyed by the formula H₂O.

Miscellaneous Exercises Long Answer Type

Answer:

Information conveyed by formula [H₂O] —

One molecule of water (H₂O) 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 H₂O is 2H + O = 2 + 16 = 18g.

Question 2

(a) What do you mean by stoichiometry?

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

Miscellaneous Exercises Long Answer Type

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.

N₂	2N
It means 1 molecule of nitrogen.	It means two atoms of nitrogen.
It can exist independently.	It cannot exist independently.

Question 3

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

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

Miscellaneous Exercises Long Answer Type

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

(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 ?

Miscellaneous Exercises Long Answer Type

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 × 10²³

Miscellaneous Exercises Multiple Choice Type

17 questions

Question 1

Which of the following weighs the least?

2 g atom of N
3 x 10²⁵ atoms of carbon
1 mole of sulphur
7 g of silver

Miscellaneous Exercises Multiple Choice Type

Answer:

7 g of silver

Reason —

Weight of 1 g atom of N = 14 g

∴ weight of 2 g atom of N = 28 g

6.022 × 10²³ atoms of C weigh = 12 g

∴ 3 × 10²⁵ atoms will weigh = $\dfrac{12}{6.022 \times 10^{23}}$ × 3 × 10²⁵ = 597.7 g

1 mole of sulphur weighs = 32 g
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 2

Four grams of caustic soda contains:

6.02 x 10²³ atoms of it,
4 g atom of sodium,
6.02 x 10²² molecules,
4 moles of NaOH.

Miscellaneous Exercises Multiple Choice Type

Answer:

6.02 × 10²² molecule

Reason

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

40 g of NaOH contains 6.022 × 10²³ molecules

∴ 4 g of NaOH contains

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

= 6.02 × 10²² molecules

Question 3

The number of molecules in 4.25 g of ammonia is :

1.0 × 10²³
1.5 × 10²³
2.0 × 10²³
3.5 × 10²³

Miscellaneous Exercises Multiple Choice Type

Answer:

1.5 × 10²³

Reason

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

The number of molecules in 17 g of ammonia = 6.022 × 10²³

∴ No. of molecules in 4.25 g of ammonia

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

= 1.5 × 10²³

Question 4

A gas cylinder of capacity of 20 dm³ is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:

Miscellaneous Exercises Multiple Choice Type

Answer:

Reason

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{10}{2} \\[0.5em] = 5 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 5 = 10 g

Hence, relative molecular mass of gas is 10 g

Question 5

Twice the vapour density gives:

actual vapour density
relative vapour density
molecular mass
molar volume

Miscellaneous Exercises Multiple Choice Type

Answer:

molecular mass

Reason

Relative V.D. = $\dfrac{\text{Mass of vol. 'v' of the gas under similar conditions}}{\text{Mass of vol. 'v' of hydrogen gas under similar condition}}$

According to Avogadro's Law, volumes at the same temperature and pressure may be substituted by molecules. Hence,

Relative V.D. = $\dfrac{\text{mass of 1 molecule of gas or vapour}}{\text{mass of two atoms of hydrogen }}$

(Molecule of hydrogen contains 2 atoms) By multiplying both sides by 2

2 x Relative V.D = Relative molecular mass of a gas or vapour

So, twice the vapour density gives relative molecular mass.

Question 6

The empirical formula of the compound is CH₂O, the possible molecular formula can be:

C₃H₆O₃
C₂H₄O
C₄H₃O₂
C₄H₆O₂

Miscellaneous Exercises Multiple Choice Type

Answer:

C₃H₆O₃

Reason — 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 given empirical formula is CH₂O

The simplest ratio of given compound, is 1:2:1

So, C₃H₆O₃ has simplest ratio in multiples of 3
i.e., 3 x CH₂O = C₃H₆O₃

Whereas C₂H₄O, C₄H₃O₂ and C₄H₆O₂ has simplest ratio 2:4:1, 4:3:2 and 2:3:1 respectively. This cannot be derived from the simplest ratio 1:2:1 by multiplying it with any whole number value of n.

Question 7

If relative molecular mass of (C₄H₁₀) is 58, then its vapour density will be:

Miscellaneous Exercises Multiple Choice Type

Answer:

Reason

Given relative molecular mass of (C₄H₁₀) is 58

2 x Relative V.D. = $\dfrac{\text{mass of 1 molecule of gas or vapour}}{\text{mass of one atom of hydrogen }}$

2 x Relative V.D = Relative molecular mass of a gas or vapour.

Rearranging the above formula

Relative V.D = $\dfrac{\text{Relative molecular mass of a gas or vapour}}{\text{2}}$

Relative V.D = $\dfrac{58}{2}$ = 29

Hence the vapour density of (C₄H₁₀) is 29

Question 8

If the empirical mass of the formula PQ₂ is 10 and the relative molecular mass is 30, then the molecular formula will be:

PQ₂
P₃Q₂
P₆Q₃
P₃Q₆

Miscellaneous Exercises Multiple Choice Type

Answer:

P₃Q₆

Reason

Molecular formula = empirical formula x n

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{30}{10} = 3$

Hence the molecular formula is,
= 3 x empirical formula
= 3 x PQ₂ = P₃Q₆

Question 9

The ratio between the number of molecules in 2 g of hydrogen and 32 g of oxygen is:
[Atomic mass: H=1, o=16]

1:2
1:0.01
1:1
0.01:1

Miscellaneous Exercises Multiple Choice Type

Answer:

1:1

Reason

Hydrogen (H₂):

Molecular mass of H₂ = 2 g/mol

Moles of H₂ = $\dfrac{2}{2}$ = 1 mole

Molecules = 1 × Avogadro’s number = 6.022 × 10²³

Oxygen (O₂):

Molecular mass of O₂ = 32 g/mol

Moles of O₂ = $\dfrac{32}{32}$ = 1 mole

Molecules = 1 × Avogadro’s number = 6.022 × 10²³

Hence the ratio is 1:1

Question 10

One mole of sulphur dioxide represents which of the following?

P 22.4 litres at STP

Q 6.02 x 10²³ atoms

R 6.02 x 10²³ molecules

Only P
Only Q
Both P and Q
Both P and R

Miscellaneous Exercises Multiple Choice Type

Answer:

Both P and R

Reason —

One mole of sulphur dioxide (SO₂) contains 6.02 x 10²³ molecules and is equivalent to its gram molecular mass.

At Standard Temperature and Pressure (STP),
1 mole of any ideal gas = 22.4 litres

So, 1 mole of sulphur dioxide (SO₂) gas occupies 22.4 litres at STP.

Question 11

Assertion (A): According to gas equation,

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

Reason (R): On combining Boyle's law and Charle's law, 'volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature."

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

Both A and R are true and R is the correct explanation of A.

Explanation — On combining Boyle's law and Charle's law we conclude that 'volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature'.

V= $\dfrac{T}{P}$ × constant

or $\dfrac{PV}{T}$ = K (constant)

If the volume of a given mass of a gas changes from V₁ to V₂ pressure from P₁ to P₂ and temperature from T₁ to T₂ then,

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$ =K (constant) ⟶ Gas Equation

Hence Both assertion(A) and reason(R) are true and R is the correct explanation of A.

Question 12

Assertion (A): The absolute scale of temperature is the Kelvin scale.

Reason (R): Kelvin scale starts from 0°C.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

A is true but R is false.

Explanation — A temperature scale with absolute zero (zero kelvin) as the starting point is called the absolute scale or the kelvin scale. Hence the assertion (A) is true.

Kelvin scale starts from zero kelvin or -273°C. Hence reason (R) is false.

Question 13

Assertion (A): Triatomic molecules consist of four atoms.

Reason (R): Monoatomic molecules contain only one atom.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

A is false but R is true.

Explanation — Triatomic molecule is composed of three similar atoms.

Example: Ozone gas (O₃). Hence the assertion (A) is false.

Monoatomic molecule is composed of only one atom.

Examples: Inert gases like Helium, Neon, Argon, etc.
Hence the reason (R) is true.

Question 14

Assertion (A): One litre of hydrogen weighs the same as one litre of oxygen.

Reason (R): One litre of hydrogen contains the same number of molecules as one litre of oxygen.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

A is false but R is true.

Explanation — At standard temperature and pressure (STP), one litre of any ideal gas contains the same number of molecules (Avogadro’s law). But the masses differ because the molar masses of the gases are different. Hence the assertion (A) is false.

Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules". This means that one litre of hydrogen contains the same number of molecules as are present in one litre of oxygen or any other gas. Hence the reason (R) is true

Question 15

Assertion (A): Atomic mass is expressed in atomic mass unit.

Reason (R): Atomic mass is defined as $\dfrac{1}{12}$ times the mass of carbon-12.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

Both A and R are true but R is not the correct explanation of A.

Explanation — Atomic mass is expressed in atomic mass units [a.m.u.].
The relative atomic mass or atomic weight 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.
Relative atomic mass = $\dfrac{\text{mass of 1 atom of the element}}{\dfrac{1}{12}\text{ times the mass of one C-12 atom}}$

Therefore, both assertion (A) and reason (R) are true.

Atomic mass is expressed in atomic mass unit because atoms being extremely small cannot be weighed directly and conventional units of mass like kilograms or grams are not suitable for measuring them.

Hence, both assertion (A) and reason (R) are true but reason (R) does not correctly explain assertion (A).

Question 16

Assertion (A): Sulphur is an octatomic molecule.

Reason (R): The number of atoms in a molecule of an element is known as atomicity.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

Both A and R are true but R is not the correct explanation of A.

Explanation — Assertion (A) is true because sulphur molecule (S₈) is composed of eight similar atoms, so it is called Octatomic molecule.

Atomicity is defined as the number of atoms in a molecule of an element.

Therefore, both assertion (A) and reason (R) are true.

