The ratio of the rate of diffusion of SO_2 to | vedclass

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(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas

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$1: 2$

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(B) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.The molar mass of $SO_2$ is $32 + (16 	imes 2) = 64 \ g/mol$.The molar mass of $CH_4$ is $12 + (1 	imes 4) = 16 \ g/mol$.Therefore,the ratio of the rate of diffusion of $SO_2$ to $CH_4$ is:$\frac{r_{SO_2}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{SO_2}}} = \sqrt{\frac{16}{64}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,the ratio is $1: 2$.

The ratio of the rate of diffusion of $SO_2$ to $CH_4$ is

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