A) \[6.71\]
B) \[14.49\]
C) \[16.94\]
D) \[20\]
Correct Answer: B
Solution :
Mole fraction of urea in its solution \[=\frac{\frac{12}{60}}{\frac{12}{60}+\frac{140.4}{18}}\Rightarrow 0.025\]| Mole fraction of glucose \[=\frac{\frac{18}{180}}{\frac{18}{180}+\frac{178.2}{18}}\Rightarrow 0.01\] |
| \[\because \]Mole fraction of glucose is less so vapour pressure above the glucose solution will be higher than the pressure above urea solution, so some \[{{\operatorname{H}}_{2}}O\]molecules will transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilibrium, let \[x\]moles \[{{\operatorname{H}}_{2}}O\]transferred |
| \[\therefore \]\[\frac{0.2}{0.2+7.8+x}=\frac{0.1}{0.1+9.9-x}\Rightarrow x=4\] |
| Now mass of glucose solution \[\Rightarrow 196.2-4\times 18\Rightarrow 124.2\] |
| wt. % of glucose\[=\frac{18}{124.2}\times 100\Rightarrow 14.49\] |
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