However, the reason (R) does not explain why Sulphur is an octatomic molecule.

Hence, both assertion (A) and reason (R) are true but reason (R) does not correctly explain assertion (A).

Question 17

Assertion (A): One mole of a gas occupies 24.4 litres at S.T.P.

Reason (R): The mass of one mole of a gas is equal to its molecular mass.

Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Miscellaneous Exercises Multiple Choice Type

Answer:

A is false but R is true.

Explanation — At S.T.P. (Standard Temperature and Pressure), one mole of a gas actually occupies 22.4 litres, not 24.4 litres, so Assertion (A) is false.

Reason (R) is true because by definition, the mass of one mole of a substance equals its molecular mass in grams.

Miscellaneous Exercises Short Answer Type

4 questions

Question 1

Define or explain the terms:

(a) Vapour density,

(b) Molar volume,

(d) Relative molecular mass,

(e) Avogadro's number,

(f) Gram atom,

(g) Mole.

Miscellaneous Exercises Short Answer Type

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 dm³.

(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 x10²³ 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 2

State:

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

(b) Avogadro's law

Miscellaneous Exercises Short Answer Type

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 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."

Miscellaneous Exercises Short Answer Type

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., nH₂ = 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.

Question 4

Explain the terms empirical formula and molecular formula.

Miscellaneous Exercises Short Answer Type

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.

Miscellaneous Exercises Very Short Answer Type

3 questions

Question 1

Complete the following blanks in the equation as indicated.

CaH₂ (s) + 2H₂O (aq) ⟶ Ca(OH)₂ (s) + 2H₂ (g)

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

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

Miscellaneous Exercises Very Short Answer Type

Answer:

(a) Moles: 1 mole + 2 mole ⟶ 1 mole + 2 mole

(b) Grams: 42g + 36g ⟶ 74g + 4g

Question 2

Correct the statements, if required

(a) One mole of chlorine contains 6.023 × 10²³ 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 [C¹²].

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

Miscellaneous Exercises Very Short Answer Type

Answer:

(a) One mole of chlorine contains 6.022 × 10²³ 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 [C¹²].

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

Question 3

The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?

Miscellaneous Exercises Very Short Answer Type

Answer:

According to Avogadro's law:

Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal numbers of molecules.

So, 1 mole of each gas contains = 6.02 × 10²³ molecules

Mol. Mass of :

H₂ = 2 g,

O₂ = 32 g,

CO₂ = 12 + 2(16) = 44 g,

SO₂ = 32 + 2(16) = 64 g,

Cl₂ = 2(35.5) = 71 g

(i) 2 g of hydrogen contains molecules = 6.02 × 10²³

So, 8 g of hydrogen contains molecules

= $\dfrac{6.02 × 10^{23}}{2}$ x 8

= 4 × 6.02 × 10²³molecules

= 24.08 × 10²³molecules

(ii) 32 g of oxygen contains molecules = 6.02 × 10²³

So, 8 g of oxygen contains molecules

= $\dfrac{6.02 × 10^{23}}{32}$ x 8

= 1.505 × 10²³molecules

(iii) 44 g of carbon dioxide contains molecules = 6.02 × 10²³

So, 8 g of carbon dioxide contains

= $\dfrac{6.02 × 10^{23}}{44}$ x 8

= 1.09 × 10²³molecules

(iv) 64 g of sulphur dioxide contains molecules = 6.02 × 10²³

So, 8g of sulphur dioxide contains

= $\dfrac{6.02 × 10^{23}}{64}$ x 8

= 0.75 × 10²³molecules

(v) 71 g of chlorine contains molecules = 6.02 × 10²³

So, 8g of chlorine contains

= $\dfrac{6.02 × 10^{23}}{71}$ x 8

= 0.67 × 10²³molecules

Thus Cl₂ < SO₂ < CO₂ < O₂ < H₂

(i) Least number of molecules in Cl₂

(ii) Most number of molecules in H₂

Numerical Problems

72 questions

Question 1

From the equation for burning of hydrogen and oxygen

2H₂ + O₂ ⟶ 2H₂O (Steam)

Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

Numerical Problems

Answer:

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

1 mole of oxygen gives 2 moles of steam

∴ 0.5 mole of oxygen will give $\dfrac{2}{1}$ × 0.5

= 1 mole of steam

Question 2

From the equation

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(At. mass Cu=64, H=1, N=14, O=16)

Calculate:

(a) Mass of copper needed to react with 63 g of HNO₃

(b) Volume of nitric oxide at S.T.P. that can be collected.

Numerical Problems

Answer:

$\begin{matrix} 3\text{Cu} & + &8\text{HNO}_3 & \longrightarrow & 3\text{Cu(NO}_3)_2 + 4\text{H}_2\text{O} + 2\text{NO} \\ 3(64) && 8(1 + 14 + 3(16)) \\ = 192 \text{g} & & = 8(63) \\ &&= 504 \text{g} \\ \end{matrix}$

(a) 504 g nitric acid reacts with 192 g of copper

∴ 63 g of nitric acid reacts with $\dfrac{192}{504}$ x 63 = 24 g of copper

Hence, 24 g of copper is required.

(b) 504 g of nitric acid gives 2 × 22.4 litre volume of NO

∴ 63 g of nitric acid gives $\dfrac{2 × 22.4}{504}$ x 63

= 5.6 litre of NO

5.6 L of NO is collected.

Question 3

(a) Calculate the number of moles in 7 g of nitrogen.

(b) What is the volume at S.T.P. of 7.1 g of chlorine?

Numerical Problems

Answer:

(a) Gram molecular mass of N₂ = 2 x 14 = 28 g

28 g of nitrogen = 1 mole

∴ 7 g of nitrogen = $\dfrac{1}{28}$ x 7

= 0.25 moles

(b) Gram molecular mass of Cl₂ = 2 x 35.5 = 71 g

71 g of chlorine at S.T.P. occupies 22.4 litres

∴ 7.1 g of chlorine will occupy $\dfrac{22.4}{71}$ x 7.1

= 2.24 litre = 2.24 dm³

28 g of carbon monoxide at S.T.P. occupies 22,400 cm³ volume

So, 22,400 cm³ volume have mass = 28 g

∴ 56 cm³ volume will have mass $\dfrac{28}{22,400}$ x 56 = 0.07 g

Question 4

Some of the fertilizers are sodium nitrate NaNO₃, ammonium sulphate (NH₄)₂SO₄ and urea CO(NH₂)₂. Which of these contains the highest percentage of nitrogen?

Numerical Problems

Answer:

(i) Molar mass of NaNO₃ = 23 + 14 + 3(16) = 23 + 14 + 48 = 85 g

Nitrogen in 85 g NaNO₃ = 14 g

∴ Percentage of Nitrogen in NaNO₃ = $\dfrac{14}{85}$ x 100 = 16.47%

(ii) Molar mass of (NH₄)₂SO₄ = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132 g

Nitrogen in 132 g of (NH₄)₂SO₄ = 28 g

∴ Percentage of Nitrogen in (NH₄)₂SO₄ = $\dfrac{28}{132}$ x 100 = 21.21%

(iii) Molar mass of CO(NH₂)₂ = 12 + 16 + 2[14 + (2 x 1)]

= 28 + 2(16)

= 28 + 32

= 60 g

Nitrogen in 60 g of CO(NH₂)₂ = 2 x 14 = 28 g

∴ Percentage of Nitrogen in CO(NH₂)₂ = $\dfrac{28}{60}$ x 100 = 46.66%

So, the highest percentage of Nitrogen is in Urea.

Question 5

200 cm³ of CO₂ 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.

2C₂H₂ (g) + 5O₂ (g) ⟶ 4CO₂ (g)+ 2H₂O (g)

Numerical Problems

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 CO₂ is collected with 2 Vol. of C₂H₂

So, 200 cm³ CO₂ will be collected with

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

Similarly, 4 Vol of CO₂ is produced by 5 Vol of O₂

So, 200 cm³ CO₂ will be produced by = $\dfrac{5}{4}$ x 200 = 250 cm³

Hence, acetylene = 100 cm³ and oxygen = 250 cm³

Question 6

Urea (CO(NH₂)₂) is an important nitrogenous fertilizer, and is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Numerical Problems

Answer:

Molar mass of urea [CO(NH₂)₂] = 12 + 16 + 2[14 + (2 x 1)]

= 28 + 2(16)

= 28 + 32

= 60 g

50 kg of urea = 50,000 g of urea

Nitrogen in 60 g of CO(NH₂)₂ = 2 x 14 = 28 g

∴ Nitrogen in 50,000 g urea = $\dfrac{28 \times 50,000 }{60}$

= 23,333 g = 23.3 kg

Question 7

Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	80	12	$\dfrac{80}{12}$ = 6.66	$\dfrac{6.66}{6.66}$ = 1
Hydrogen	20	1	$\dfrac{20}{1}$ = 20	$\dfrac{20}{6.66}$ = 3

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

Hence, empirical formula is CH₃

Empirical formula weight = 12 + 3(1) = 15

V.D. = 15

Molecular weight = 2 x V.D. = 2 x 15

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 15}{15} = 2$

∴ Molecular formula = n[E.F.] = 2[CH₃] = C₂H₆

Question 8

The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.

0.145 g of X was heated with dry copper (II) oxide and 224 cm³ of carbon dioxide was collected at S.T.P.

(a) Which elements does X contain?

(b) What was the purpose of copper (II) oxide?

(c ) Calculate the empirical formula of X by the following steps:

(i) Calculate the number of moles of carbon dioxide gas.

(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.

(iii) Calculate the mass of hydrogen in sample X.

(iv) Deduce the ratio of atoms of each element in X (empirical formula).

Numerical Problems

Answer:

(a) X contains carbon and hydrogen as it is a hydrocarbon.

