Question

The latent heat of fusion of ice is 80 calories per gram at ${{O}^{o}}C$ . what is the freezing point of a solution of KCl in water containing 7.45 grams of solute 500 grams of water, assuming the salt is dissociated to the extent of 95%?
(A) ${{T}_{f}}=-{{0.73}^{o}}C$
(B) ${{T}_{f}}=-{{0.92}^{o}}C$
(C) ${{T}_{f}}=-{{1.46}^{o}}C$
(D) ${{T}_{f}}=-{{1.84}^{o}}C$

Answer 1

Hint: The freezing point of a substance is the dynamic equilibrium of a solid phase with the liquid phase. It is defined as the temperature at which the vapor pressure in the liquid phase is equal to vapor pressure in the solid phase. If a solute is added to the solvent, vapor pressure decreases, simultaneously freezing point also decreases.

Complete step by step answer:
The Vapour pressure of the solution decreases when a volatile solute is added to that volatile solvent. There are many properties of solutions depending on this decrease in vapor pressure.
Those are,
(1) Relative lowering of the vapor pressure of the solvent
(2) Depression of freezing point of the solvent
(3) Elevation of the boiling point of the solvent
(4) Osmotic pressure

Let us discuss depression and freezing point.
The lowering of the vapour pressure of solution changes to a low freezing point compared to the pure solvent.
According to Raoult’s law, when a non-volatile solute is added to the pure solvent, the freezing point of the solvent decreases.

Consider the freezing point of the pure solvent and freezing points of a solution when non-volatile is added are ${{T}^{o}}_{f}\ and {{T}_{f}}$ respectively.
The depression in the freezing point is, $\Delta T={{T}^{o}}_{f}-{{T}_{f}}$
As mentioned above, this property is proportional to the molality of the solution. Thus,
$\begin{align}
& \Delta T\alpha m \\
& \Delta T={{K}_{f}}m \\
\end{align}$
Where ${{K}_{f}}$ = freezing point depression constant or cryoscopic constant.

This constant dependence upon the nature of the solvent then the above equation can be expressed as,
${{K}_{f}}({{H}_{2}}O)=\dfrac{R{{T}^{2}}_{f}{{M}_{solvent}}}{\Delta {{H}_{f}}mX1000}$

Substitute the given values of latent heat of fusion = 80 cal per gram, R =2 cal/K-mole
${{K}_{f}}({{H}_{2}}O)=\dfrac{2X{{(273.15)}^{2}}X18}{80X18X1000}=1.865K.Kg/mol$
The depression in the freezing point is, $\Delta T={{T}^{o}}_{f}-{{T}_{f}}$
$\begin{align}
& {{T}^{o}}_{f}-{{T}_{f}}=i{{K}_{f}}m \\
& {{T}_{f}}={{T}^{o}}_{f}-i{{K}_{f}}m={{0}^{o}}C-1.865X1.95(\dfrac{\dfrac{7.45}{74.5}}{0.5Kg})=-{{0.73}^{o}}C \\
\end{align}$
Therefore, the freezing of point of water when KCl is added, ${{T}_{f}}=-{{0.73}^{o}}C$
So, the correct answer is “Option A”.

Note: All of these properties mentioned are known as colligative properties which are the number of solute particles irrespective of their nature related to the total number of particles present in the solution. These properties are used to determine the molar mass of the solutes.