Question

180 ml of hydrocarbon diffuses through a porous membrane in 15 minutes while 120 ml of $S{O}_2$ under identical conditions diffuses in 20 minutes. What is the molecular mass of the hydrocarbon?

• A. 8
• B. 16
• C. 24
• D. 32

Answer 1

Answer: (B)

16

According to Graham's law of diffusion, the rates of diffusion of gases are inversely proportional to the square root of their molecular weights.

$\frac{r_1}{r_2}=\sqrt{\frac{M_{S{O}_2}}{M_{hydrocarbon}}}$

The rate of diffusion of hydrocarbon is $r_1=\frac{\text{volume diffused}}{\text{time required for diffusion}}=\frac{180}{15}$.

The rate of diffusion of $S{O}_2$ is $r_2=\frac{\text{volume diffused}}{\text{time required for diffusion}}=\frac{120}{20}$.

The molecular weight of $S{O}_2$ is 64 g/mol.

Substitute values in the above expression.

$\frac{\frac{180}{15}}{\frac{120}{20}}=\sqrt{\frac{64}{M}}$

Thus, $\sqrt{\frac{64}{M}}=2$ or $M=16$.

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