(b) The purpose of copper (II) oxide is to act as an oxidizing agent.

∴ 224 cm³ CO₂ will have mass = $\dfrac{44}{22400}$ x 224 = 0.44 g

Molar mass of CO₂ = 12 + 2(16) = 12 + 32 = 44 g

Moles of CO₂ = $\dfrac{\text{Mass}}{\text{{Molecular mass}}}$ = $\dfrac{0.44}{44}$ = 0.01 moles

(ii) Mass of carbon in 44 g CO₂ = 12 g

∴ Mass of carbon in 0.44 g CO₂ = $\dfrac{12}{44}$ x 0.44 = 0.12 g

As X contains carbon and hydrogen, so sample X has 0.12 g of carbon

(iii) Mass of Hydrogen in X = 0.145 - 0.12 = 0.025 g

(iv) Ratio of moles of C : H

= $\dfrac{0.01}{0.01}$ : $\dfrac{0.025}{0.01}$

= 1 : $\dfrac{25}{10}$

= 1 : $\dfrac{5}{2}$

= 2 : 5

Hence, ratio of C : H = 2 : 5

so, the empirical formula of hydrocarbon is C₂H₅

Question 9

Compound is formed by 24 g of X and 64 g of oxygen. If atomic mass of X = 12 and O = 16, calculate the simplest formula of compound.

Numerical Problems

Answer:

Element	Mass (g)	Atomic mass	Moles	Simplest ratio
Element X	24	12	$\dfrac{24}{12}$ = 2	$\dfrac{2}{2}$ = 1
Oxygen	64	16	$\dfrac{64}{16}$ = 4	$\dfrac{4}{2}$ = 2

Simplest ratio of whole numbers = X : O = 1 : 2

Hence, simplest formula of compound is XO₂

Question 10

A gas cylinder filled with hydrogen holds 5 g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate:

(a) Vapour density of gas X.

(b) Molecular weight of gas X.

Numerical Problems

Answer:

(a ) V.D. = $\dfrac{\text{mass of certain volume of gas at S.T.P.}}{\text{mass of equal volume of H}_2\text{ at S.T.P.}}$ = $\dfrac{85}{5}$ = 17

(b) Molecular weight = 2 x V.D. = 2 x 17 = 34 a.m.u.

Question 11

(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :

CO₂ + C ⟶ 2CO

What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?

(b) 60 cm³ of oxygen was added to 24 cm³ of carbon monoxide and mixture ignited. Calculate:

(i) volume of oxygen used up and

(ii) Volume of carbon dioxide formed.

Numerical Problems

Answer:

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

[By Gay Lussac's law]

1 Vol. of C produces 2V of CO

12 g of C produces 2 x 22.4 L of CO

∴ 3 g of C will produce $\dfrac{2 \times 22.4}{12}$ x 3

= 11.2 L of CO

(b)

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

(i) 2 Vol. of CO requires 1 Vol. of oxygen

∴ 24 cm³ CO will require $\dfrac{1}{2}$ x 24

= 12 cm³ of oxygen

(ii) 2 Vol. of CO gives 2 vol. of CO₂

∴ 24 cm³ of CO will give 24 cm³ of CO₂

Question 12

How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO₂ evolved:

2Ca(NO₃)₂ ⟶ 2CaO + 4NO₂ + O₂

Numerical Problems

Answer:

$\begin{matrix} 2\text{Ca(NO}_3)_2 & \longrightarrow & 2\text{CaO} & + &\space 4\text{NO}_2 \space + & \text{O}_2 \\ 2[40 + 2[14 + 3(16)]] & & 2[40 + 16] \\ = 2[40 + 28 + 96] & & = 2(56) \\ = 328 \text{ g} & & = 112 \text{ g} \end{matrix}$

328 g of calcium nitrate produces 112 g of CaO

∴ 82 g of calcium nitrate will produce $\dfrac{112}{328}$ x 82

= 28 g of CaO

328 g of calcium nitrate produces 4 x 22.4 L of NO₂

∴ 82 g of calcium nitrate will produce $\dfrac{4 \times 22.4}{328}$ x 82

= 22.4 L of NO₂

Question 13

The equation for the burning of octane is:

2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 18H₂O

(i) How many moles of carbon dioxide are produced when one mole of octane burns?

(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?

(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?

(iv) What is the empirical formula of octane?

Numerical Problems

Answer:

$\begin{matrix} 2\text{C}_8\text{H}_{18} & + &25\text{O}_2 & \longrightarrow & 16\text{CO}_2 & + & 18\text{H}_2\text{O} \\ 2\text{ Vol.}&& 25\text{ Vol.} && 16\text{ Vol.}&& 18\text{ Vol.} \\ &&&&16[12 + 2(16)] \\ &&&&16[12 + 32] \\ &&&&16[44] = 704 \text{g} \\ \end{matrix}$

(i) 2 moles of octane produces 16 moles of CO₂

∴ 1 mole octane produces $\dfrac{16}{2}$ x 1

= 8 moles of CO₂

(ii) 1 mole CO₂ occupies volume = 22.4 L

As 2 moles of octane produces 8 moles of carbon dioxide which will occupy volume $\dfrac{22.4}{1}$ x 8

= 179.2 dm³ of CO₂

(iii) 1 mole CO₂ has mass = 44 g

∴ 16 moles will have mass $\dfrac{44}{1}$ x 16

= 704 g of CO₂

(iv) Molecular formula is C₈H₁₈

∴ Ratio of C and H is 8 : 18

Simple ratio is 4 : 9

Hence, empirical formula = C₄H₉

Question 14

Ordinary chlorine gas has two isotopes ³⁵₁₇Cl and ³⁷₁₇Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.

Numerical Problems

Answer:

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

Question 15

Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.

Numerical Problems

Answer:

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 SiCl₃

Question 16

An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and it's molecular formula, given that it's relative molecular mass is 162.

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Phosphorus	38.27	31	$\dfrac{38.27 }{31}$ = 1.234	$\dfrac{ 1.234}{1.234 }$ = 1
Hydrogen	2.47	1	$\dfrac{2.47 }{1}$ = 2.47	$\dfrac{2.47 }{1.234 }$ = 2
Oxygen	59.26	16	$\dfrac{59.26}{16}$ = 3.70	$\dfrac{3.70}{ 1.234 }$ = 3

Simplest ratio of whole numbers = P : H : O = 1 : 2 : 3

Hence, empirical formula is H₂PO₃

Empirical formula weight = 31 + 2(1) + 3(16) = 31 + 2 + 48 = 81

Molecular weight = 162

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{162}{81} = 2$

So, molecular formula = 2(H₂PO₃) = H₄P₂O₆

Question 17

(a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular mass of 'A' is 60.

(b) A gas occupied 360 cm³ at 87°C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find it's relative molecular mass.

Numerical Problems

Answer:

(a) Given, molecular mass of 'A' is 60

V₁ = 10 L

T₁ = 27 + 273 K = 300 K

P₁ = 700 mm

T₂ = 273 K

P₂ = 760 mm

V₂ = ?

Using formula:

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

Substituting in the formula,

$\dfrac{700 \times 10}{300}$ = $\dfrac{760 \times\text{V}_2}{273}$

V₂ = $\dfrac{700 \times10 \times 273}{300 \times760}$ = $\dfrac{1911}{228}$ = 8.38 L

As, 22.4 L of A weighs 60 g

∴ 8.38 L of A will weigh $\dfrac{60}{22.4}$ x 8.38

= 22.446 = 22.45 g

(b) V₁ = 360 cm³

T₁ = 87 + 273 K = 360 K

P₁ = 380 mm Hg pressure

T₂ = 273 K

P₂ = 760 mm Hg pressure

V₂ = ?

Using formula:

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

Substituting in the formula,

$\dfrac{380\times 360}{360}$ = $\dfrac{760 \times\text{V}_2}{273}$

V₂ = $\dfrac{380 \times 273}{760}$ = $\dfrac{10,374}{760}$ = 136.5 cm³

136.5 cm³ of gas weigh = 0.546 g

22400 cm³ of gas weigh $\dfrac{0.546}{136.5}$ x 22400

= 89.6 amu

Question 18

A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.

(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?

(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result.

Numerical Problems

Answer:

(a) Given, cylinder can hold = 1 kg of hydrogen

Molar mass of hydrogen = 2[1] = 2g

Number of moles of hydrogen present in cylinder = $\dfrac{1000}{2}$ = 500

According to avogadro's law : cylinder will hold 500 moles of carbon dioxide gas

Molar mass of CO₂ = C + 2(O) = 12 + 2(16) = 44 g

1 mole of carbon dioxide = 44 g

∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.

Hence, Weight of carbon dioxide in cylinder = 22 kg

(b) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

∴ molecules in the cylinder of carbon dioxide = X

Avogadro's law helped to arrive at this result.

Question 19

Following questions refer to one mole of chlorine gas.

(a) What is the volume occupied by this gas at S.T.P.?

(b) What will happen to the volume of gas, if pressure is doubled?

(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?

Numerical Problems

Answer:

(a) According to Avogadro's law : The volume occupied by 1 mole of chlorine is 22.4 dm³

(b) According to Boyle's Law: PV = constant

Hence, if pressure is doubled then volume will become half i.e. $\dfrac{22.4}{2}$ = 11.2 dm³

T₁ = 273 K

T₂ = 273 + 273 K = 546 K

V₂ = ?

According to charles law:

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

Substituting to we get,

$\dfrac{ 22.4}{273}$ = $\dfrac{\text{V}_2}{546}$

Hence, V₂ = $\dfrac{ 22.4}{273}$ x 546 = 44.8 dm³

Question 20

(a) A hydrate of calcium sulphate CaSO4.xH₂O contains 21% water of crystallisation. Find the value of x.

(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.

