If the molar ratio of CH_4 and O_2 is 1:2,the | vedclass

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(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propt

Vedclass · Accepted Answer

$1.414:1$

Vedclass · Answer

(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.Given the molar ratio $n_{CH_4} : n_{O_2} = 1:2$.The rate of diffusion is also proportional to the number of moles $(n)$: $r \propto n$.Therefore,the ratio of rates is $\frac{r_{O_2}}{r_{CH_4}} = \frac{n_{O_2}}{n_{CH_4}} 	imes \sqrt{\frac{M_{CH_4}}{M_{O_2}}}$.Substituting the values: $M_{CH_4} = 16 \ g/mol$ and $M_{O_2} = 32 \ g/mol$.$\frac{r_{O_2}}{r_{CH_4}} = \frac{2}{1} 	imes \sqrt{\frac{16}{32}} = 2 	imes \sqrt{\frac{1}{2}} = 2 	imes \frac{1}{1.414} = \frac{1.414}{1}$.Thus,the ratio is $1.414:1$.

If the molar ratio of $CH_4$ and $O_2$ is $1:2$,then the ratio of the rate of diffusion of $O_2$ and $CH_4$ is respectively:

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