$300 \mathrm{~mL}$ of a gas ' $x$ ' of molar mass $32 \mathrm{~g} \mathrm{~mol}^{-1}$ is effused in 25…
$300 \mathrm{~mL}$ of a gas ' $x$ ' of molar mass $32 \mathrm{~g} \mathrm{~mol}^{-1}$ is effused in 25 seconds. What volume of methane would effuse in the same time?
$624 \mathrm{~mL}$
$524 \mathrm{~mL}$
$32 \mathrm{~mL}$
$424 \mathrm{~mL}$
Solution
Let, the volume of methane effused in $25 \mathrm{~s}$ be $x \mathrm{~mL}$.
Now, from the Graham's law of diffusion,
Rate of effusion $\propto \frac{1}{\sqrt{\text { Molecular mass }}}$...(i)
Also, rate of effusion $=\frac{\text { Volume diffused }}{\text { Time taken }}$...(ii)
From Eq. (i) and (ii) we get,
$\begin{aligned} & \frac{r_x}{r_{\mathrm{CH}_4}}=\sqrt{\frac{M_{\mathrm{CH}_4}}{M_x}} \\ & \frac{\frac{300}{25}}{\frac{x}{25}}=\sqrt{\frac{16}{32}} \Rightarrow \frac{300}{x}=\sqrt{\frac{1}{2}}\end{aligned}$
$\Rightarrow \quad x=300 \sqrt{2}=424.26 \mathrm{~mL} \simeq 424 \mathrm{~mL}$