(d) What would be the mass of CO₂ occupying a volume of 44 litres at 25°C and 750 mm pressure?

(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.

AgNO₃ (aq) + NaCl (aq) ⟶ AgCl (s) + NaNO₃

Calculate the percentage of NaCl in the mixture.

Numerical Problems

Answer:

Relative molecular mass of CuSO₄.xH₂O

= 40 + 32 + (4×16) + [x(2+16)]

= 40 + 32 + 64 + 18x

= 136 + 18x

∴ 21% water of crystallization = 18 x

$\Rightarrow \dfrac{18x}{136 + 18x } = \dfrac{21}{100} \\[1em] 1800x = 21(136 + 18x) \\[1em] 1800x = 2856 + 378x \\[1em] 1800x - 378x = 2856 \\[1em] 1422x = 2856 \\[1em] x = \dfrac{2856}{1422} \\[1em] x = 2$

Hence, water of crystallization = 2

(b) Molar mass of H₂O = 2(1) + 16 = 18 g

For 18 g water, vol. of hydrogen needed = 22.4 litre

∴ For 1.8 g, vol. of H₂ needed = $\dfrac{22.4}{18}$ x 1.8 = 2.24 L

According to Gay Lussac's Law, 2 vol of hydrogen requires 1 vol of oxygen

When 2 vol of hydrogen in 1.8 g H₂O is 2.24 L, then one vol. of oxygen will be:

$\dfrac{2.24}{2}$ = 1.12 L

1 mole of O₂ weighs 32 g and occupies 22.4 lit. vol.

∴ 2 g of O₂ occupies = $\dfrac{22.4}{32} \times 2 = 1.4 \text{ lit.}$

Volume occupied by 2 g of O₂ gas at 27°C and 740 mm pressure:

S.T.P.	Given Values
P₁ = 760 mm of Hg	P₂ = 740 mm of Hg
V₁ = 1.4 lit	V₂ = x lit
T₁ = 273 K	T₂ = 27 + 273 K

Using the gas equation,

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

Substituting the values we get,

$\dfrac{760 \times 1.4}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 1.4 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{3,19,200}{2,02,020} \\ \\[0.5em] x = 1.58 \text{ lit}$

Hence, the volume occupied by 2 g of O₂ gas at 27°C and 740 mm pressure is 1.58 lit.

(d) Gram molecular mass of CO₂ = 12 + 2(16) = 12 + 32 = 44 g

Given Values	S.T.P.
P₁ = 750 mm of Hg	P₂ = 760 mm of Hg
V₁ = 44 lit	V₂ = x lit
T₁ = 25 + 273 K = 298 K	T₂ = 273 K

Using the gas equation,

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

Substituting the values we get,

$\dfrac{750 \times 44}{298} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{750 \times 44 \times 273}{760 \times 298 } \\[0.5em] x = \dfrac{9,009,000}{226,480} \\ \\[0.5em] x = 39.78 \text{ lit}$

22.4 litre of CO₂ at S.T.P. has mass = 44 g

39.78 litre of CO₂ at S.T.P. has mass $\dfrac{44}{22.4}$ x 39.78

= 78.14 g

(e)

$\begin{matrix} \text{AgNO}_3 & + &\text{NaCl} & \longrightarrow & \text{AgCl} & + & \text{NaNO}_3 \\ && 23 + 35.5 && 108 + 35.5 \\ && = 58.5 \text{g}&& = 143.5\text{g} \\ \end{matrix}$

(i) 143.5 g AgCl is formed by 58.5 g NaCl

∴ 1.435 g of AgCl will be formed by $\dfrac{58.5}{143.5}$ x 1.435 = 0.582 g

Percentage of NaCl = $\dfrac{0.582}{1}$ x 100 = 58.5%

Hence, percentage of NaCl is 58.5%

Question 21(a)

From the equation:

C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂

Calculate:

(i) The mass of carbon oxidized by 49 g of sulphuric acid.

(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.

Numerical Problems

Answer:

$\begin{matrix} \text{C} & + &2\text{H}_2\text{SO}_4 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O}& + & 2\text{SO}_2 \\ 12 \text{g} && 2[2(1) + 32 + 4(16)] \\ && 2[2 + 32 + 64] \\ && 196 \text{g} \\ \end{matrix}$

(i) 196 g of sulphuric acid oxidizes 12 g carbon

∴ 49 g of sulphuric acid will oxidize = $\dfrac{12}{196}$ x 49 = 3 g

Hence, 3 g of carbon is oxidized.

(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO₂

∴ 3 g of carbon will liberate $\dfrac{2 \times 22.4}{12}$ x 3 = 11.2 lit of SO₂.

Hence, 11.2 lit of SO₂ is liberated.

Question 21(b)

(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)

(ii) The relative molecular mass of this compound is 168, so what is it's molecular formula?

Numerical Problems

Answer:

(i)

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	14.4	12	$\dfrac{14.4}{12}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
Hydrogen	1.2	1	$\dfrac{1.2}{1}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
chlorine	84.5	35.5	$\dfrac{84.5}{35.5}$ = 2.38	$\dfrac{2.38}{1.2}$ = 1.98 = 2

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

Hence, empirical formula is CHCl₂

Empirical formula weight = 12 + 1 + 2(35.5) = 84 g

Relative molecular mass = 168

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{168}{84} = 2$

Molecular formula = n[E.F.] = 2[CHCl₂] = C₂H₂Cl₄

∴ Molecular formula = C₂H₂Cl₄

Question 22

Find the percentage of

(a) oxygen in magnesium nitrate crystals [Mg(NO₃)₂.6H₂O].

(b) boron in Na₂B₄O₇.10H₂O. H=1,B=11,O=16,Na=23.

Numerical Problems

Answer:

Relative molecular mass of [Mg (NO₃)₂.6H₂O]

= 24 + 2[14 + 3(16)] + 12(1) + 6(16)
= 24 + 2[14 + 48] + 12 + 96
= 24 + 28 + 96 + 12 + 96
= 256 g

Molar mass of oxygen in [Mg (NO₃)₂.6H₂O] = 96 + 96 = 192 g

Since, 256 g of [Mg(NO₃)₂.6H₂O] contains 192 g of oxygen

∴ 100 g of [Mg(NO₃)₂.6H₂O] contains

$\dfrac{192}{256}$ x 100 = 75% of oxygen

(b) Relative molecular mass of Na₂B₄O₇.10H₂O
= 2(23) + 4(11) + 7(16) + 20(1) + 10(16)
= 46 + 44 + 112 + 20 + 160
= 382 g

Molar mass of boron in Na₂B₄O₇.10H₂O = 44 g

382 g of Na₂B₄O₇.10H₂O contains 44 g of boron

∴ 100 g Na₂B₄O₇.10H₂O contains $\dfrac{44}{382}$ x 100 = 11.5%

Molar mass of phosphorus in Ca(H₂PO₄)₂ = 62 g

234g of Ca(H₂PO₄)₂ contains 62 g of phosphorus

∴ 100 g of Ca(H₂PO₄)₂ contains $\dfrac{62}{234}$ x 100 = 26.5%

Question 23

What mass of copper hydroxide is precipitated by using 200 g of sodium hydroxide?

2NaOH + CuSO₄ ⟶ Na₂SO₄ + Cu(OH)₂ ↓

[Cu = 64, Na = 23, S = 32, H = 1]

Numerical Problems

Answer:

$\begin{matrix} 2\text{NaOH} & + & \text{CuSO}_4 & \longrightarrow & \text{Na}_2\text{SO}_4 & + & \text{Cu(OH)}_2\downarrow \\ 2[23 + 16 & & & & & & 64 + 2[16 + 1] \\ + 1] & & & & & & = 64 + 34 \\ = 80 \text{ g} & & & & & & = 98 \text{g} \\ \end{matrix}$

80 g of sodium hydroxide precipitates 98 g of copper hydroxide.

∴ 200 g of sodium hydroxide will precipitate $\dfrac{98}{80}$ x 200 = 245 g of copper hydroxide.

Hence, 245 g. of copper hydroxide is precipitated.

Question 24

Solid ammonium dichromate decomposes as under:

(NH₄)₂Cr₂O₇ ⟶ N₂ + Cr₂O₃ + 4H₂O

If 63 g of ammonium dichromate decomposes. Calculate

(a) the quantity in moles of (NH₄)₂Cr₂O₇

(b) the quantity in moles of nitrogen formed.

(d) the loss of mass

(e) the mass of chromium (III) oxide formed at the same time.

Numerical Problems

Answer:

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$

(a) 252 g of (NH₄)₂Cr₂O₇ = 1 mole

∴ 63 g of (NH₄)₂Cr₂O₇ = $\dfrac{1}{252}$ x 63 = 0.25 moles

Hence, no. of moles = 0.25 moles

(b)

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & : & \text{N}_2 \\ 1 \text{ mol.} & : & 1 \text{ mol.} \\ 0.25 \text{ mol.} & : & x \\ \end{matrix}$

Hence, 0.25 moles of (NH₄)₂Cr₂O₇ will produce 0.25 moles of nitrogen.

∴ 0.25 moles of N₂ will occupy = 22.4 x 0.25 = 5.6 lit.

Hence, volume of N₂ evolved at s.t.p = 5.6 lit.

(d) 252 g of (NH₄)₂Cr₂O₇ decomposes to give 152 g of solid Cr₂O₃, loss in mass = 252 - 152 = 100 g

If 63 g of (NH₄)₂Cr₂O₇ decomposes then loss in mass is

$\dfrac{100}{252}$ x 63 = 25 g

(e) 1 mole of Cr₂O₇ = 152 g.

∴ 0.25 moles of Cr₂O₇ = 152 x 0.25 = 38 g.

Hence, mass in gms of Cr₂O₇ formed = 38 g.

