Question

$20{\text{dm}}^3$ of $S{O}_2$ diffuses through a porous partition in $60\text{s}.$ What volume (${\text{in dm}}^3$) of $O_2$ will diffuse under similar conditions in $30\text{s}$?

Answer 1

Let the volume of $O_2$ diffused in $30\text{s}=V{\text{dm}}^3$

Rate of diffusion of $S{O}_2=\frac{20}{60}{\text{dm}}^3{\text{s}}^{-1}$

Rate of diffusion of $O_2=\frac{V}{30}{\text{dm}}^3{\text{s}}^{-1}$

According to Graham's law of diffusion,
$\frac{r_{O_2}}{r_{S{O}_2}}=\sqrt{\frac{M_{S{O}_2}}{M_{O_2}}}$
$\frac{\left(V/30\right)}{\left(20/60\right)}=\sqrt{\frac{M_{S{O}_2}}{M_{O_2}}}=\sqrt{\frac{64}{32}}$
$V=14.1{\text{dm}}^3$