Question

Two beakers $A$ and $B$ present in a closed vessel. Beaker $A$ contains $152.4g$ aqueous solution of urea, containing $12g$ of urea. Beaker $B$ contains $196.2g$ glucose solution, containing $18g$ of glucose. Both the solutions allowed to attain the equilibrium. Determine weight $\%$ of glucose in its solution is at equilibrium.

• A. $25\%$
• B. $10\%$
• C. $18\%$
• D. $15.5\%$

Answer 1

Answer: (D)

$15.5\%$

The molar masses of glucose and urea are 180 g/mol and 60 g/mol respectively.
The number of moles of urea are $\frac{12}{60}=0.2$
The number of moles of glucose are $\frac{18}{180}=0.1$
When equilibrium is achieved, the mass ($196.2g$) of glucose solution will reduce by x g and the mass of urea ($152.4g$) will increase by x g. Due to this, the concentration (ratio of number of moles of solute to the mass of solution) of urea solution will become equal to the concentration of glucose solution.
$\frac{0.2}{152.4+x}=\frac{0.1}{196.2-x}152.4+x=392.4-2x3x=240x=80$
Total mass of glucose solution $=196.2-80=116.2$ g
weight percent of glucose $=\frac{18}{116.2}\times 100=15.5$ %