According to law of diffusion rate $(r) \alpha \frac{1}{\sqrt{M}}$ (molar mass) and $r$ (rate) $\alpha \frac{\text { volume }(V)}{\text { time }(t)}$
i.e. $\quad \frac{r_{\left(\mathrm{SO}_2\right)}}{r_{(z)}}=\frac{(V / t)_{\left(\mathrm{SO}_2\right)}}{(V / t)_{(z)}}=\frac{[\mathrm{l} / \sqrt{M}]_{\left(\mathrm{SO}_2\right)}}{(\mathrm{l} / \sqrt{M})_{C_z}}$ or, $\quad \frac{V_{\left(\mathrm{SO}_2\right)} \times t_{(z)}}{V_{(z)} \times t_{\left(\mathrm{SO}_2\right)}}=\sqrt{\frac{M_{(z)}}{M_{\left(\mathrm{SO}_2\right)}}}$
Given,
$$
\begin{aligned}
V_{\left(\mathrm{SO}_2\right)} & =12 \mathrm{~cm}^3 \\
t_{\left(\mathrm{SO}_2\right)} & =1 \text { minute }=1 \times 60=60 \mathrm{sec} \\
V_{(z)} & =120 \mathrm{~cm}^3 \\
t_{(z)} & =5 \text { minutes }=5 \times 60=300 \mathrm{sec}
\end{aligned}
$$
and molar mass $(M)$ of $\mathrm{SO}_2=32+(2 \times 16)=64$
Thus, $\frac{12 \times 300}{120 \times 60}=\sqrt{\frac{M_z}{64}}$ or, $\frac{1}{4}=\frac{M_z}{4}$
$\therefore \quad M_z=\frac{64}{4}=16$


Hence, molar mass of given gas is $=16 \mathrm{~g} \mathrm{~mol}^{-1}$