Gay Lussac's Law — Avogadro's Law — Mole Concept

Lussac's Law

Question 1

Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with 6 litres of hydrogen. All volumes measured at s.t.p.

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.

$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 6 & : & x \end{matrix}$

Therefore,

$\dfrac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ lts}$

Hence, volume of ammonia gas formed is 4 lts

Question 2

2500 cc of oxygen was burnt with 600 cc of ethane [C₂H₆]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.

Answer

[By Lussac's law]

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

To calculate the volume of unused oxygen.

$\begin{matrix} \text{C}_2\text{H}_6 & : &\text{O}_2 \\ 2 & : & 7 \\ 600 & : & x \end{matrix}$

Therefore,

$\dfrac{7}{2} \times 600 = x \\[0.5em] \Rightarrow x = 2100 \text{cc}$

Therefore, volume of unused oxygen = 2500 - 2100 = 400 cc

To calculate the volume of carbon dioxide formed

$\begin{matrix} \text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 & : & 4 \\ 600 & : & x \end{matrix}$

Therefore,

$\dfrac{4}{2} \times 600 = x \\[0.5em] \Rightarrow x = 1200 \text{cc}$

Therefore, volume of carbon dioxide formed is 1200 cc

Question 3

20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure. What will be the volume and composition of the mixture of the gases when they are cooled to room temperature.

Answer

Given,

20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide

[By Lussac's law]

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

To calculate the amount of CO₂ produced,

$\begin{matrix} \text{CO}& : & \text{CO}_2 & \\ 2 \text{ vol.} & : & 2 \text{ vol.} & \\ 10 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$

Therefore, CO₂ produced is 10 ml

To calculate the amount of O₂ used,

$\begin{matrix} \text{CO}& : & \text{O}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \\ 10 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$

$\dfrac{1}{2} \times 10 = x \\[0.5em] \Rightarrow x = 5 \text{ ml}$

From relation,

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

$\begin{matrix} \text{H}_2 & : & \text{O}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \\ 20 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$

$\dfrac{1}{2} \times 20 = x \\[0.5em] \Rightarrow x = 10 \text{ ml}$

Therefore, total vol. of oxygen used = 10 + 5 = 15 ml

Hence, oxygen left = 20 - 15 = 5 ml

Therefore, oxygen left is 5 ml and CO₂ produced is 10 ml.

Question 4

224 cm³ of ammonia undergoes catalytic oxidation in presence of Pt to given nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measured at room temperature and pressure.

Answer

[By Lussac's law]

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

To calculate the volume of oxygen required.

$\begin{matrix} \text{NH}_3 & : & \text{O}_2 & \\ 4 & : & 5 \\ 224 & : & x \end{matrix}$

Therefore,

$\dfrac{5}{4} \times 224 = x \\[0.5em] \Rightarrow x = 280 \text{ cm}^3$

Therefore, volume of oxygen required is 280 cm³.

Question 5

Acetylene [C₂H₂] burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn 50 cm³ of acetylene. [Assume air contains 20% oxygen].

Answer

[By Lussac's law]

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

To calculate the volume of air required,

$\begin{matrix} \text{C}_2 \text{H}_2 & : & \text{O}_2 & \\ 2 & : & 5 \\ 50 & : & x \end{matrix}$

Therefore, volume of oxygen (x),

$\dfrac{5}{2} \times 50 = x \\[0.5em] \Rightarrow x = 125 \text{ cm}^3$

When oxygen is 20% then air is 100%
Therefore when, oxygen is 125 cm³ then air is

$\dfrac{100}{20} \times 125 = x \\[0.5em] \Rightarrow x = 625 \text{ cm}^3$

Therefore, volume of air required is 625 cm³.

Question 6

On igniting a mixture of acetylene [C₂H₂] and oxygen, 200 cm³ of CO₂ is collected at s.t.p. Calculate the volume of acetylene & O₂ at s.t.p. in the original mixture.

Answer

[By Lussac's law]

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

To calculate the volume of acetylene :

$\begin{matrix} \text{CO}_2 & : & \text{C}_2 \text{H}_2 & \\ 4 & : & 2 \\ 200 & : & x \end{matrix}$

Therefore,

$\dfrac{2}{4} \times 200 = x \\[0.5em] \Rightarrow x = 100 \text { cm}^3$

Hence, volume of acetylene in the original mixture is 100 cm³.

