Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm^3 of SO_2 diffuses through the porous partition in 60 seconds…

Question

Sulphur dioxide and oxygen were allowed​ to diffuse through a porous partition. 20 dm^3 of SO_2 diffuses through the porous partition in 60 seconds. The volume of O_2 in dm^3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur= 32 u);

MARKS Content Team · Accepted Answer

Correct answer: (C) 14.1. As per Graham's law, (V_(SO 2) / t_(SO 2))/(V_(O 2) / t_(O 2)) = (p_(SO 2))/(p_(O 2)) √(M_(O 2) M_(SO 2)) = √(M_(O 2) M_(SO 2)) (2 0 dm^3 / 6 0 sec)/(V_(O 2) / 3 0 sec) = √(3 2 6 4) = 1/(1 · 4 1 4) (2 0 × 3 0)/(6 0 × V_(O 2)) = 1/(1 · 4 1 4) ⇒ V_(O 2) = 1 0 × 1 · 4 1 4 = 14.14 dm^3

Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20 {dm}^{3}$ of ${SO}_{2}$ diffuses through the porous partition in $60$ seconds. The volume of $O_{2}$ in ${dm}^{3}$ which diffuses under the similar condition in $30$ seconds will be (atomic mass of sulphur $= 32 u$ );

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Sulphur dioxide and oxygen were allowed​ to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur= 32 u);

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