The molecular weight of a gas which diffuses through a porous plug at $\frac{1}{6}th$ of the speed of hydrogen under identical conditions is

Let us assume the rate of diffusion of hydrogen gas ${\mathrm{isr}}_1$

The rate of diffusion of other gas with molecular mass ${\mathrm{M}}_2 \mathrm{to} \mathrm{be} {\mathrm{r}}_2$

Then ${\mathrm{r}}_2=\frac{1}{6}{\mathrm{r}}_1$

According to Graham’s law of diffusion

$\mathrm{r} \mathrm{\alpha }\frac{1}{\sqrt{\mathrm{M}}}$

According to Graham’s law of diffusion

$\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\sqrt{\frac{M}{M_{H_2}}}$

$\Rightarrow \frac{\mathrm{r}}{\frac{1}{6}\mathrm{r}}=\sqrt{\frac{M}{2}}$

$\Rightarrow 6=\sqrt{\frac{M}{2}}$

Take square on both sides of the above equation

$\Rightarrow \frac{\mathrm{M}}{2}=36$

M = 72