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 $KCl$. $K_{f}\left( H _{2} O \right)=1.86\, K\, kg\, mol ^{-1}$.

• 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 $100 g$ of solution. It will contain $5 g$ urea, $1.0 g KCl$ and $10 g$ glucose.

The molar masses of urea, $KCl$ and glucose are $60 g / mol , 74.5 g / mol$ and 180 $g / mol$ respectively.
The number of moles of urea $=\frac{5 g }{60 g / mol }$
The number of moles of $KCl =\frac{1 g }{74.5 g / mol }$
The number of moles of glucose $=\frac{10 g }{180 g / mol }$
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 $100-[5+1+10]=84 g$
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 $273.15 K$.
The freezing point of aqueous solution is
$273.15-3.66=269.48 K$