Q17 / 155 · WBJEE 2025 · Chemistry
$360 \mathrm{~cm}^3$ of a hydrocabon diffuses in 30 minutes, while under the same conditions $360 \mathrm{~cm}^3$ of $\mathrm{SO}_2$ gas diffuses in one hour. The molecular formula of the hydrocabon is
- Correct answerA$\mathrm{CH}_4$
- B$\mathrm{C}_2 \mathrm{H}_6$
- C$\mathrm{C}_2 \mathrm{H}_4$
- D$\mathrm{C}_2 \mathrm{H}_2$
Solution
According to Graham's Law: (MW = Molecular Weight)
$\begin{aligned}
& \frac{\mathrm{R}_{\mathrm{d}}(\text { hydrocarbon })}{\mathrm{R}_{\mathrm{d}}\left(\mathrm{SO}_2\right)}=\sqrt{\frac{\mathrm{MW}_{\mathrm{SO}_2}}{\mathrm{MW}_{\text {hydrocarbon }}}} \\
& \Rightarrow \frac{360 / 30}{360 / 60}=\sqrt{\frac{64}{\mathrm{MW}_{\text {hydrocarbon }}}} \\
& \Rightarrow 4=64 / \mathrm{MW}_{\text {hydrocabon }} \\
& \Rightarrow \mathrm{MW}_{\text {hydrocarbon }}=\frac{64}{4}=16
\end{aligned}$
$\mathrm{CH}_4$ is the hydrocarbon with possible molecular weight 16.
Hence, the correct answer is option: 'A'.