Question

360 cm$^3$ of a hydrocarbon diffuses in 30 minutes, while under the same conditions 360 cm$^3$ of SO$_2$ gas diffuses in one hour. The molecular formula of the hydrocarbon is

• A. CH$_4$
• B. C$_2$H$_6$
• C. C$_2$H$_4$
• D. C$_2$H$_2$

Hint:

Using a simple frame or just bolding for the box Key Points: Graham's Law: $\frac{r_1{r_2 = \sqrt{\frac{M_2{M_1$ (at constant T, P). Rate ($r$) = Volume ($V$) / Time ($t$). Ensure time units are consistent (e.g., both in minutes). Calculate the unknown molar mass and match it with the options. Molar Mass: SO$_2$ = 64 g/mol, CH$_4$ = 16 g/mol, C$_2$H$_6$ = 30 g/mol, C$_2$H$_4$ = 28 g/mol, C$_2$H$_2$ = 26 g/mol.