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

Molar mass of oxygen, ${\mathrm{Mo}}_2=32$

Molar mass of hydrogen,${\mathrm{M}}_{{\mathrm{H}}_2}=2$

The relation between rate of diffusion of oxygen and hydrogen is follows below

$\frac{{{\mathrm{r}}_{\mathrm{O}}}_2}{{\mathrm{r}}_{{\mathrm{H}}_2}}=\sqrt{\frac{M_{H_2}}{M_{O_2}}}$

$\Rightarrow \frac{{{\mathrm{r}}_{\mathrm{O}}}_2}{{\mathrm{r}}_{{\mathrm{H}}_2}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{6}}=\frac{1}{4}\to \mathrm{eqn}\left(1\right)$

$32\mathrm{g} \mathrm{of} {\mathrm{O}}_2=1 \mathrm{mole} \mathrm{of} {\mathrm{O}}_2$

$4 \mathrm{g} \mathrm{of} {\mathrm{O}}_2=\frac{1 \mathrm{mol}}{32\mathrm{g}}\times 4\mathrm{g}$

$=\frac{1}{8} mol diffuse out$

$\mathrm{from} \mathrm{eq} \left(1\right)\Rightarrow \mathrm{Rate} \mathrm{of} 1 \mathrm{mole} {\mathrm{O}}_2=\mathrm{Rate} \mathrm{of} 4 \mathrm{mole} {\mathrm{H}}_2$$\frac{1}{8}mole {\mathrm{O}}_2=\frac{1}{8}\times 4=\frac{1}{2}\mathrm{mole} {\mathrm{H}}_2$

$Mass of \frac{1}{2}\mathrm{mole} {\mathrm{H}}_2 \mathrm{is} \mathrm{equal} \mathrm{to} 1\mathrm{g} \mathrm{of} \mathrm{hydrogen}$