Question

In a $0.2$ molal aqueous solution of weak acid $HX$, the degree of ionisation is $0.3$. Taking ${K_f}$ for water as $1.85$, the freezing point of the solution will be nearest to:
A. $ - 0.360^\circ C$
B. $ - 0.260^\circ C$
C. $ + 0.480^\circ C$
D. $ - 0.480^\circ C$

Answer 1

Hint: Freezing point of the solution can be defined as the lowering of the freezing point of the solvents on the addition of solutes. It is categorized as one of the colligative properties of a solution. It is directly related to the molality of solute added in the solution. The formula can be written as: $\Delta {T_f} = i \times {K_f} \times m$
Here, m represents molality
${K_f}$ represent cryoscopic constant
i represents Van’t Hoff factor
$\Delta {T_f}$ represent freezing point depression in the solution

Complete step by step answer:
In the question, we are given with the $0.2$molal aqueous solution of a weak acid and the value of the degree of ionisation is $0.3$
If we ionise a weak acid, it can be represented as follows:
$HX \rightleftharpoons {H^ + } + {X^ - }$
$(1 - 0.3)$ $(0.3)$ $(0.3)$
For this reaction, i can be written as:
$i = \dfrac{{1 - 0.3 + 0.3 + 0.3}}{1}$
$i = 1.3$
Now, we have the value of Van’t Hoff factor i.e. $1.3$
As mentioned the formula used to determine the freezing point depression of the solution is
$\Delta {T_f} = i \times {K_f} \times m$
So, we are given ${K_f}$ for water is $1.85$
After substituting the values in the formula we have
$\Delta {T_f} = 1.3 \times 1.85 \times 0.2$
$\Delta {T_f} = {0.480^\circ }C$
Here, we have to calculate the freezing point of the solution that will be as follows:
${T_f} = 0^\circ C - 0.480^\circ C$
${T_f} = - 0.480^\circ C$
Thus, we can conclude that the freezing point of the solution will be nearest to the $ - 0.480^\circ C$

So, the correct answer is Option D.

Note: There are several reasons for the occurrence of freezing point depression mentioned as follows:
There will be an equilibrium condition between the liquid state and the solid- state at the freezing point, so we can say that vapour pressure of solid and liquid are equal.
When there is the addition of non-volatile solute, the vapour pressure of the solution is determined to be lower than the vapour pressure of the solvent present in the solution. So, the equilibrium point will be attained at a lower temperature between the solid and the solution.