Question 25

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:

2H₂S + 3O₂ ⟶ 2H₂O + 2SO₂

Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at S.T.P. which will complete the combustion of hydrogen sulphide determined in (litres).

Numerical Problems

Answer:

$\begin{matrix} 2\text{H}_2\text{S} & + &3\text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} & + & 2\text{SO}_2 \\ 2\text{ Vol.} && 3\text{ Vol.} && && 2\text{ Vol.} \\ 2[2(1) + 32] && 3[2(16)] &&&& 2[32 + 2(16)] \\ 68 \text{ g} && 96 \text{ g} &&&& 128 \text{ g} \end{matrix}$

128 g of SO₂ has volume 2 × 22.4 litres

∴ 12.8 g of SO₂ has volume =

$\dfrac{2 × 22.4}{128}$ x 12.8 = 4.48 L

Volume of oxygen = ?

2 × 22.4 L H₂S requires = 3 × 22.4 litre of oxygen

∴ 4.48 L H₂S will require

$\dfrac{3 × 22.4}{2 × 22.4 }$ x 4.48 = 6.72 L of oxygen.

Question 26

Ammonia burns in oxygen and combustion, in the presence of a catalyst, may be represented by

2NH₃ + 2 $\dfrac{1}{2}$ O₂ ⟶ 2NO + 3H₂O

[H = 1, N = 14, O = 16]

What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

Numerical Problems

Answer:

$\begin{matrix} 2\text{NH}_3 & + &2\dfrac{1}{2}\text{O}_2 & \longrightarrow & 2\text{NO} & + &3\text{H}_2\text{O} \\ &&&&2[14 + 16] && 3[2(1) + 16] \\ &&&& 60 \text{ g} && 54 \text{ g} \\ \end{matrix}$

When 60 g NO is formed then, 54 g mass of steam is produced

∴ when 1.5 g NO is formed, mass of steam produced

= $\dfrac{54}{60}$ x 1.5 = 1.35 g

Question 27

If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate Ca(NO₃)₂ would be required to replace the nitrogen in a 10 hectare field ?

Numerical Problems

Answer:

Molecular mass of Ca(NO₃)₂

= 40 + [2(14) + 6(16)]
= 40 + 28 + 96
= 164 g

Mass of N₂ in Ca(NO₃)₂ = 28 g

In 1 hectare of soil, 20 kg of N₂ is removed

∴ In 10 hectare field N₂ removed is 20 x 10 = 200 kg

28 g N₂ is present in 164 g of Ca(NO₃)₂

So, 200 kg or 200,000 g of N₂ is present in $\dfrac{164}{28}$ x 200,000

= 1171428 g = 1171.4 kg

Question 28

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5HNO₃ ⟶ H₃PO₄+ 5NO₂ + H₂O

If 6.2 g of phosphorus was used in the reaction calculate:

(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.

(b) mass of nitric acid consumed at the same time?

Numerical Problems

Answer:

$\begin{matrix} \text{P}& + &5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & 5\text{NO}_2& + &\text{H}_2\text{O} \\ 31 \text{g} && 5[1 + 14 + 3(16)] && 3(1) + 31 + 4(16) \\ && 315 \text{ g} && 98 \text{ g} \end{matrix}$

(a) 31 g of P = 1 mole

∴ 6.2 g of P = $\dfrac{1}{31}$ x 6.2 = 0.2 mole of P

Mass of phosphoric acid formed = ?

31 g of P produces 98 g of phosphoric acid

∴ 6.2 g of P will form $\dfrac{98}{31}$ x 6.2 = 19.6 g

(b) 31 g P reacts with 315 g HNO₃

∴ 6.2 g P will react with $\dfrac{315}{31}$ x 6.2 = 63 g HNO₃

∴ 6.2 g P produces $\dfrac{1}{31}$ x 6.2 = 0.2 moles

Volume of steam produced at STP = 0.2 × 22.4 = 4.48 litre

V₁ = 4.48 litre

T₁ = 273 K

P₁ = 760 mm Hg pressure

T₂ = 273 + 273 = 546 K

P₂ = 760 mm Hg pressure

V₂ = ?

Using formula:

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

Substituting in the formula,

$\dfrac{760\times 4.48}{273}$ = $\dfrac{760 \times\text{V}_2}{546}$

V₂ = 2 x 4.48 = 8.96 L

Hence, volume of steam produced = 8.96 L

Question 29

112 cm³ of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [ F = 19, P = 31]

Numerical Problems

Answer:

Given,

112 cm³ of gaseous fluoride has mass = 0.63 g

so, 22400 cm³ of gaseous fluoride will have mass = $\dfrac{0.63}{112}$ x 22400

= 126 g

∴ Relative molecular mass of fluoride = 126 g

The molecular mass = At mass P + At. mass of F

126 = 31 + At. Mass of F

∴ At. Mass of F = 126 - 31 = 95 g

However, At. mass of F = 19

∴ $\dfrac{95}{19}$ = 5

So, 5 atoms of F, hence, the molecular formula = PF₅

Question 30

Washing soda has formula Na₂CO₃.10H₂O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda?

Numerical Problems

Answer:

$\begin{matrix} \text{Na}_2\text{CO}_3.10\text{H}_2\text{O} & \xrightarrow{\Delta} & \text{Na}_2\text{CO}_3 & + & 10\text{H}_2\text{O} \\ 2(23) + 12 + 3 (16) & & 2(23) + 12 & & 10[2(1) \\ + 10[2(1) + 16] & & + 3(16) & & + 16] \\ = 46 + 12 + 48 & & = 46 + 12 & & = 180 \text{ g} \\ + 10(18) & & + 48 \\ = 286 \text{ g} & & = 106 \text{ g} \\ \end{matrix}$

286 g of washing soda had 106 g of anhydrous sodium carbonate

∴ 57.2 g will have = $\dfrac{106}{286}$ x 57.2 = 21.2 g anhydrous sodium carbonate.

Hence, 21.2 g anhydrous sodium carbonate is left.

Question 31

A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M = 56).

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Metal M	34.5	56	$\dfrac{34.5}{56}$ = 0.616	$\dfrac{0.616}{0.616}$ = 1
chlorine	65.5	35.5	$\dfrac{65.5}{35.5}$ = 1.84	$\dfrac{1.84}{0.616}$ = 2.98 = 3

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

Hence, empirical formula is MCl₃

Empirical formula weight = 56 + 3(35.5) = 162.5 g

Molecular weight = 2 x V.D. = 2 x 162.5 = 325

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{ 325}{162.5} = 2$

Molecular formula = n[E.F.] = 2[MCl₃] = M₂Cl₆

∴ Molecular formula = M₂Cl₆

Question 32

A compound X consists of 4.8% carbon and 95.2% bromine by mass.

(i) Determine the empirical formula of this compound working correct to one decimal place (C = 12; Br = 80)

(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
carbon	4.8	12	$\dfrac{4.8}{12}$ = 0.4	$\dfrac{0.4}{0.4}$ = 1
bromine	95.2	80	$\dfrac{95.2}{80}$ = 1.19	$\dfrac{1.19}{0.4}$ = 2.9 = 3

Simplest ratio of whole numbers = C : Br = 1 : 3

Hence, empirical formula is CBr₃

(ii) Empirical formula weight = 12 + 3(80) = 252 g

Molecular weight = 2 x V.D. = 2 x 252 = 504

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{ 504}{252} = 2$

Molecular formula = n[E.F.] = 2[CBr₃] = C₂Br₆

∴ Molecular formula = C₂Br₆

Question 33

The reaction: 4N₂O + CH₄ ⟶ CO₂ + 2H₂O + 4N₂ takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N₂O) required to give 150 cm³ of steam.

Numerical Problems

Answer:

$\begin{matrix} 4\text{N}_2\text{O} & + &\text{CH}_4 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} & + &4\text{N}_2 \\ 4 \text{ Vol} && 1 \text{ Vol} && 1 \text{ Vol} && 2 \text{ Vol} && 4\text{ Vol} \\ \end{matrix}$

2 x 22400 litre steam is produced by 4 × 22400 cm³ N₂O

∴ 150 cm³ steam will be produced by

= $\dfrac{4 × 22400}{2 × 22400}$ x 150 = 300 cc of N₂O

Question 34

Samples of the gases O₂, N₂, CO₂ and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure

What is the volume occupied by:

(a) x molecules of N₂

(b) 3x molecules of CO

(d) In answering the above questions, which law have you used?

Numerical Problems

Answer:

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

∴ If gases under the same conditions have same number of molecules then they must have the same volume.

(i) So, X molecules of N₂ occupy 1V litres.

(ii) 3X molecules of CO will occupies 3V litres.

(iii) Molar mass of CO₂ = C + 2(O) = 12 + 2(16) = 44 g

1 mole of O₂ weighs 32 g and occupy 22.4 lit. volume

∴ 8 g of O₂ will occupy $\dfrac{22.4}{32}$ x 8 = 5.6 lit. vol. = Molar volume

If 8 g of O₂ occupies 5.6 lit. vol.

And 1 mole of CO₂ occupy 22.4 lit and weighs = 44 g at s.t.p.

∴ 5.6 lit. vol. of CO₂ will weigh = $\dfrac{44}{22.4}$ x 5.6 = 11 g

Hence, mass of CO₂ = 11 g

(iv) Avogadro's Law is used above.

Question 35

The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound?

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
sodium	42.1	23	$\dfrac{42.1}{23}$ = 1.83	$\dfrac{1.83}{0.609}$ = 3
phosphorus	18.9	31	$\dfrac{18.9}{31}$ = 0.609	$\dfrac{0.609}{0.609}$ = 1
oxygen	39	16	$\dfrac{39}{16}$ = 2.43	$\dfrac{2.43}{0.609}$ = 4

Simplest ratio of whole numbers = Na : P : O = 3 : 1 : 4

Hence, empirical formula is Na₃PO₄

Question 36

What volume of oxygen is required to burn completely a mixture of 22.4 dm³ of methane and 11.2 dm³ of hydrogen into carbon dioxide and steam?

