180ml of hydrocarbon having the molecular weight 16 diffuses in 1.5 min. Under similar conditions time taken by 120ml of SO₂ to diffuse is

According to Graham's law, $\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\sqrt{\frac{{\mathrm{M}}_2}{{\mathrm{M}}_1}}$

where r₁ and r₂ are the rates of diffusion of gases 1 and 2 respectively and, M₁ and M₂ are the molecular weights of gases 1 and 2 respectively

$$\begin{gathered} \mathrm{rate}=\frac{\mathrm{volume} \mathrm{of} \mathrm{gas}}{\mathrm{time} \mathrm{required} \mathrm{for} \mathrm{diffusion}} \\ {\mathrm{r}}_1=\frac{180}{1.5}=120 \\ \frac{120}{{\mathrm{r}}_2}=\sqrt{\frac{64}{16}} \\ {\mathrm{r}}_2=60 \\ {\mathrm{r}}_2=\frac{120}{{\mathrm{t}}_2} \\ {\mathrm{t}}_2=\frac{120}{60} \\ {\mathrm{t}}_2=2\min \end{gathered}$$