Question

An aqueous solution containing 5% by weight urea and 10% by weight glucose. What will be its freezing point?$({K_f} = {1.86^{\circ}})$

Answer 1

At freezing point liquid and solid are in equilibrium. At the freezing point the vapour pressure of liquid and solid will be the same. Additions of a non-volatile solute, decreases the vapour pressure of pure liquid. This phenomenon is called depression in freezing point. $T_0$ and $T_1$ represent freezing points of solvent and solution respectively, and $\Delta {T_f}$, represents depression in freezing can be written
As

\Delta {T_f} = {\text{ }}{K_f}\; \times {\text{ }}m\;\;\;\; $$ $$\Rightarrow {K_f} = \dfrac{{{\text{MRT}}_{\text{f}}^{\text{2}}}}{{{\text{1000}}\,\Delta {H_v}}}

${{{K}}_{{f}}}$ is defined as the depression in freezing point of a solvent when 1 mole of a non-volatile solute is dissolved in 1000 gm of solvent. ${{{K}}_{{f}}}$ is also known as cryoscopic constant.

Formula Used:
1. $\Delta {T_f} = \dfrac{{{K_f} \times w \times 1000}}{{MW}}$
where w = Mass of solute taken
M = Molecular weight of solute
W = Mass of solvent taken in grams
2. $\Delta {T_f} = \left( {{T_0} - {T_1}} \right)$
3. $\Delta {T_f} \propto m$

Complete answer:
Urea =$5g$ ; Glucose =$10{\text{ }}g$
Weight of solute = $15{\text{ }}g$
Weight of solvent =$100 - 15{\text{ }} = 85{\text{ }}g$
${m_{urea}} = \dfrac{{{W_2} \times 1000}}{{M{W_2} \times {W_1}}} = \dfrac{{5 \times 1000}}{{60 \times 85}} = 0.98$
${m_{glu\cos e}} = \dfrac{{10 \times 1000}}{{180 \times 85}} = 0.653$
$\Delta {T_f} = {K_f} \times m$ = $\text{1.86(0.98+0.653)=3.037}$
$\text{Freezing point = 273-3.037=270.11 K}$

Therefore Freezing point of solution will be 270.11 K

Note:
Colligative properties are those which depend entirely upon the number of particles of the solute contained in a known volume of a given solvent and not at all upon mass and the nature (i.e., chemical composition or constitution) of the solute. Here we shall consider four properties of solution containing non volatile solutes.