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The rate of diffusion of a gas $\mathrm{A}$ is $\sqrt{5}$ times more than that of gas $\mathrm{B}$. If the molar mass of $\mathrm{A}$ is $\mathrm{x} \mathrm{g}$ $\mathrm{mol}^{-1}$, the molar mass of $\mathrm{B}\left(\mathrm{in} \mathrm{g} \mathrm{mol}^{-1}\right)$ is
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$5 x$
$\mathrm{r}_{\mathrm{A}}=\sqrt{5}{ }_{\mathrm{B}}^{\mathrm{r}}, \mathrm{M}_{\mathrm{A}}=\mathrm{xgmol}^{-1}$
$\begin{aligned} & \frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \Rightarrow\left(\frac{r_A}{r_B}\right)^2=\frac{M_B}{M_A} \\ & \Rightarrow\left(\frac{\sqrt{5} r_B}{r_B}\right)^2=\frac{M_B}{x} \\ & \Rightarrow 5=\frac{M_B}{x} \Rightarrow M_B=5 x .\end{aligned}$
$\begin{aligned} & \frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \Rightarrow\left(\frac{r_A}{r_B}\right)^2=\frac{M_B}{M_A} \\ & \Rightarrow\left(\frac{\sqrt{5} r_B}{r_B}\right)^2=\frac{M_B}{x} \\ & \Rightarrow 5=\frac{M_B}{x} \Rightarrow M_B=5 x .\end{aligned}$
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