Question

A 0.1 molal solution of a monobasic acid is 45% ionized. Calculate the depression in freezing point:
(Molecular weight of acid = 300g/mol, ${{K}_{f}}=1.86Kmo{{l}^{-1}}Kg$ )
(A) $-{{0.199}^{o}}C$
(B) ${{2.00}^{o}}C$
(C) ${{0}^{o}}C$
(D) ${{0.269}^{o}}C$

Answer 1

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

Complete 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 vapour pressure.
Those are,
(i) Relative lowering of the vapour pressure of the solvent
(ii) Depression of freezing point of the solvent
(iii) Elevation of the boiling point of the solvent
(iv) Osmotic pressure
Let us discuss depression and freezing point.
The lowering of 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 pure solvent, the freezing point of the solvent decreases.
Consider the freezing point of the pure solvent and freezing points of a solution when a non-volatile is added are ${{T}_{f}}^{o}\And {{T}_{f}}$ respectively.
The depression in the freezing point is, $\Delta T={{T}_{f}}^{o}-{{T}_{f}}$
As mentioned above, this property is proportional to the molality of the solution. Thus,
$\Delta {{T}_{f}}={{K}_{f}}m$ --- (1)
Where ${{K}_{f}}$ = freezing point depression constant
Let us consider monobasic acid given,
$HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}$
Given that this monobasic acid ionized = 45%
Then $\alpha =0.45$
Molality of monobasic acid = 0.1m
${{K}_{f}}=1.86Kmo{{l}^{-1}}Kg$
Substitute the above values in the equation (1)
$\Delta {{T}_{f}}=0.1\times 1.86={{0.186}^{o}}C$
Experimental depression freezing point, $\Delta {{T}_{\exp }}={{T}_{f}}(1+\alpha )=0.186(1+0.45)={{0.269}^{0}}C$

Hence, the correct answer is option D.

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.