Question

Two beaker A and B present in a closed vessel. Beaker A contains $152.4 \, g$ aqueous solution of urea, containing $12 \, g$ of urea. Beaker B contains 196.2 g glucose solution, containing 18 g of glucose. Both solutions allowed to attain the equilibrium. Determine wt. % of glucose in it's solution at equilibrium:

• A. 6.71
• B. 14.49
• C. 16.94
• D. 20

Answer 1

Answer: (B)

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

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

Now mass of glucose solution

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

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