A vessel contains $H_2$ and $O_2$ in the molar ratio of 8 : 1. This mixture of gases is allowed to diffuse through a hole. Find the ratio of the composition of $H_2$ to that of $O_2$ coming out through the hole.

The correct option is B 32 : 1
$\text{Rate of effusion of a gas}\left(r\right)\propto \frac{P}{\sqrt{M}}$ Where, $P$ is the pressure and $M$ is the molar mass of the gas.
From the ideal gas equation,
$P\propto n$
$\therefore \text{Rate of effusion of a gas}\left(r\right)\propto \frac{n}{\sqrt{M}}$.
Given, $n_{H_2}:n_{O_2}=8:1$
$\frac{r_{H_2}}{r_{O_2}}=\frac{n_{H_2}}{n_{O_2}}\sqrt{\frac{M_{O_2}}{M_{H_2}}}$
$\frac{r_{H_2}}{r_{O_2}}=\frac{8}{1}\times \sqrt{\frac{32}{2}}=\frac{32}{1}$
$\Rightarrow \frac{\left(\frac{Numberofmolesof{H}_{2}comingout}{\Delta t}\right)}{\left(\frac{Numberofmolesof{O}_{2}comingout}{\Delta t}\right)}=\frac{32}{1}$
Hence, the ratio of composition of $H_2$ to that of $O_2$ coming out through the hole $=32:1$