To calculate the volume of oxygen :

$\begin{matrix} \text{CO}_2 & : & \text{O}_2 & \\ 4 & : & 5 \\ 200 & : & x \end{matrix}$

Therefore,

$\dfrac{5}{4} \times 200 = x \\[0.5em] \Rightarrow x = 250 \text { cm}^3$

Hence, volume of oxygen in the original mixture is 250 cm³.

Question 7

Ammonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitable conditions. Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia formed if only 10% conversion has taken place.

Answer

[By Lussac's law]

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

To calculate the volume of ammonia formed when only 10% conversion takes place,

$\therefore 10\% \text{ of 2 lit} = \dfrac{10}{100} \times 2 = 0.2 \text{ lit}$

Hence, volume of ammonia formed is 0.2 lit or 20% or 1/5th of vol of N₂ and H₂

Question 8

100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temp. Calculate the composition of the resultant mixture. [Water gas contains CO and H₂ in equal ratio]

Answer

Given,

Vol of CO = 50 cc and H₂ = 50 cc [As Water gas contains CO and H₂ in equal ratio] and O₂ = 100 cc

[By Lussac's law]

$\underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{CO}}} \underset{:}{\text{\scriptsize{+}}} \underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{H}}_2} \underset{:}{\text{\scriptsize{+}}} \underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{O}}_2} \longrightarrow \underset{\text{\scriptsize{1 vol}}}{\text{\scriptsize{CO}}_2} \underset{:}{\text{\scriptsize{ +}}} \underset{\text{\scriptsize{1 vol}}}{\text{\scriptsize{ H}}_2 \text{\scriptsize{O}}}$

To calculate the amount of O₂ used,

$\begin{matrix} \text{CO}& : & \text{O}_2 & \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \\ 50 \text{ cc} & : & x \text{ cc} & \\ \end{matrix}$

As ratio between CO and O₂ is same so, O₂ used is 50 cc.

Hence, remaining O₂ = 100 - 50 = 50 cc.

To calculate the amount of CO₂ produced,

$\begin{matrix} \text{CO}& : & \text{CO}_2 & \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \\ 50 \text{ cc} & : & y \text{ cc} & \\ \end{matrix}$

1 vol of CO produces 1 vol. of CO₂
Hence, CO₂ produced = 50 cc.

Therefore, the resultant mixture has 50 cc of O₂ + 50 cc of CO₂

Mole Concept — Avogadro's Law and No.

Question 1

Calculate the mass of 2.8 litres of CO₂. [C = 12, O = 16]

Answer

[1 mole = 1 gm mol. wt. and occupies 22.4 lit at s.t.p.]

gm mol. wt. of CO₂ = 12 + (16 x 2) = 44 g

1 mole of CO₂ = 1 gm mol. wt.

22.4 lit of CO₂ has mass 44 g [s.t.p.]

Therefore, 2.8 lit of CO₂ has mass

$= \dfrac{44}{22.4} \times 2.8 = 5.5 \text{ g}$

Therefore, mass of 2.8 litres of CO₂ is 5.5 g

Question 2

Calculate the volume occupied by 53.5 g of Cl₂. [Cl = 35.5]

Answer

[1 mole = 1 gm mol. wt. and occupies 22.4 lit at s.t.p.]

gm mol. wt. of Cl₂ = (35.5 x 2) = 71 g

1 mole of Cl₂ = 1 gm mol. wt. and occupies 22.4 lit.

71 g of Cl₂ occupies 22.4 lit [s.t.p.]

∴ Vol. occupied by 53.5 g of Cl₂

$= \dfrac{22.4}{71} \times 53.5 = 16.87 \text{ lit}$

Therefore, volume occupied by 53.5 g of Cl₂ is 16.87 lit

Question 3

Calculate the number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]

Answer

gm mol. wt. of HCl = 1 + 35.5 = 36.5 g

At s.t.p.,

36.5 g of HCl = 6.023 x 10²³ number of molecules    [Avogadro's law]

Therefore, 109.5 g of HCl

$= \dfrac{6.023 \times 10^{23}}{36.5} \times 109.5 \\[0.5em] = \dfrac{109.5}{36.5} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]$

Hence, number of molecules in 109.5 g of HCl is 3 x 6.023 x 10²³

Question 4

Calculate the number of :

(i) molecules [S = 32]

(ii) atoms in 192 g. of sulphur. [S₈]

Answer

(i) 1 mole of any substance contains 6.023 x 10²³ number of molecules.

