A) 14.1 \[d{{m}^{3}}\]
B) 16.1 \[d{{m}^{3}}\]
C) 18.1 \[d{{m}^{3}}\]
D) 20.1 \[d{{m}^{3}}\]
Correct Answer: A
Solution :
From Graham law of diffusion, \[\frac{\left( \frac{dV}{dt} \right)}{\left( \frac{dV}{dt} \right)}=\sqrt{\frac{{{M}_{S{{O}_{2}}}}}{{{M}_{{{O}_{2}}}}}}\] \[\frac{\frac{20}{60}}{V}=\sqrt{\frac{64}{32}}=\sqrt{2}\] \[\frac{20}{60}\times \frac{60}{V}=\sqrt{2}\] \[\Rightarrow \] \[\frac{20}{V}=\sqrt{2}\] \[V=\frac{20}{\sqrt{2}}=141\,d{{m}^{3}}\]
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