Question

180 mL of hydrocarbon having the molecular weight 16 diffuses in 1.5 minutes. Under the same conditions, the time taken by 120 mL of sulphur dioxide to diffuse is :

• A. 2.5 min
• B. 3.0 min
• C. 2.0 min
• D. 1.5 min

Answer 1

Answer: (C)

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$ or $\frac{V_1/t_1}{V_2/t_2}=\sqrt{\frac{M_2}{M_1}}$ or $\frac{V_1}{t_1}\times \frac{t_2}{V_2}=\sqrt{\frac{M_2}{M_1}}$ or $\frac{V_{Hydrocarbon}}{t_{Hydrocarbon}}\times \frac{t_{S{O}_2}}{V_{S{O}_2}}=\sqrt{\frac{M_{S{O}_2}}{M_{Hydrocarbon}}}$ $M_{S{O}_2}=32+32=64$ $\frac{180}{1.5}\times \frac{t_{S{O}_2}}{120}=\sqrt{\frac{64}{16}}$ $\frac{180\times t_{S{O}_2}}{1.5\times 120}=2$ $t_{S{O}_2}=\frac{2\times 1.5\times 120}{180}=2\min .$