A) 7.09
B) 14.1
C) 10.0
D) 28.2
Correct Answer: B
Solution :
[b] According to Graham's Law Diffusion: \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] Since rate of diffusion \[=\frac{Vol.\text{ }of\text{ }gas\text{ }diffused\left( V \right)}{Time\,taken\,for\,diffusion(t)}\] \[\therefore \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}\] Or \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] \[=\frac{20/60}{{{V}_{2}}/30}=\sqrt{\frac{16/2}{32/2}}=\sqrt{\frac{1}{2}}\] \[\therefore \] Mol. \[wt=2\times V.D\] \[\therefore V.D=\frac{Mol.wt}{2}\] On calculating, \[{{V}_{2}}=14.1\]You need to login to perform this action.
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