$60 \mathrm{~cm}^3$ of $\mathrm{SO}_2$ gas diffused through a porous membrane in ' $x$ ' min. Under similar…
$60 \mathrm{~cm}^3$ of $\mathrm{SO}_2$ gas diffused through a porous membrane in ' $x$ ' min. Under similar conditions $360 \mathrm{~cm}^3$ of another gas (molar mass $4 \mathrm{~g} \mathrm{~mol}^{-1}$ ) diffused ' $y$ ' min. The ratio of $x$ and $y$ is
$3: 2$
$2: 3$
$1: 3$
$3: 1$
Solution
$\begin{aligned}
\mathrm{V}_1 & =60 \mathrm{~cm}^3, \mathrm{M}_1=64, \mathrm{t}_1=\mathrm{x} \\
\mathrm{~V}_2 & =360 \mathrm{~cm}^3, \mathrm{M}_2=4, \mathrm{t}_2=\mathrm{y}
\end{aligned}$
rate of diffusion,
$\begin{aligned}
& r=\frac{V}{t}=\frac{\text { volume of gas diffused }}{\text { Time taken for diffusion }} \\
& \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}=\frac{v_1 t_2}{v_2 t t_1} \\
& \therefore \sqrt{\frac{4}{64}}=\frac{60}{360} \times \frac{y}{x} \\
& \text { or, } \frac{1}{4} \times 6=\frac{y}{x} \\
& \therefore \frac{y}{x}=\frac{3}{2} \therefore \frac{x}{y}=\frac{2}{3}
\end{aligned}$