Question

25 ml of an aqueous solution of KCl was found to require 20 ml of 1M $AgN{{O}_{3}}$solution when titrated using a ${{K}_{2}}Cr{{O}_{4}}$as indicator. The depression in freezing point of KCl solution with 100% ionization will be:
A. ${{5.0}^{o}}$
B. ${{3.2}^{o}}$
C. ${{1.6}^{o}}$
D. ${{0.8}^{o}}$

Answer 1

Depression in freezing-point depression is the decrease in freezing point of a solvent on an addition of a non-volatile solute. For example salt in water and addition of two solids such as impurities into a finely purified powdered drug.
We can calculate the depression in freezing point using the below formula.
$\Delta {{T}_{f}}=m\times {{K}_{f}}\times i$
Where ${{K}_{f}}$ = molal freezing point depression constant
i = Vant’s Hoff factor
m = Molality
$\Delta {{T}_{f}}$ = Depression in freezing point

Complete step by step answer:
In the question it is given that 25 ml of an aqueous solution of KCl requires to neutralize 20 ml of 1 M $AgN{{O}_{3}}$.
In the question the data given is as follows.
Volume of KCl solution ${{V}_{1}}$ = 25 ml
Volume of $AgN{{O}_{3}}$solution ${{V}_{2}}$= 20 ml
Molarity of $AgN{{O}_{3}}$solution ${{M}_{2}}$= 1 M
Molarity of KCl solution ${{M}_{1}}$ =?
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
Substitute all the known values in the above equation to calculate the molarity of KCl solution (${{M}_{1}}$)
$\begin{aligned}& {{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}} \\\ & {{M}_{1}}=\frac{{{M}_{2}}{{V}_{2}}}{{{V}_{1}}} \\\ & {{M}_{1}}=\frac{1\times 20}{25} \\\ & \text{ = }0.8M \\\ \end{aligned}$
Molarity of KCl solution ${{M}_{1}}$ = 0.8M
In the case of KCl the molarity of the solution is equal to the molality of the solution.
(The molality of KCl solution and water solution is the same (1.3 m). The solution is very dilute so the molality is almost equal to the molarity of the solution)
Therefore m = 0.8 M.
Now substitute the molality of KCl in the following to calculate the depression in freezing point of the KCl
$\Delta {{T}_{f}}=m\times {{K}_{f}}\times i$
Where ${{K}_{f}}$ = molal freezing point depression constant
i = Vant’s Hoff factor (for$KCl\to {{K}^{+}}+C{{l}^{-}}$, i = 2, because two ions are formed after ionization)
m = Molality
$\Delta {{T}_{f}}$ = Depression in freezing point

& \Delta {{T}_{f}}=m\times {{K}_{f}}\times i \\\ & \text{ = 0}\text{.8}\times 2\times 2 \\\ & \text{ = 3}\text{.2} \\\ \end{aligned}$$ Therefore the depression in freezing point of KCl solution = 3.2. **So, the correct option is B.** **Note:** Don’t be confused with Molarity and molality. Both are not the same. Molarity: number of gram molecular weight of the solute present in the 1 liter of the solvent. Molality: number of gram molecular weight of the solute present in the 1 kg of the solvent.