Question

Rate of diffusion of $NH_3$ 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, $r_{x}=\left(\frac{r_{NH_3}}{2}\right)$ (given)
Molecular mass of $NH_{3}, M_{NH_{3}}=(N)+3(H)$
$=14+3(1)=17$
Molecular mass of $X, M_{X}$= ?
By Graham’s law of diffusion
$\therefore \frac{r_{NH_3}}{r_{x}}=\sqrt{\frac{M_{X}}{M_{NH_3}}}$ or $2=\sqrt{\frac{M_{x}}{M_{NH_3}}}$
$\therefore 4\times M_{NH_3}=M_{X}$
$\Rightarrow M_{x}=4\times17=68$