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O

2H₂ + O₂ ⟶ 2H₂O

Numerical Problems

Answer:

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

From equation:

22.4 dm³ of methane requires oxygen = 2 x 22.4 dm³ of O₂ = 44.8 dm³

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

From equation,

[2 x 22.4] dm³ hydrogen requires oxygen = 22.4 dm³

∴ 11.2 dm³ hydrogen will require oxygen = $\dfrac{22.4}{2 \times 22.4}$ x 11.2 =

= 5.6 dm³

Total volume of oxygen required = 44.8 + 5.6 = 50.4 dm³

Question 37

An Experiment showed that in a lead chloride solution, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chlorine? (Pb = 207; Cl = 35.5)

Numerical Problems

Answer:

Element	Mass (g)	Atomic mass	Moles	Simplest ratio
Pb	6.21	207	$\dfrac{6.21}{207}$ = 0.03	$\dfrac{0.03}{0.03}$ = 1
Cl	4.26	35.5	$\dfrac{4.26}{35.5 }$ = 0.12	$\dfrac{0.12}{0.03}$ = 4

Simplest ratio of whole numbers Pb : Cl = 1 : 4

Hence, empirical formula is PbCl₄

Question 38

10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na₂SO₄ + BaCl₂ ⟶ BaSO₄ + 2NaCl

Calculate the percentage of sodium sulphate in the original mixture.

Numerical Problems

Answer:

$\begin{matrix} \text{Na}_2\text{SO}_4 & + & \text{BaCl}_2 & \longrightarrow & \text{BaSO}_4 \downarrow & + & 2\text{NaCl} \\ 2(23) + 32 & & & & 137 + 32 \\ + 4(16) & & & & + 4(16) \\ = 46 + 32 & & & & = 137 + 32 \\ + 64 & & & & + 64 \\ = 142 \text{ g} & & & & = 233 \text{ g} \\ \end{matrix}$

233 g of BaSO₄ is obtained from 142 g of Na₂SO₄

6.99 g of BaSO₄ will be obtained from $\dfrac{142}{233}$ x 6.99 = 4.26 g of Na₂SO₄.

∴ In 10 g mixture $\dfrac{4.26}{10}$ x 100 = 42.6% Na₂SO₄ is present.

Hence, 42.6% Na₂SO₄ is present in the original mixture

Question 39

When heated, potassium permanganate decomposes according to the following equation:

2KMnO₄ ⟶ K₂MnO₄ + MnO₂ + O₂

(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.

(b) Given that the molecular mass of potassium permanganate is 158. What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres)

Numerical Problems

Answer:

2KMnO₄ ⟶ K₂MnO₄ + MnO₂ + O₂

Loss in mass = 1.32 g = 1 lit of oxygen

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{1.32}{0.0825} \\[0.5em] = 16 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 16 = 32 g

Hence, relative molecular mass of oxygen is 32 g

(b) Molar mass of 2KMnO₄ = 2[39 + 55 + 4(16)] = 2[39 + 55 + 64] = 316 g

316 g of KMnO₄ gives oxygen = 24 litres

∴ 15.8 g of KMnO₄ will give

= $\dfrac{24}{316}$ x 15.8 = 1.2 L

Question 40

A flask contains 3.2 g of sulphur dioxide. Calculate the following:

(a) The moles of sulphur dioxide present in the flask.

(b) The number of molecules of sulphur dioxide present in the flask.

Numerical Problems

Answer:

(a) Molar mass of sulphur dioxide = 32 + 2(16) = 64 g

64 g of sulphur dioxide = 1 mole

So, 3.2 g = $\dfrac{1}{64}$ x 3.2 = 0.05 moles

(b) 1 mole of SO₂ = 6.02 × 10²³ molecules

So, in 0.05 moles, no. of molecules = 6.02 × 10²³ × 0.05 = 3 × 10²²

3.2 g of SO₂ will be occupied by volume

$\dfrac{22.4}{64}$ x 3.2 = 1.12 L

Question 41

The volume of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure .

(i) Which sample of gas contains the maximum number of molecules?

(ii) If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?

(iii) If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?

(iv) If the volume of A is actually 5.6 dm³ at S.T.P., calculate the number of molecules in the actual Volume of D at S.T.P. (Avogadro's number is 6 × 10²³).

(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N₂O)

Numerical Problems

Answer:

(i) Volume is directly proportional to the number of molecules, hence gas D will have maximum no. of molecules as its volume is maximum.

(ii) If number of molecules of gas A is doubled, the volume will also be doubled i.e. 2V.

(iii) Gay Lussac's Law is observed.

(iv) 1 mole contains 6 x 10²³ number of molecules and occupies 22.4 lit. vol.

Given, volume of 'A' is 5.6 dm³ at s.t.p.

∴ vol. of D will be 4 × 5.6 = 22.4 lit.
No. of molecules in 22.4 lit. of D = 6 x 10²³ (Avogadro no.)

(v) As D is 1 mole hence, mass of 1 mole of D (N₂O) = 2(14) + 16 = 28 + 16 = 44 g

Hence, mass of N₂O = 44 g

Question 42

The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na₂CO₃ = 106]

(i) NaCl + NH₃ + CO₂ + H₂O ⟶ NaHCO₃ + NH₄Cl

(ii) 2NaHCO₃ ⟶ Na₂CO₃ + H₂O + CO₂

Equations (1) and (2) are based on the production of 21.2 g. of sodium carbonate.

(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate

(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.

Numerical Problems

Answer:

$\begin{matrix} 2\text{NaHCO}_3 & \longrightarrow & \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{O} + \text{CO}_2 \\ 2 \text{mol.} & & 1 \text{mol.} \\ 2 \times 84 & & 106\text{ g.} \\ = 168\text{ g.} & & \end{matrix}$

106 g of Na₂CO₃ is obtained from 168 g of NaHCO₃

∴ 21.2 g. of Na₂CO₃ is obtained from $\dfrac{168}{106}$ x 21.2 = 33.6 g of NaHCO₃

(ii)

$\begin{matrix} \text{NaCl} & + & \text{NH}_3 & + & \text{CO}_2 & + & \text{H}_2\text{O} \\ & & & & 1\text{mol.} & \\ & & & & 22.4 \text{lit.} \\ \longrightarrow & \text{NaHCO}_3 & + & \text{NH}_4\text{Cl} \\ & 1 \text{mol.} \\ & 84\text{ g.} \end{matrix}$

84 g of NaHCO₃ is obtained from 22.4 lit of CO₂

∴ 33.6 g. of NaHCO₃ is obtained from $\dfrac{22.4}{84}$ x 33.6 = 8.96 lit.

Hence, 8.96 lit of CO₂ is required.

Question 43

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)

NH₄NO₃ ⟶ N₂O + 2H₂O

(i) What volume of di nitrogen oxide is produced at the same time as 8.96 litres of steam.

(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam [Relative molecular mass of NH₄NO₃ is 80]

(iii) Determine the percentage of oxygen in ammonium nitrate [O = 16]

Numerical Problems

Answer:

$\begin{matrix} \text{NH}_4\text{NO}_3 &\xrightarrow{\Delta} & \text{N}_2\text{O} & + & 2\text{H}_2\text{O} \\ 1\text{ vol.} = 80 \text{g} & & 1 \text{ vol.} & & 2\text{ vol.} \\ \end{matrix}$

(i) Given,

1 vol. of di nitrogen produced at the same time as 8.96 litres of steam (2 vol)

Hence, Vol of di nitrogen = $\dfrac{8.96}{2}$ = 4.48 lit.

Hence, volume of di nitrogen oxide produced = 4.48 lit.

(ii) 2 vol = (2 x 22.4) lit steam is produced from 80 g NH₄NO₃

∴ 8.96 lit of steam will be produced by $\dfrac{80}{2 \times 22.4}$ x 8.96 = 16 g

Hence, 16 g of ammonium nitrate is required to be heated.

(iii) % of oxygen in ammonium nitrate = $\dfrac{3 \times 16}{80}$ x 100 = 60 %

Question 44

Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide? The equation for the reaction is:

3CuO + 2NH₃ ⟶ 3Cu + 3H₂O + N₂

Numerical Problems

Answer:

$\begin{matrix} 3\text{CuO} & + & 2\text{NH}_3 \\ 3\text{ mol.} & & 2 \text{ mol.} \\ = 3 \times 80 \text{ g.} & & = 2 \times 22.4 \text{ lit.} \\ = 240 \text{ g.} & & = 44.8\text{ lit.} \\ \end{matrix}$

$\longrightarrow 3\text{Cu} + 3\text{H}_2\text{O} + \text{N}_2$

240 g of CuO is reduced by 44.8 lit of NH₃

∴ 120 g of CuO is reduced by $\dfrac{44.8}{240}$ x 120 = 22.4 lit.

Hence, 22.4 lit of NH₃ is required

Question 45

(a) Calculate the number of moles and the number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?

(b) What is the vapour density of ethylene?

Numerical Problems

Answer:

Gram molecular mass of C₂H₄
= 2(12) + 4(1)
= 24 + 4 = 28 g

As,

28 g of C₂H₄ = 1 mole
∴ 1.4 g of C₂H₄ = $\dfrac{1}{28}$ x 1.4 = 0.05 moles

1 mole = 6 × 10²³ molecules
∴ 0.05 moles = 6 × 10²³ x 0.05 = 3 x 10²² molecules

Hence, no. of moles is 0.05 and no. of molecules is 3 x 10²².

