Question

One litre of a gaseous mixture of two gases effuses in 311 seconds while 2 liters of oxygen takes 20 minutes. The vapour density of the gaseous mixture containing ${CH}_4$ and $H_2$:

• A. $4$
• B. $4.3$
• C. $3.4$
• D. $2.15$

Answer 1

Answer: (D)

$2.15$

We know that, the rate of effusion of gas, $r=\frac{V}{t}$

where, V = volume of gas effused and t = time taken for effusion.

For gaseous mixture of $C{H}_4$ and $H_2$,

Volume, $V_1=$ 1 litre

Time, $t_1=$ 311 sec

For Oxygen ($O_2)$ gas,

Volume, $V_2=$ 2 litre

Time, $t_2=20min=$ 20 x 60 sec

Therefore,

$\frac{r_1}{r_2}=\frac{\frac{V_1}{t_1}}{\frac{V_2}{t_2}}=\frac{V_1}{t_1}\times \frac{t_2}{V_2}=\frac{1}{311}\times \frac{20\times 60}{2}=\frac{600}{311}$

According to Graham's Law of Effusion, for two gases having molecular masses, $M_1$ and $M_2$, the ratio of their effusion rates $r_1$ and $r_2$ is given by,

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Here,

$r_1$ - rate of effusion of gaseous mixture

$r_2$ - rate of effusion of oxygen

$M_1$ - molecular mass of gaseous mixture

$M_2$ -molecular mass ofoxygen

We know that, for oxygen ($O_2)$, molecular mass, $M_2$ = 32

So, we can write as,

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}=>\frac{600}{311}=\sqrt{\frac{32}{M_1}}$

$=>\sqrt{\frac{32}{M_1}}=\frac{600}{311}$

$=>\frac{32}{M_1}=\frac{{600}^2}{{311}^2}=3.72$

$=>M_1=\frac{32}{3.72}=8.6$

That is, the molecular mass of the gaseous mixture = 8.6 g

We have, Vapour density = Molecular mass / 2

So. vapour density of gaseous mixture = $\frac{8.6}{2}$ = 4.3

Therefore. option B. 4.3 is the correct answer.