Question

Rate of diffusion of $N{H}_3$ is twice that of $X$. What is the molecular mass of $X$ ?

• A. $68$
• B. $48$
• C. $12$
• D. $8$

Answer 1

Answer: (A)

Rate of diffusion, $r_x=\left(\frac{r_{N{H}_3}}{2}\right)$ (given)
Molecular mass of $N{H}_3,M_{N{H}_3}=(N)+3(H)$
$=14+3(1)=17$
Molecular mass of $X,M_X$= ?
By Graham’s law of diffusion
$\therefore \frac{r_{N{H}_3}}{r_x}=\sqrt{\frac{M_X}{M_{N{H}_3}}}$ or $2=\sqrt{\frac{M_x}{M_{N{H}_3}}}$
$\therefore 4\times M_{N{H}_3}=M_X$
$\Rightarrow M_x=4\times 17=68$