Question

$C{H}_4$ diffuses two times faster than a gas $X$. The number of molecules present in $32g$ of gas $X$ is :

($N$ is Avogadro number)

• A. $N$
• B. $\frac{N}{2}$
• C. $\frac{N}{4}$
• D. $\frac{N}{16}$

Answer 1

The correct option is B $\frac{N}{2}$

Using graham's law of diffusion,

$\frac{r_{C{H}_4}}{r_{gas}}=\sqrt{\frac{m_{gas}}{m_{C{H}_4}}}$

$2=\sqrt{\frac{m_{gas}}{16}}$

$m_{gas}=64$

Number of molecules $=\frac{w}{molarmass}\times N$

$=\frac{32}{64}\times N$

$=\frac{N}{2}$.

Hence, the correct option is $\text{B}$