Chemistry - Liquid Solution Question with Solution | TestHub

Mole fraction of urea in its solution

$=\frac{\frac{12}{60}}{\frac{12}{60}+\frac{140.4}{18}}=0.025$

Mole fraction of glucose

$=\frac{\frac{18}{180}}{\frac{18}{180}+\frac{178.2}{18}}=0.01$

Mole fraction of glucose is less so vapour pressure above the glucose solution will be higher than the pressure above urea solution, so some${\text{H}}_2\text{O}$molecules will transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilibrium, let$\text{x}$moles${\text{H}}_2\text{O}$transferred

$\therefore \frac{0.2}{0.2+7.8+\mathrm{\text{x}}}=\frac{0.1}{0.1+9.9-\mathrm{\text{x}}}\Rightarrow \mathrm{\text{x}}=4$

Now mass of glucose solution

$\Rightarrow 196.2-4\times 18=124.2$

wt% of glucose$=\frac{18}{124.2}\times 100=14.49$