$\mathrm{CH}_4$ diffuses two times faster than a gas $X$. The number of molecules present in $32…
$\mathrm{CH}_4$ diffuses two times faster than a gas $X$. The number of molecules present in $32 \mathrm{~g}$ of gas $X$ is ( $N$ is Avogadro number)
$N$
$\frac{N}{2}$
$\frac{N}{4}$
$\frac{N}{16}$
Solution
From Graham's law of diffusion,
$\frac{r_{\mathrm{CH}_4}}{r_X}=\sqrt{\frac{M_X}{M_{\mathrm{CH}_4}}}$
(given, $r_{\mathrm{CH}_4}=2 \cdot r_X$ )
$2=\sqrt{\frac{M_X}{16}}$
$M_X=16 \times 4=64$
Thus, the molecular mass of gas $X$ is 64 . Number of molecules of gas $X$ in $32 \mathrm{~g}$ gas
$=\frac{32}{64} \times N=\frac{N}{2}$