180ml of a hydrocarbon having the molecular weight 16 diffuses in 1.5 min. Under similar conditions time taken by 120 ml of $S{O}_2$ to diffuse is ______________ sec.

According to Graham's law,  

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$, where $r_1$ and $r_2$ are the rates of diffusion of gases 1 and 2 respectively.

and, $M_1$and $M_2$​ are the molecular weights of gases 1 and 2 respectively 

Rate=time required for diffusion volume of gas​

$$\begin{gathered} r_1=\frac{180}{1.5}=120 \\ {r}_2=\frac{t_2}{120} \\ {t}_2=\frac{120}{60}=2 \end{gathered}$$

Therefore, the required time is 2 minute = 120 sec.