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Certain volume of oxygen gas diffuses through a porous pot in 20 seconds. Same volume of another gas (X) diffuses in $\mathrm{Y}$ seconds as that of oxygen, then $(\mathrm{X})$ and $\mathrm{Y}$ respectively are
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Verified Answer
The correct answer is:
$\mathrm{H}_2, 5$
From Graham's law :-
$$
\begin{aligned}
& \frac{R_2}{R_1}=\sqrt{\frac{M_1}{M_2}}=\frac{t_1}{t_2} \Rightarrow \frac{M_1}{M_2}=\frac{t_1^2}{t_2^2} \\
& M_{O_2}=32, M_x=?, t_{o_2}=20 \text { seconds, } \\
& \Rightarrow M_1 t_2{ }^2=M_2 t_1{ }^2 \Rightarrow \frac{M_1}{t_1{ }^2}=\frac{M_2}{t_2{ }^2}
\end{aligned}
$$
Thus, $\frac{\mathrm{M}_{\mathrm{o}_2}}{\mathrm{t}_{\mathrm{o}_2}^2}=\frac{\mathrm{M}_{\mathrm{x}}}{\mathrm{t}_{\mathrm{x}}^2}$
$$
\Rightarrow \frac{32}{(20)^2}=\frac{32}{400}=\frac{4}{50}=\frac{M_x}{t_x{ }^2}
$$
For $\mathrm{H}_2$ gas, $\mathrm{M}_{\mathrm{x}}=2$ and if $\mathrm{t}_{\mathrm{x}}=5$ seconds.
$$
\begin{aligned}
& \frac{R_2}{R_1}=\sqrt{\frac{M_1}{M_2}}=\frac{t_1}{t_2} \Rightarrow \frac{M_1}{M_2}=\frac{t_1^2}{t_2^2} \\
& M_{O_2}=32, M_x=?, t_{o_2}=20 \text { seconds, } \\
& \Rightarrow M_1 t_2{ }^2=M_2 t_1{ }^2 \Rightarrow \frac{M_1}{t_1{ }^2}=\frac{M_2}{t_2{ }^2}
\end{aligned}
$$
Thus, $\frac{\mathrm{M}_{\mathrm{o}_2}}{\mathrm{t}_{\mathrm{o}_2}^2}=\frac{\mathrm{M}_{\mathrm{x}}}{\mathrm{t}_{\mathrm{x}}^2}$
$$
\Rightarrow \frac{32}{(20)^2}=\frac{32}{400}=\frac{4}{50}=\frac{M_x}{t_x{ }^2}
$$
For $\mathrm{H}_2$ gas, $\mathrm{M}_{\mathrm{x}}=2$ and if $\mathrm{t}_{\mathrm{x}}=5$ seconds.
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