A) 7.09
B) 14.1
C) 10.0
D) 28.2
Correct Answer: B
Solution :
According to Graham?s Law Diffusion: \[\frac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] Since rate of diffusion = \[\text{=}\frac{\text{Vol}\text{. of gas diffused }\left( \text{V} \right)}{\text{Time}\,\text{taken}\,\text{for}\,\text{diffusion}\,\text{(t)}}\] \[\therefore \] \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}\]or \[\frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{V}_{1}}/{{t}_{1}}}{{{V}_{2}}/{{t}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] \[=\frac{20/60}{{{V}_{2}}/30}=\sqrt{\frac{16/2}{32/2}}=\sqrt{\frac{1}{2}}\] \[\because \]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.
You will be redirected in
3 sec