The rate of diffusion of two gases x and y is | 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 $Mw$: $r \propto \

Vedclass · Accepted Answer

$16:1$

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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 $Mw$: $r \propto \frac{1}{\sqrt{Mw}}$.Therefore,the ratio of rates is given by: $\frac{r_{x}}{r_{y}} = \sqrt{\frac{(Mw)_{y}}{(Mw)_{x}}}$.Given $r_{x} = 10 \ mL/sec$ and $r_{y} = 40 \ mL/sec$,we have: $\frac{10}{40} = \sqrt{\frac{(Mw)_{y}}{(Mw)_{x}}}$.Squaring both sides: $(\frac{1}{4})^2 = \frac{(Mw)_{y}}{(Mw)_{x}}$,which gives $\frac{1}{16} = \frac{(Mw)_{y}}{(Mw)_{x}}$.Thus,the ratio of molar masses $(Mw)_{x} : (Mw)_{y}$ is $16:1$.

The rate of diffusion of two gases $x$ and $y$ is $10 \ mL/sec$ and $40 \ mL/sec$ respectively. The ratio of their molar masses $(Mw)_{x} : (Mw)_{y}$ is:

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