Discussion Forum

Diffusion of $CH_{4}(g) $ and $O_{2}$  (g) occurs under similar conditions, then the ratio of their rates of diffusion is 

Answer:

Explanation:

From Grahjam 's law of diffusion:

$\frac{r_{CH_{4}}}{r_{O_{2}}}=\sqrt{\frac{M_{O_{2}}}{M_{CH_{4}}}}$

here, $r_{CH_{4}}$= rate of diffusion of $ CH_{4}$ gas

      $r_{O_{2}}$= rate of diffusion of $O_{2}$ gas

$M_{CH_{4}}$= molecular mass of $CH_{4}$=16

and   $M_{O_{2}}$ = molecular mass of $O_{2}$ gas=32

Hence,   $\frac{r_{CH_{4}}}{r_{O_{2}}}=\sqrt{\frac{32}{16}}=\sqrt{2}=1.414$