Q5 / 200 · WBJEE 2014 · Chemistry
At a certain temperature the time required for the complete diffusion of $200 \mathrm{mL}$ of $\mathrm{H}_{2}$ gas is 30 min. The time required for the complete diffusion of $50 \mathrm{mL}$ of $\mathrm{O}_{2}$ gas at the same temperature will be
- A$60 \mathrm{min}$
- Correct answerB$30 \mathrm{min}$
- C$45 \mathrm{min}$
- D$15 \mathrm{min}$
Solution
According to Graham's law of diffusion.
Rate of diffusion, $r \propto \frac{1}{\sqrt{M}}=\frac{V}{t}$
Here, $M=$ molecular mass, $V=$ volume and $t=$ timel Thus, for $\mathrm{H}_{2}$ gas,
$$
\frac{200}{30}=\frac{1}{\sqrt{2}}
$$
for $O_{2}$ gas,
$$
\frac{50}{t}=\frac{1}{\sqrt{32}}
$$
Eqs. (i) and (ii), we get From
$$
\begin{aligned}
\frac{200 \times t}{30 \times 50} &=\frac{\sqrt{32}}{\sqrt{2}} \\
t &=\frac{\sqrt{16} \times 30 \times 50}{200} \\
&=\frac{4 \times 30}{4}=30 \mathrm{min}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.