S₈ = 8 atoms = 8 x 32 = 256.

256 g = 6.023 x 10²³ number of molecules

So, 192 g will have

$= \dfrac{6.023 \times 10^{23}}{256} \times 192 \\[0.5em] = \dfrac{192}{256} \times 6.023 \times 10^{23} \\[0.5em] = 0.75 \times 6.023 \times 10^{23} \text { molecules} \\[0.5em]$

(ii) Gram atomic mass of S = 32 g

32 g of S has 6.023 x 10²³ number of atoms

Therefore, 192 g will have

$= \dfrac{6.023 \times 10^{23}}{32} \times 192 \\[0.5em] = \dfrac{192}{32} \times 6.023 \times 10^{23} \\[0.5em] = 6 \times 6.023 \times 10^{23} \text { atoms} \\[0.5em]$

Question 5

Calculate the mass of Na which will contain 6.023 × 10²³ atoms. [Na = 23]

Answer

As, a mole of atoms contain 6.023 × 10²³ atoms [Avogadro number] and has weight [mass] equal to gram atomic mass of the element.

Therefore, a mole of Na will contain 6.023 × 10²³ atoms and will have mass equal to gram atomic mass of Na = 23 g

Question 6

Calculate the no. of atoms of potassium present in 117 g. of K. [K = 39]

Answer

Gram atomic mass of K = 39

At s.t.p.,

39 g of K = 6.023 x 10²³ atoms of K    [Avogadro's law]

Therefore, 117 g of K

$= \dfrac{6.023 \times 10^{23}}{39} \times 117 \\[0.5em] = \dfrac{117}{39} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]$

Hence, number of atoms in 117 g of K = 3 x 6.023 x 10²³

Question 7

Calculate the number of moles and molecules in 19.86 g. of Pb(NO₃)₂. [Pb = 207, N = 14, O = 16]

Answer

Gram molecular mass of Pb(NO₃)₂
= Pb + 2 [N x 3(O)]
= Pb + 2N + 6O
= 207 + (2 x 14) + (6 x 16)
= 207 + 28 + 96 = 331

As,

331 g of Pb(NO₃)₂ = 1 mole
Therefore, 19.86 g

$= \dfrac{1}{331} \times 19.86 = 0.06 \text { moles} \\[0.5em]$

Hence, number of moles in 19.86 g. of Pb(NO₃)₂ = 0.06 moles

As,

1 mole of Pb(NO₃)₂ weighs 331 g and has 6.023 x 10²³ molecules

Therefore, 19.86 g of Pb(NO₃)₂ will have

$= \dfrac{ 6.023 \times 10^{23}}{331} \times 19.86 \\[0.5em] = \dfrac{19.86}{331} \times 6.023 \times 10^{23} \\[0.5em] = 0.06 \times 6.023 \times 10^{23} \text{molecules}$

Hence, number of molecules in 19.86 g. of Pb(NO₃)₂ = 0.06 x 6.023 x 10²³ moles

Question 8

Calculate the mass of an atom of lead [Pb = 202].

Answer

As,

1 mole of Pb weighs 202 g and has 6.023 x 10²³ atoms.

So, 6.023 x 10²³ atoms of Pb has mass = 202 g

Therefore, 1 atom of Pb will have mass

$= \dfrac{202}{6.023 \times 10^{23}} = 33.53 \times 10^{-23} \text{g}$

Hence, mass of an atom of lead is 33.53 x 10^-23 g

Question 9

Calculate the number of molecules in 1½ litres of water. [density of water 1.0 g./cc. ∴ mass of water = volume × density]

Answer

Given,

Vol of water = 1.5 lit = 1500 mL =1500 cc

Density of water 1.0 g./cc..

∴ Mass of water = volume × density

Hence, mass of water = 1500 x 1 = 1500 g

Gram molecular mass of water

= 2H + O

= (2 x 1) + 16 = 18 g

So, 18 g of water = 6.023 x 10²³ molecules

Therefore, 1500 g of water will have

$\dfrac{6.023 \times 10^{23}}{18} \times 1500 \\[0.5em] = \dfrac{1500}{18} \times 6.023 \times 10^{23} \\[0.5em] = 83.33 \times 6.023 \times 10^{23}$

Hence, the number of molecules in 1½ litres of water is 83.33 x 6.023 x 10²³ molecules.

