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):
(a) 7.09
(b) 14.1
(c) 10.0
(d) 28.2
Correct answer: (b) 14.1
Explanation:
According to Graham’s Law Diffusion:
\(\frac{\eta}{r_2}\) = \(\sqrt{\frac{d_2}{d_1}}\)
Since rate of diffusion
= \(\frac{Vol. \;of\;gas\;diffused(V)}{Time\;taken\;for\;diffusion(t)}\)
∴ \(\frac{\eta}{r_2}\) = \(\frac{V_1/t_1}{V_2/t_2}\)
or \(\frac{r_1}{r_2}\) = \(\frac{V_1/t_1}{V_2/t_2}\) = \(\sqrt{\frac{d_2}{d_1}}\)
= \(\frac{20/60}{V_2/30}\) = \(\sqrt{\frac{\frac{16}{2}}{\frac{32}{2}}}\)
= \(\sqrt{\frac{1}{2}}\)
∵ Mol. wt = 2 × V.D
∴ V.D = \(\frac{Mol.wt}{2}\)
On calculating,
V2 = 14.1