The ratio of rates of diffusion of oxygen and | vedclass

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(C) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $\frac{r_

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$H_2$

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(C) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $\frac{r_{O_2}}{r_{gas}} = \frac{1}{4}$ and $M_{O_2} = 32 \ g/mol$.Substituting the values in the formula:$\frac{1}{4} = \sqrt{\frac{M_{gas}}{32}}$.Squaring both sides:$\frac{1}{16} = \frac{M_{gas}}{32}$.Solving for $M_{gas}$:$M_{gas} = \frac{32}{16} = 2 \ g/mol$.The gas with a molar mass of $2 \ g/mol$ is Hydrogen $(H_2)$.

The ratio of rates of diffusion of oxygen and an unknown gas is $1:4$. The unknown gas is

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