The ratio of rates of diffusion of SO_2,O_2 a | vedclass

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(A) 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 \f

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$1:\sqrt{2}:2$

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.For $SO_2$ $(M = 64)$,$O_2$ $(M = 32)$,and $CH_4$ $(M = 16)$:$r_{SO_2} : r_{O_2} : r_{CH_4} = \frac{1}{\sqrt{64}} : \frac{1}{\sqrt{32}} : \frac{1}{\sqrt{16}}$$= \frac{1}{8} : \frac{1}{4\sqrt{2}} : \frac{1}{4}$Multiplying the entire ratio by $8$ to simplify:$= 1 : \frac{8}{4\sqrt{2}} : 2$$= 1 : \frac{2}{\sqrt{2}} : 2$$= 1 : \sqrt{2} : 2$

The ratio of rates of diffusion of $SO_2$,$O_2$ and $CH_4$ is

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