Question 10

Calculate the gram-atoms in 88.75 g of chlorine [Cl = 35.5]

Answer

Gram atoms is the relative atomic mass of an element expressed in grams.

$\text{Gram atoms} = \dfrac{\text{Mass in grams}}{\text{Relative At. mass [At. wt.]}} \\[0.5em] = \dfrac{88.75}{35.5} \\[0.5em] = 2.5 \text { gram atoms} \\[0.5em]$

Hence, gram-atoms in 88.75 g of chlorine is 2.5 g. atoms

Question 11

Calculate the number of hydrogen atoms in 0.25 mole of H₂SO₄.

Answer

1 mole of hydrogen atom has 2 x 6.023 x 10²³ atoms.

∴ 0.25 mole will have

$= \dfrac{2 \times 6.023 \times 10^{23}}{1} \times 0.25 \\[0.5em] = 0.5 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, number of hydrogen atoms in 0.25 mole of H₂SO₄ = 0.5 x 6.023 x 10²³ particles

Question 12

Calculate the gram molecules in 21 g of nitrogen [N = 14]

Answer

Gram molecules is the relative molecular mass of a substance expressed in grams.

Relative molecular mass of N₂ = 2 x 14 = 28 g

$\text{Gram molecules} = \dfrac{\text{Mass in grams [of nitrogen]}}{\text{Rel. molecular mass [Mol. wt.]}} \\[0.5em] = \dfrac{21}{28} \\[0.5em] = 0.75 \text { gram molecules}$

Hence, gram molecules in 21 g of nitrogen = 0.75 gram molecules

Question 13

Calculate the number of atoms in 10 litres of ammonia [N = 14, H = 1]

Answer

1 mole of NH₃ = N + 3 H = 4 atoms = 4 x 6.023 x 10²³ atoms and occupies 22.4 lit at s.t.p.

If 22.4 lit of NH₃ has 4 x 6.023 x 10²³ atoms,

then, 10 lit will have

$= \dfrac{4 \times 6.023 \times 10^{23}}{22.4} \times 10 \\[0.5em] = \dfrac{40}{22.4} \times 6.023 \times 10^{23} \\[0.5em] = 1.786 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, number of atoms in 10 litres of ammonia = 1.786 x 6.023 x 10²³ atoms

Question 14

Calculate the number of atoms in 60 g of neon [Ne = 20]

Answer

Gram atomic mass of Ne = 20 g

As, 1 mole of Ne weighs 20 g and has 6.023 x 10²³ atoms

So, 60 g of Ne will have

$\dfrac{ 6.023 \times 10^{23}}{20} \times 60 \\[0.5em] = \dfrac{60}{20} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, number of atoms in 60 g of Ne = 3 x 6.023 x 10²³ atoms

Question 15

Calculate the number of moles of 'X' atoms in 93 g of 'X' [X is phosphorus = 31]

Answer

Gram atomic mass of X (phosphorus) = 31 g

As, 31 g of phosphorus = 1 mole

So, 93 g of phosphorus = $\dfrac{1}{31} \times 93 = 3 \text{ moles}$

Hence, number of moles in 93 g of X is 3 moles

Question 16

Calculate the volume occupied by 3.5 g of O₂ gas at 27 °C and 740 mm pressure. [O = 16]

Answer

Gram molecular mass of O₂ = 2 x 16 = 32 g

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

∴ 3.5 g of O₂ occupies = $\dfrac{22.4}{32} \times 3.5 = 2.45 \text{ lit.}$

Volume occupied by 3.5 g of O₂ gas at 27°C and 740 mm pressure:

Using the gas equation,

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

Substituting the values we get,

$\dfrac{760 \times 2.45}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 2.45 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{5,58,600}{2,02,020} \\ \\[0.5em] x = 2.76 \text{ lit}$

Hence, the volume occupied by 3.5 g of O₂ gas at 27°C and 740 mm pressure is 2.76 lit.

Question 17

Calculate the moles of sodium hydroxide contained in 160 g of it. [Na = 23, O = 16, H = 1]

Answer

Gram molecular mass of sodium hydroxide
= Na + O + H
= 23 + 16 + 1
= 40 g

As, 40 g of sodium hydroxide = 1 mole

∴ 160 g of sodium hydroxide = $\dfrac{ 1}{40} \times 160 = 4 \text{ moles}$

Hence, number of moles in 160 g of sodium hydroxide is 4

Question 18

Calculate the weight in g. of 2.5 moles of ethane [C₂H₆]. [C = 12, H = 1]

Answer

Gram molecular mass of C₂H₆ = (2 x 12) + (6 x 1)
= 24 + 6
= 30 g

1 mole of C₂H₆ weighs 30 g

∴ 2.5 moles weighs = $\dfrac{ 30}{1} \times 2.5 = 75 \text{ g}$

Hence, weight of 2.5 moles of ethane [C₂H₆] is 75 g.

