Solved Sample Paper 2 — Question 21

(a) Given, 40% of air contains oxygen

∴ 40% of 1000 cm³ = $\dfrac{40}{100}$ x 1000 = 400 cm³

[By Lussac's law]

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

To calculate the volume of propane consumed :

$\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 400 \text{ cm.}^3 & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{1}{5} \times 400 = 80 \text{ cm.}^3$

Hence, volume of propane consumed is 80 cm³

(b) The mass of 22.4 L of a gas at S.T.P. is equal to its gram molecular mass.

360 cm³ of N₂ at S.T.P. weighs 0.45 g

∴ 22,400 cm³ of N₂ will weigh $\dfrac{0.45}{360} \times 22,400$ = 28 g

Hence, Gram Molecular Mass of N₂ = 28 g.