Question

Calculate the freezing point of an aqueous solution containing $10\%$ by mass of glucose, $5\%$ by mass of urea and $1\%$ by mass of $KCl$. $K_f\left(H_{2}O\right)=1.86Kkgmo{l}^{-1}$.

• A. $-2.25^{\circ }C$
• B. $0^{\circ }C$
• C. $-3.67^{\circ }C$
• D. $-1.63^{\circ }C$

Answer 1

Answer: (C)

The aqueous solution contains $5\%$ by mass urea, $1.0\%$ by mass $KCl$ and $10\%$ by mass of glucose.

Suppose we have $100g$ of solution. It will contain $5g$ urea, $1.0gKCl$ and $10g$ glucose.

The molar masses of urea, $KCl$ and glucose are $60g/mol,74.5g/mol$ and 180 $g/mol$ respectively.
The number of moles of urea $=\frac{5g}{60g/mol}$
The number of moles of $KCl=\frac{1g}{74.5g/mol}$
The number of moles of glucose $=\frac{10g}{180g/mol}$
The depression in the freezing point depends on the number of solute particles. Glucose and urea are non electrolytes. So they do not decompose.
$KCl$ is a strong electrolyte. It dissociates in aqueous solution in $K+$ ions and $Cl-$ ions. So number of moles of $KCl$ should be multiplied by 2 .
Mass of solvent is $100-[5+1+10]=84g$
The molality of the solution is the ratio of the total number of moles of solute particles to mass of solvent in $kg$.
The depression in the freezing point $\Delta T_f=K_{f}m=1.86\times 1.97=3.66$
The freezing point of water is $273.15K$.
The freezing point of aqueous solution is
$273.15-3.66=269.48K$