Question 19

Calculate the molecular weight of 2.6 g of a gas which occupies 2.24 lits. at 0°C and 760 mm press.

Answer

Given,

2.24 lit volume of gas weighs 2.6 g

∴ 22.4 lit volume of gas weighs = $\dfrac{2.6}{2.24} \times 22.4 = 26 \text{ g}$

As 22.4 lit of gas at s.t.p. weighs 26 g
∴ Molecular weight = 26 g

Question 20

Calculate the gram atoms in 46 g of sodium [Na = 23]

Answer

Gram atoms is the relative atomic mass of an element expressed in grams.

$\text{Gram atoms} = \dfrac{\text{Mass in grams [of sodium]}}{\text{Rel. atomic mass [At. wt.]}} \\[0.5em] = \dfrac{46}{23} = 2 \text { gram atoms}$

Hence, gram-atoms in 46 g of sodium is 2 g. atoms

Question 21

Calculate the number of moles of KClO₃ that will be required to give 6 moles of oxygen.

Answer

2KClO₃ ⟶ 2KCl + 3O₂

$\begin{matrix} \text{O}_2 & : & \text{KClO}_3 & \\ 3 & : & 2 \\ 6 & : & x \end{matrix}$

$\dfrac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ moles}$

Hence, 4 moles of KClO₃ are required to give 6 moles of oxygen.

Question 22

Calculate the weight of the substance if it's molecular weight is 70 and in the gaseous form occupies 10 lits. at 27°C and 700 mm pressure.

Answer

Using the gas equation,

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

Substituting the values we get,

$\dfrac{700 \times 10}{300} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{700 \times 10 \times 273}{300 \times 760 } \\[0.5em] x = \dfrac{1911}{228 } \\[0.5em] x = 8.38 \text{ lit}$

1 gram molecular weight of the gas occupies 22.4 lit. at s.t.p.

∴ 70 g of the gas occupies 22.4 lit. at s.t.p.

If $y$ g of the gas occupies 8.38 lits, then

$y = \dfrac{70}{22.4} \times 8.38 \\[0.5em] = 26.18 \text{ g}$

Hence, weight of substance is 26.18 g

Question 23

State which has higher number of moles : 5 g. of N₂O or 5 g. of NO [N = 14, O = 16]

Answer

Gram molecular mass of N₂O
= (2 x 14) + 16
= 44 g

44 g of N₂O = 1 mole

∴ 5 g of N₂O = $\dfrac{1}{44} \times 5 = 0.11$ moles

Gram molecular mass of NO

= 14 + 16
= 30 g

30 g of NO = 1 mole

∴ 5 g of NO = $\dfrac{1}{30} \times 5 = 0.16$ moles

Hence, 5 g. of NO has higher moles than 5 g. of N₂O

Question 24

State which has higher mass : 1 mole of CO₂ or 1 mole of CO [C = 12, O = 16]

Answer

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

1 mole of CO₂ = 44 g

Gram molecular mass of CO
= 12 + 16
= 28 g

1 mole of CO = 28 g

Therefore, 1 mole of CO₂ has higher mass.

Question 25

State which has higher no. of atoms : 1 g of O₂ or 1 g of Cl₂ [O = 16, Cl = 35.5]

Answer

Molecular mass of O₂ = 2 x 16 = 32 g

32 g of O₂ = 2 x 6.023 x 10²³ atoms

∴ 1 g of O₂

$= \dfrac{2 \times 6.023 \times 10^{23}}{32} \\[0.5em] = 0.0625 \times 6.023 \times 10^{23} \text{ atoms}$

Molecular mass of Cl₂ = 2 x 35.5 = 71 g

71 g of Cl₂ = 2 x 6.023 x 10²³ atoms

∴ 1 g of Cl₂

$= \dfrac{2 \times 6.023 \times 10^{23}}{71} \\[0.5em] = 0.0281 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, 1 g of O₂ has more number of atoms.

