Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. \(20\ \mathrm{dm^3}\) of \(\mathrm{SO_2}\) diffuses through the porous partition in 60 seconds. The volume of \(\mathrm{O_2}\) in \(\mathrm{dm^3}\) which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u) :
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- 7.09
- 14.1
- 10.0
- 28.2
The Correct Option is B
Solution and Explanation
Graham's law: \(\dfrac{r_{O_2}}{r_{SO_2}}=\sqrt{\dfrac{64}{32}}=\sqrt2\).
\(r_{O_2}=\sqrt2/3=0.4714\ \mathrm{dm^3\,s^{-1}}\).
Volume in 30 s \(=0.4714\times30=14.14\ \mathrm{dm^3}\approx14.1\).
Answer: (2) 14.1
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