The ratio of the rates of diffusion of SO_2,O | 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

Vedclass · Accepted Answer

$1:\sqrt{2}:2$

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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 $M$: $r \propto \frac{1}{\sqrt{M}}$.The molar masses are: $M(SO_2) = 64 \ g/mol$,$M(O_2) = 32 \ g/mol$,and $M(CH_4) = 16 \ g/mol$.The ratio of rates is $r(SO_2) : r(O_2) : r(CH_4) = \frac{1}{\sqrt{64}} : \frac{1}{\sqrt{32}} : \frac{1}{\sqrt{16}}$.This simplifies to $\frac{1}{8} : \frac{1}{4\sqrt{2}} : \frac{1}{4}$.Multiplying by $8$,we get $1 : \frac{8}{4\sqrt{2}} : \frac{8}{4} = 1 : \frac{2}{\sqrt{2}} : 2$.Since $\frac{2}{\sqrt{2}} = \sqrt{2}$,the ratio is $1 : \sqrt{2} : 2$.

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

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