Vol. occupied by 1 mole = 22.4 lit
∴ Vol. occupied by 0.05 moles = 22.4 x 0.05 = 1.12 lit.

Hence, vol. occupied is 1.12 lit

(b)
$\text{Vapour density} = \dfrac{\text{Molecular weight}}{2} = \dfrac{28}{2} = 14$

Hence, vapour density is 14

Question 46

(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na₃AlF₆) correct to the nearest whole number.

(F = 19; Na =23; Al = 27)

(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows :

2CO + O₂ ⟶ 2CO₂

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Numerical Problems

Answer:

Molecular weight of sodium aluminium fluoride (Na₃AlF₆)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 69 g of Na

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{69}{210}$ x 100 = 32.85% = 33%

Hence, percentage of sodium in sodium aluminium fluoride (Na₃AlF₆) is 33%

(b)

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

1 mole of O₂ has volume = 22400 ml

Volume of oxygen used by 2 × 22400 ml CO = 22400 ml

∴ Vol. of O₂ used by 560 ml CO

= $\dfrac{22400}{2 × 22400}$ x 560

= 280 ml

Volume of CO₂ formed by 2 × 22400 ml CO = 2 x 22400 ml

∴ Vol. of CO₂ formed by by 560 ml CO

= 560 ml

Question 47(a)

Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm³ of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.

2C₂H₂(g) + 5O₂(g) ⟶4CO₂(g) + 2H₂O(g)

Numerical Problems

Answer:

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 \longrightarrow & 4\text{CO}_2 & + &2\text{H}_2\text{O} \\ 2\text{ mol.} & & 5 \text{ mol.} & 4\text{ mol.}\ \end {matrix}$ According to Gay-Lussac's law,

2 volume of acetylene requires 5 volume of oxygen

1 volume of acetylene requires $\dfrac{5}{2}$ = 2.5 volume of oxygen

∴ 200 cm³ of acetylene requires $\dfrac{2.5}{1}$ x 200

= 500 cm³ of oxygen

2 volume of acetylene produces 4 volume of CO₂

1 volume of acetylene produces $\dfrac{4}{2}$ = 2 volume of CO₂

∴ 200 cm³ produces $\dfrac{2}{1}$ x 200 = 400 cm³ of CO₂

Question 47(b)

A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37. (N = 14, H = 1).

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Nitrogen	100 - 12.5 = 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 NH₂

Empirical formula weight = 14 + 2(1) = 16

Relative molecular mass = 37

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{37}{16} = 2.31 = 2$

∴ Molecular formula = n[E.F.] = 2[NH₂] = N₂H₄

Question 47(c)

A gas cylinder contains 24 × 10²⁴ molecules of nitrogen gas. If Avogadro's number is 6 × 10²³ and the relative atomic mass of nitrogen is 14, calculate :

(1) Mass of nitrogen gas in the cylinder.

(2) Volume of nitrogen at STP in dm³

Numerical Problems

Answer:

(1) Gram molecular mass of N₂ = 2(14) = 28 g

6 × 10²³ molecules of N₂ weighs 28 g

∴ 24 × 10²⁴ molecules will weigh $\dfrac{28}{6 \times 10^{23}}$ x 24 × 10²⁴ = 1120 g

(2) 6 × 10²³ molecules of N₂ occupies 22.4 dm³

∴ 24 × 10²⁴ molecules of N₂ will occupy $\dfrac{22.4}{6 \times 10^{23}}$ x 24 × 10²⁴ = 896 dm³

Question 47(d)

Commercial sodium hydroxide weighing 30g has some sodium chloride in it. The Mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is

NaCl + AgNO₃ ⟶ AgCl + NaNO₃

(Relative molecular mass of NaCl = 58; AgCl = 143)

Numerical Problems

Answer:

$\underset{58 \text{g}}{\text{NaCl}} + \text{AgNO}_3 \longrightarrow \underset{143 \text{ g}}{\text{AgCl}} + \text{NaNO}_3$

(i) 143 g AgCl is formed by 58 g NaCl

∴ 14.3 g of AgCl will be formed by $\dfrac{58}{143}$ x 14.3 = 5.8 g

Percentage of NaCl = $\dfrac{5.8}{30}$ x 100 = 19.33%

Hence, percentage of NaOH is 19.33%

Question 47(e)

A certain gas 'X' occupies a volume of 100 cm³ at S.T.P. and weighs 0.5 g. Find its relative molecular mass

Numerical Problems

Answer:

100 cm³ weighs 0.5 g

∴ 22400 cm³ weighs $\dfrac{0.5}{100}$ x 22400 = 112 g

Hence, relative molecular mass of the gas = 112 g.

Question 48(a-i)

LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

Numerical Problems

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. C₃H₈ produces carbon dioxide = 3 Vol

So, 6 litres C₃H₈ will produce carbon dioxide = 3 x 6 = 18 litres

2 Vol. C₄H₁₀ produces carbon dioxide = 8 Vol

So, 4 litres C₄H₁₀ will produce carbon dioxide = $\dfrac{8}{2}$ x 4 = 16 litres

Hence, 34 (i.e., 18 + 16) litres of CO₂ is produced.

Question 48(a-ii)

Calculate the percentage of nitrogen and oxygen in ammonium nitrate. (Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16).

Numerical Problems

Answer:

Given,

molecular mass of ammonium nitrate = 80

mass of nitrogen in 1 mole of ammonium nitrate (NH₄NO₃) is 2(14) = 28 g

percentage of nitrogen in ammonium nitrate = $\dfrac{28}{80}$ x 100 = 35%

mass of oxygen in 1 mole of ammonium nitrate (NH₄NO₃) is 3(16) = 48 g

percentage of oxygen in ammonium nitrate = $\dfrac{48}{80}$ x 100 = 60%

Hence, percentage of nitrogen is 35% and percentage of oxygen is 60%

Question 48(b)

4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.

(i) Write the equation for the reaction.

(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)

(iii) What is the volume of carbon dioxide liberated at STP?

(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).

(v) How many moles of HCl are used in this reaction?

Numerical Problems

Answer:

(i) CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

(ii) Given,

1 mole of CaCO₃ = molecular mass of CaCO₃ = 100 g

∴ 4.5 moles of CaCO₃ weighs $\dfrac{100}{1}$ x 4.5 = 450 g

(iii) 1 mole of CaCO₃ produces 1 mole of CO₂ and 1 mole occupies 22.4 l of volume.

∴ 4.5 moles of CaCO₃ will produce 4.5 moles of CO₂ and 4.5 moles will occupy 22.4 x 4.5 = 100.8 L

(iv) 1 mole CaCO₃ produces 111 g. CaCl₂

∴ 4.5 moles of CaCO₃ will produce = 111 x 4.5 = 499.5 g.

(v)

$\begin{matrix}\text{CaCO}_3 & : & \text{HCl} & \\ \text{ 1 mole} & : & 2 \text{ moles} \\ \text{ 4.5 mole} & : & x \text{ moles} \\ \end{matrix}$

∴ number of moles of HCl used = 2 x 4.5 = 9 moles.

Hence, moles of HCl used = 9 moles.

Question 49(a)

i. Calculate the volume of 320 g of SO₂ at STP. (Atomic mass: S = 32 and O = 16).

ii. State Gay-Lussac's Law of combining volumes

iii. Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈). (Atomic mass: C = 12, O = 16, H = I, Molar Volume = 22.4 dm³ at stp.)

Numerical Problems

Answer:

(i) Gram molecular mass of SO₂ = 32 + 2(16) = 32 + 32 = 64 g

64 g of SO₂ occupy 22.4 lit of vol.

320 g of SO₂ will occupy = $\dfrac{22.4}{64}$ x 320 = 112 lit.

Hence, volume of 320 g of SO₂ = 112 lit.

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

(iii)

$\begin{matrix} \text{C}_3\text{H}_8 & + &5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 3(12) + 8(1) & & 5 \text{ vol} \\ = 44 \text{ g}& & 5 (22.4) \text{ lit} \\ \end{matrix}$

(i) 44 g propane requires 5 x 22.4 lit of oxygen

∴ 8.8 g of propane will require $\dfrac{5 \times 22.4}{44}$ x 8.8 = 22.4 lit.

Hence, 22.4 lit of Oxygen is required.

Question 49(b-i)

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula. (Atomic mass : C = 12, H = 1, Br = 80)

Numerical Problems

Answer:

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
C	12.67	12	$\dfrac{12.67}{12}$ = 1.05	$\dfrac{1.05}{1.05}$ = 1
H	2.13	1	$\dfrac{2.13}{1}$ = 2.13	$\dfrac{2.13}{1.05}$ = 2.02 = 2
Br	85.11	80	$\dfrac{85.11}{80}$ = 1.06	$\dfrac{1.06}{1.05}$ = 1

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

Hence, empirical formula is CH₂Br

Empirical formula weight = 12 + 2(1) + 80 = 94

Vapour density (V.D.) = 94

Molecular weight = 2 x V.D. = 2 x 94

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 94}{94} = 2$

∴ Molecular formula = n[E.F.] = 2[CH₂Br] = C₂H₄Br₂

Question 49(b-ii)

Calculate the mass of:

10²² atoms of sulphur.
0.1 mole of carbon dioxide.

Numerical Problems

Answer:

(i) gram molecular mass of S = 32

6 × 10²³ atoms weigh = 32 g

10²² atoms will weigh = $\dfrac{32}{6 \times 10^{23}}$ x 10²² = 0.533 g

(ii) 1 mole of carbon dioxide weighs = C + 2(O) = 12 + 2(16) = 12 + 32 = 44 g

∴ 0.1 mole of carbon dioxide weighs = 44 x 0.1 = 4.4 g.

Question 50(a)

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5HNO₃ [conc.] ⟶ H₃PO₄ + H₂O + 5NO₂.

