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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 its molar mass $(M)$ and inver

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$H_2, 5$

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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 its molar mass $(M)$ and inversely proportional to the time taken $(t)$ for the same volume to diffuse: $r \propto \frac{1}{\sqrt{M}}$ and $r \propto \frac{1}{t}$. Therefore, $\frac{r_X}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_X}} = \frac{t_{O_2}}{t_X}$. Squaring both sides: $\frac{M_{O_2}}{M_X} = \frac{t_{O_2}^2}{t_X^2} \Rightarrow \frac{M_X}{t_X^2} = \frac{M_{O_2}}{t_{O_2}^2}$. Given $M_{O_2} = 32 \ g/mol$ and $t_{O_2} = 20 \ s$: $\frac{M_X}{t_X^2} = \frac{32}{20^2} = \frac{32}{400} = 0.08$. Checking option $A$: For $H_2$, $M_X = 2$ and $t_X = 5 \ s$. $\frac{2}{5^2} = \frac{2}{25} = 0.08$. Since the ratio matches, the correct pair is $H_2$ and $5 \ s$.

$A$ certain volume of oxygen gas diffuses through a porous pot in $20 \ s$. If the same volume of another gas $(X)$ diffuses in $Y \ s$, then $(X)$ and $Y$ respectively are:

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