the rate of diffusion of methane is twice that of gas x , when diffuse through same hole under all identical conditions. the molecular mass ofgasx is
1)64
2) 80 3) 332

Dear User,
Let rate of diffusion of gas x, r₁ = a
Therefore, rate of diffusion of methane,r₂ = 2 a
According to Graham's Law of Diffusion
$$\begin{gathered} \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\sqrt{\frac{{\mathrm{M}}_2}{{\mathrm{M}}_1}} \\ {\mathrm{M}}_1 = \mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{gas} \mathrm{x} \\ {\mathrm{M}}_2 = \mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{Methane} = 16 \mathrm{g} \\ \mathrm{Therefore}, \\ \frac{\mathrm{a}}{2\mathrm{a}}= \sqrt{\frac{16}{{\mathrm{M}}_2}} \\ \mathrm{Squaring} \mathrm{both} \mathrm{the} \mathrm{sides}, \\ \frac{1}{4}=\frac{16}{{\mathrm{M}}_2} \\ \mathrm{or}, {\mathrm{M}}_2 = 16\times 4=64 \mathrm{g} \end{gathered}$$
Hence, the correct answer is 1) 64 g