If 9.3 g of phosphorus was used in the reaction, calculate :

(i) Number of moles of phosphorus taken.

(ii) The mass of phosphoric acid formed.

(iii) The volume of nitrogen dioxide produced at STP.

Numerical Problems

Answer:

$\begin{matrix} \text{P} & + & 5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & \text{H}_2\text{O} & + & 5\text{NO}_2 \\ 31 \text{ g} & & [conc.] & & 3(1) + 31 \\ & & & & + 4(16) \\ & & & & = 98 \text{ g} \\ \end{matrix}$

(i)

31 g of P = 1 mole

∴ 9.3 g of P = $\dfrac{1}{31}$ x 9.3 = 0.3 moles.

Hence, 0.3 moles of phosphorous was taken for the reaction.

(ii)

31 g of P forms 98 g of phosphoric acid

∴ 9.3 g will form $\dfrac{ 98}{31}$ x 9.3 = 29.4 g.

Hence, 29.4 g. of phosphoric acid is formed

(iii)

31 g of P produces 5 vol = 5 x 22.4 lit.

∴ 9.3 g will produce = $\dfrac{5 \times 22.4}{31}$ x 9.3 = 33.6 lit.

Question 50(b-i)

67.2 litres of hydrogen combines with 44.8 litres of nitrogen to form ammonia :

N₂(g) + 3H₂(g) ⟶ 2NH₃(g).

Calculate the vol. of ammonia produced. What is the substance, if any, that remains in the resultant mixture ?

Numerical Problems

Answer:

[By Lussac's law]

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

To calculate the volume of ammonia gas formed.

$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 67.2 & : & x \end{matrix}$

$\therefore \dfrac{2}{3} \times 67.2 = 44.8 \text{ lit}$

Hence, volume of NH₃ formed is 44.8 lit.

We know,

$\begin{matrix}\text{H}_2 & : & \text{N}_2 \\ 3 & : & 1 \\ 67.2 & : & x \end{matrix}$

$\therefore \dfrac{1}{3} \times 67.2 = 22.4 \text{ lit}$

∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.

Question 50(b-ii)

The mass of 5.6 dm³ of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

Numerical Problems

Answer:

5.6 dm³ weighs 12 g

∴ 22.4 dm³ weighs $\dfrac{12}{5.6}$ x 22.4 = 48 g

Hence, relative molecular mass of the gas = 48 amu

Question 50(b-iii)

Find the total percentage of Magnesium in magnesium nitrate crystals, Mg(NO₃)₂.6H₂O.

(Mg = 24, N = 14, 0 = 16 and H = 1)

Numerical Problems

Answer:

Molecular weight of Mg(NO₃)₂.6H₂O = 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6[18]
= 24 + 2(62) + 108
= 24 + 124 + 108 = 256 g

256 g of magnesium nitrate crystals contains 24 g of magnesium

∴ 100 g of magnesium nitrate crystals will contain = $\dfrac{24}{256}$ x 100 = 9.375% = 9.38%

Hence, percentage of magnesium in magnesium nitrate crystals is 9.38%

Question 51(a-i)

What volume of oxygen is required to burn completely 90 dm³ of butane under similar conditions of temperature and pressure?

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

Numerical Problems

Answer:

[By Lussac's law]

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

To calculate the volume of oxygen

$\begin{matrix}\text{C}_4\text{H}_{10} & : & \text{O}_2 \\ 2 & : & 13 \\ 90 & : & x \end{matrix}$

∴

$\dfrac{13}{2} \times 90 = 585 \text{ dm}^3$

Hence, vol of oxygen = 585 dm³

Question 51(a-ii)

The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?

Numerical Problems

Answer:

Given, V.D. = 8

Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.

16 g occupies 22.4 lit.

∴ 24 g. will occupy = $\dfrac{22.4}{16}$ x 24 = 33.6 lit.

Hence, volume occupied by gas = 33.6 lit.

Question 51(a-iii)

A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

Numerical Problems

Answer:

According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N₂ = Number of molecules of H₂ = X

Question 51(b)

O₂ is evolved by heating KClO3 using MnO₂ as a catalyst.

2KClO₃ $\xrightarrow{\text{MnO}_2}$ 2KCl + 3O₂

(i) Calculate the mass of KClO₃ required to produce 6.72 litre of O₂ at STP.

(atomic masses of K = 39, Cl = 35.5, 0 = 16).

(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

(iii) Calculate the volume occupied by 0.01 mole of CO₂ at STP.

Numerical Problems

Answer:

$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{MnO_2} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & 2[39 & & 3(22.4) \\ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \\ = 245 \text{ g} & & = 149 \text{g} \\ \end{matrix}$

(i)

67.2 lit. of O₂ is produced by 245 g of KClO₃

∴ 6.72 lit of O₂ will be obtained from $\dfrac{245}{67.2}$ x 6.72 = 24.5 g

(ii)

22.4 lit = 1 mole
∴ 6.72 lit = $\dfrac{1}{22.4}$ x 6.72 = 0.3 moles

1 mole = 6.023 x 10²³ molecules
∴ 0.3 moles = 0.3 x 6.023 x 10²³ = 1.806 x 10²³ molecules.

(iii)

Volume occupied by 1 mole of CO₂ at STP = 22.4 litres

So, volume occupied by 0.01 mole of CO₂ at STP
= 22.4 × 0.01
= 0.224 litres

Question 52(a-i)

Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:

2CO₂H₂ + 5O₂ ⟶ 4CO₂ + 2H₂O

What volume of ethyne gas at s.t.p. is required to produce 8.4 dm³ of carbon dioxide at STP ? [H = 1, C = 12, 0 = 16]

Numerical Problems

Answer:

[By Lussac's law]

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

To calculate the volume of ethyne gas

$\begin{matrix}\text{CO}_2 & : & \text{C}_2\text{H}_2 \\ 4 & : & 2 \\ 8.4 & : & x \end{matrix}$

∴

$\dfrac{2}{4} \times 8.4 = 4.2 \text{ dm}^3$

Hence, volume of ethyne gas required = 4.2 dm³.

Question 52(a-ii)

A compound made up of two elements X and Y has an empirical formula X₂ Y. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density (V.D.) 25, find it's molecular formula.

Numerical Problems

Answer:

Empirical formula is X₂Y

Empirical formula weight = 2(10) + 5 = 25

Vapour density (V.D.) = 25

Molecular weight = 2 x V.D. = 2 x 25

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 25}{25} = 2$

∴ Molecular formula = n[E.F.] = 2[X₂Y] = X₄Y₂

Question 52(b)

A cylinder contains 68 g of Ammonia gas at STP

(i) What is the volume occupied by this gas?

(ii) How many moles of ammonia are present in the cylinder?

(iii) How many molecules of ammonia are present in the cylinder?

Numerical Problems

Answer:

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(i) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.

17 g. occupies 22.4 lit. of vol.

∴ 68 g will occupy = $\dfrac{22.4}{17}$ x 68 = 89.6 lit.

Hence, volume occupied by this gas = 89.6 lit.

(ii) 17 g = 1 mole

∴ 68 g = $\dfrac{1}{17}$ x 68 = 4 moles.

1 mole = 6 × 10²³

∴ 4 moles = 4 x 6.023 × 10²³ molecules.

Hence, Moles = 4

(iii) molecules = 4 x 6.023 × 10²³ = 2.4 x 10²⁴ molecules

Question 53

In an Experiment, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of the resulting compound?

Numerical Problems

Answer:

Element	Mass (g)	Atomic mass	Moles	Simplest ratio
Pb	6.21	207	$\dfrac{6.21}{207}$ = 0.03	$\dfrac{0.03}{0.03}$ = 1
Cl	4.26	35.5	$\dfrac{4.26}{35.5 }$ = 0.12	$\dfrac{0.12}{0.03}$ = 4

Simplest ratio of whole numbers Pb : Cl = 1 : 4

Hence, empirical formula is PbCl₄

Question 54

An organic compound with vapour density 94 contains H = 2.13%, and Br = 85.11%. Find the molecular formula

Numerical Problems

Answer:

Element	% composition	Atomic weight	Relative no. of atoms	Simplest ratio
C	12.67	12	$\dfrac{12.67}{12}$ = 1.05	$\dfrac{1.05}{1.05}$ = 1
H	2.13	1	$\dfrac{2.13}{1 }$ = 2.13	$\dfrac{2.13}{1.05}$ = 2.02=2
Br	85.11	80	$\dfrac{85.11}{80 }$ = 1.06	$\dfrac{1.06}{1.05}$ = 1

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

Hence, empirical formula is CH₂Br

Empirical formula weight = 12 + 2(1) + 80 = 94

Vapour density (V.D.) = 94

Molecular weight = 2 x V.D. = 2 x 94

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 94}{94} = 2$

∴ Molecular formula = n[E.F.] = 2[CH₂Br] = C₂H₄Br₂

Question 55

A compound A consists of 4.8% carbon and 95.2% bromine by mass. Its vapour density is 252. Find its :

(i) empirical formula

(ii) molecular formula

Numerical Problems

Answer:

Element	% composition	Atomic weight	Relative no. of atoms	Simplest ratio
Carbon	4.8	12	$\dfrac{4.8}{12}$ = 0.4	$\dfrac{0.4}{0.4}$ = 1
Bromine	95.2	80	$\dfrac{95.2}{80 }$ = 1.19	$\dfrac{1.19}{0.4}$ = 2.9=3

Simplest ratio of whole numbers = C : Br = 1 : 3

Hence, empirical formula is CBr₃

(i) Empirical formula weight = 12 + 3(80) = 252 g

Molecular weight = 2 x V.D. = 2 x 252 = 504

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{504}{252} = 2$

Molecular formula = n[E.F.] = 2[CBr₃] = C₂Br₆

∴ Molecular formula = C₂Br₆