Question

$S{O}_2$ and $C{H}_4$ are introduced in a vessel in the molar ratio 1:2. The ratio of molecules of two gases present in the container when their rate of effusion becomes equal is:

• A. 1:2
• B. 2:1
• C. 4:1
• D. 1:4

Answer 1

The correct option is B 2:1

The ratio of the number of moles of $S{O}_2$ to the number of moles of $C{H}_4$ is $\frac{n_1}{n_2}=\frac{1}{2}$.
Here, 1 represents $S{O}_2$ and 2 represents $C{H}_4$.

The rate of effusion is proportional to the ratio of pressure to the square root of molecular weight.
$r\propto \frac{p}{\sqrt{M}}$

The rates of effusions of two gases are equal.
$\frac{p_1}{\sqrt{M_1}}=\frac{p_2}{\sqrt{M_2}}$

The number of moles are proportional to pressure.

Hence, $\frac{p_1}{p_2}=\frac{\sqrt{M_1}}{\sqrt{M_2}}=\frac{n_1}{n_2}$

Thus, $\frac{n_1}{n_2}=\sqrt{\frac{M_1}{M_2}}=\sqrt{\frac{64}{16}}=2:1$

Option B is correct.