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# A solution of urea in water has a boiling point of ${100.18^ \circ }C$. Calculate the freezing point of the solution. (${K_f}$ for water is $1.86KKgmo{l^{ - 1}}$ and ${K_b}$ for water is $0.512K Kgmo{l^{ - 1}}$).

Last updated date: 13th Jun 2024
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Hint: Boiling and freezing point of any substance is altered by the presence of impurities. Boiling point of water is increased due to the presence of impurities and freezing point of substance will be lower than that of pure substance.

Formula used: $\Delta T = {K_b} \times {\text{molality = }}{{\text{K}}_f} \times {\text{molality}}$
${K_f}$ for water is $1.86KKgmo{l^{ - 1}}$
${K_b}$ for water is $0.512KKgmo{l^{ - 1}}$
Where, $\Delta T$ is change in temperature.

${K_b}$ is ebullioscopic constant. This constant relates the molality of a substance with elevation in boiling point. It depends on the properties of solvent. It can be calculated by the following formula:
${K_b} = \dfrac{{RT_b^2M}}{{\Delta {H_v}}}$
Where, $\Delta {H_v}$ is heat of vaporization of per mole of the solvent, $R$ is gas constant, ${T_b}$ is boiling point of pure solvent and $M$ is molar mass of the solvent.
${K_f}$ is a cryoscopic constant. This constant relates the molality of a substance with depression in freezing point. This depends on the properties of solvent. This constant can be calculated with the following formula:
${K_f} = \dfrac{{RT_f^2M}}{{\Delta {H_{fus}}}}$
Where, $\Delta {H_{fus}}$ is enthalpy of fusion per mole of the solvent, $R$ is gas constant, ${T_f}$ is freezing point of pure solvent and $M$ is molar mass of the solvent
In this question urea is added in water, this means impurity is urea. Impurity is added in water this means the boiling point of water will increase. And this boiling point is given that is ${100.18^ \circ }C$. Boiling point of pure water is ${100^ \circ }C$. From these two temperatures we can calculate change in boiling point (subtracting boiling point of pure water from boiling point of water after adding impurities) that is,
Change in boiling point$= \Delta T = {100.18^ \circ }C - {100^ \circ }C = {0.18^ \circ }C$
Also change in boiling point is ${K_b} \times {\text{molality}}$
${K_b}$ for water is $0.512KKgmo{l^{ - 1}}$
From this we can calculate molality. So applying the formula:
$\Delta T = {K_b} \times {\text{molality}}$
${\text{molality = }}\dfrac{{\Delta T}}{{{K_b}}}$
${\text{molality = }}\dfrac{{0.18}}{{0.512}} = 0.35$
Now, we have the molality. From this we can calculate the change in freezing point with the formula,
$\Delta T = {K_f} \times {\text{molality}}$
${K_f}$ for water is $1.86KKgmo{l^{ - 1}}$ and molality we have calculated above. So,
$\Delta T = 1.86 \times {\text{0}}{\text{.35 = 0}}{\text{.651}}$
Change in freezing point is ${0.651^ \circ }C$. Freezing point of water is ${0^ \circ }C$. From these two temperatures we can calculate the freezing point of the new solution (subtracting change in freezing point from actual freezing point). So, freezing point of solution is,
Freezing point of solution$= 0 - 0.651 = - {0.651^ \circ }C$.

Note:
There is depression in freezing point after adding impurities because, on adding impurities vapour pressure becomes lower than that of pure solution. Boiling point of liquid increases on adding impurities because addition of impurities lower the number of molecules of liquid above the surface available to become vapourized during boiling.