Question

180 ml of hydrocarbon having the molecular weight 16 diffuses in 1.5 Under similar conditions time taken by 120ml of SO2 to diffuse is​

Answer 1

time taken by 120 ml of SO2 to diffuse is 2 min

explanation : applying Graham's law of diffusion

$\frac{V_1t_2}{V_2t_1}=\sqrt{\frac{M_2}{{M_1}}$, where V is volume, t is time and M is molecular weight of gases.

given, V1 = 180 ml, V2 = 120ml, t1 = 1.5 min , M1 = 16 and M2 = 64 (as molecular weight of SO2 is 64 amu )

so, (180ml × t2)/(120ml × 1.5min) = √{64/16}

⇒3t2/(2 × 1.5) = √(4) = 2

⇒t2 = 2 min

hence, time taken by 120 ml of SO2 to diffuse is 2 min

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