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 \

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$16 : 1$

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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,$\frac{r_x}{r_y} = \sqrt{\frac{Mw_y}{Mw_x}}$.Given $r_x = 10 \, mL/s$ and $r_y = 40 \, mL/s$,we have:$\frac{10}{40} = \sqrt{\frac{Mw_y}{Mw_x}}$$\frac{1}{4} = \sqrt{\frac{Mw_y}{Mw_x}}$Squaring both sides:$\frac{1}{16} = \frac{Mw_y}{Mw_x}$Thus,the ratio of their molar masses $\frac{Mw_x}{Mw_y} = \frac{16}{1}$ or $16 : 1$.

The rate of diffusion of two gases $x$ and $y$ is $10 \, mL/s$ and $40 \, mL/s$ respectively,then the ratio of their molar masses $(Mw_x : Mw_y)$ is

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