Vapour Density And Molecular Weight

Question 1

500 ml. of a gas 'X' at s.t.p. weighs 0.50 g. Calculate the vapour density and molecular weight of the gas. [1 lit. of H₂ at s.t.p. weighs 0.09 g].

Answer

Given, 500 ml. of gas 'X' at s.t.p. weighs 0.50 g

Therefore, 1000 ml of gas 'X' at s.t.p. will weigh (0.50 x 2) g

Vapour density of gas X =

$\dfrac{\text{Wt. of 1000 ml of gas at s.t.p.}}{\text{Wt. of 1000 ml of H}_2 \text{ at s.t.p.}} \\[0.5em] = \dfrac{0.50 \times 2}{0.09} \\[0.5em] = 11.1$

Hence, vapour density of gas is 11.11

Molecular weight = 2 x Vapour density
= 2 x 11.1
= 22.2 g

Hence, molecular weight of gas is 22.2 g

Question 2

A gas cylinder holds 85 g of a gas 'X'. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure. Calculate the molecular weight of 'X'.

Answer

Vapour density of gas X =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{85}{8.5} \\[0.5em] = 10$

Molecular weight = 2 x Vapour density
= 2 x 10 = 20 g

Hence, molecular weight of gas X is 20 g

Question 3

Calculate the relative molecular mass [molecular weight] of 290 ml. of a gas 'A' at 17 °C and 1520 mm pressure which weighs 2.73 g at s.t.p. [1 litre of hydrogen at s.t.p. weighs 0.09 g.]

Answer

Convert the volume to s.t.p. using gas equation

Using the gas equation,

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

Substituting the values we get,

$\dfrac{1520 \times 290}{290} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{1520 \times 273}{760} \\[0.5em] x = 546 \text{ ml}$

546 ml of gas at s.t.p. weighs 2.73 g

∴ Wt. of 1000 ml of gas = $\dfrac{2.73}{546} \times 1000 = 5 \text { g}$

Vapour density of gas X =

$\dfrac{\text{Wt. of 1 lit of gas X}}{\text {Wt. of 1 lit of H}_2} \\[0.5em] = \dfrac{5}{0.09} \\[0.5em] = 55.55$

Hence, vapour density of gas is 55.55

Molecular weight = 2 x Vapour Density
= 2 x 55.555
= 111.11 g

Hence, molecular weight of gas is 111.11 g

Question 4

State the volume occupied by 40 g of a hydrocarbon – CH₄ at s.t.p. if it's V.D. is 8.

Answer

Vapour density of hydrocarbon = 8

∴ Molecular weight = 2 x Vapour density
= 2 x 8 = 16 g

16 g of hydrocarbon at s.t.p. occupy 22.4 lit

∴ 40 g of hydrocarbon will occupy = $\dfrac{22.4}{16} \times 40 = 56 \text{ lit.}$

Hence, volume occupied by 40 g of a hydrocarbon (CH₄) at s.t.p. is 56 lit.

Question 5

Calculate the atomicity of a gas X [at. no. 35.5] whose vapour density is equal to it's relative atomic mass.

Answer

Given,

Vapour density = relative atomic mass = 35.5

Molecular weight = 2 x Vapour Density = 2 x 35.5 = 71 g

Atomicity is the number of atoms present in one molecule of that element.

Number of atoms

$= \dfrac{\text{Molecular weight}}{\text{Atomic weight}} \\[0.5em] = \dfrac{\text{71}}{\text{35.5}} \\[0.5em] = 2$

Hence, atomicity of a gas X is 2

Question 6

Calculate the relative molecular mass and vapour density of methyl alcohol [CH₃OH] if 160 g. of the alcohol on vaporization has a volume of 112 litres at s.t.p.

Answer

Given,

Weight of 112 lit of CH₃OH = 160 g [at s.t.p.]

∴ Weight of 1 lit of CH₃OH = $\dfrac{160}{112}$ = 1.4286 g

Weight of 1 lit of H₂ = 0.09 g [at s.t.p.]

Under similar temperature and pressure,

Vapour density of gas methyl alcohol =

$\dfrac{\text{Wt. of certain vol. of gas}}{\text {Wt. of same vol. of H}_2} \\[0.5em] = \dfrac{1.4286}{0.09} \\[0.5em] = 15.87 \approx 16$

Molecular weight = 2 x Vapour density
= 2 x 15.87 = 31.74 g ≈ 32 g

Hence, molecular weight of methyl alcohol is 32 g and vapour density of methyl alcohol is 16