Question

$180\text{mL}$ of a hydrocarbon diffuses through a porous membrane in 15 minutes while $120\text{mL}$ of $S{O}_2$ under identical conditions diffuses in 20 minutes. What is the molecular mass of the hydrocarbon?

Answer 1

Let the molecular mass of hydrocarbon = M
$r_1$ = rate of diffusion of hydrocarbon $=\frac{180}{15}=12{\text{mL min}}^{-1}$
$r_2=$ rate of diffusion of $S{O}_2$$=\frac{120}{20}=6{\text{mL min}}^{-1}$
$\frac{r_1}{r_2}=\sqrt{\frac{M_{S{O}_2}}{M}}$
Thus, $\frac{12}{6}=\sqrt{\frac{64}{M}}$
$2=\sqrt{\frac{64}{M}}$
So, $M=16$