The rate of diffusion of methane is twice that of the following gas under the same conditions

The mass of unknown gas is ${\mathrm{M}}_1$

The mass of methane $\left({\mathrm{CH}}_4\right),M_2=16g$

The diffusion rate of unknown gas is ${\mathrm{r}}_1=\mathrm{r}$

The diffusion rate of methane is ${\mathrm{r}}_2=2{\mathrm{r}}_1=2\mathrm{r}$

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

$\frac{\mathrm{r}}{2\mathrm{r}}=\sqrt{\frac{16}{M_1}}\to \mathrm{eqn}\left(1\right)$

Now apply square on both sides in the equation 1.

$\frac{1}{4}=\frac{16}{M_1}$

$\Rightarrow {\mathrm{M}}_1=64 \mathrm{g}$

The compound with molar mass as 64 g is ${\mathrm{SO}}_2$