Question

If $4$ g of oxygen diffuse through a very narrow hole, how much hydrogen would have diffused under identical conditions.

Answer 1

According to Graham's law of diffusion;

$\frac{Rat{e}_1}{Rat{e}_2}=\sqrt{\frac{M_2}{M_1}}$

$\Rightarrow \frac{Rat{e}_{H_2}}{Rat{e}_{O_2}}=\sqrt{\frac{32}{2}}=4$

$4g$ of $O_2=\frac{1}{8}$ mol of $O_2$

$\therefore Rat{e}_{H_2}=\frac{1/8}{t}\times 4=\left(\frac{1}{2}\right)/t$

Hydrogen has a mass of $\frac{1}{2}\times \left(2\right)=1g$

Thus $1g$ of $H_2$ would diffuse in same time as $4g$ of $O_2$ .