Question

The freezing point ( in $^{\circ }C$) of a solution containing 0.1 g of ${{K}_{3}}[Fe{{(CN)}_{6}}]$ (Mol. Wt. 329) in 100 g of water (${{K}_{f}}$ = ${{1.86}^{\circ }}C\text{ kg/mol}$) is:
(A) $-2.3\text{ x 1}{{\text{0}}^{-2}}$
(B) $-5.7\text{ x 1}{{\text{0}}^{-2}}$
(C) $-5.7\text{ x 1}{{\text{0}}^{-3}}$
(D) $-1.2\text{ x 1}{{\text{0}}^{-2}}$

Answer 1

First calculate the i for the reaction. The mass of hydrochloric acid can be calculated by the formula of depression in freezing point,$\Delta {{T}_{f}}=i\ \text{x }{{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}$ , $\Delta {{T}_{f}}$ is the depression in freezing point, ${{w}_{2}}$ is the mass of the solute in gram, ${{M}_{2}}$ is the molar mass of the solute, and ${{w}_{1}}$is mass of the solvent in gram.

Complete step by step solution:
Let us see what is the depression in freezing point.
The temperature at which there is equilibrium between the solid and the liquid form of the substance i.eThe solid and liquid form has the same vapour pressure is known as the freezing point of the substance. It is observed that the freezing point of the solution is always lower than that of pure solvent; this is known as the depression in freezing point.
The depression of freezing point depends on the solute in a solution and has found to be related to the molality’ as below:
$\Delta {{T}_{f}}={{K}_{f}}\text{ x }m$
${{K}_{f}}$ is the molal depression constant.
So, by expanding the above formula, we get:
$\Rightarrow \Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}$
Since, ${{K}_{3}}[Fe{{(CN)}_{6}}]$ionizes, the above formula will become:
$\Rightarrow \Delta {{T}_{f}}=i\ \text{x }{{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}$
Where, $\Delta {{T}_{f}}$ is the depression in freezing point,
${{w}_{2}}$ is the mass of the solute in gram,
${{M}_{2}}$ is the molar mass of the solute,
${{w}_{1}}$ is the mass of the solvent in gram.
For the value of $i$:
$\Rightarrow {{K}_{3}}[Fe{{(CN)}_{6}}]\to 3{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{3-}}$
$i$ is $4$
So, according to the question:
${{w}_{1}}$ is the mass of solute = $0.1 g$
${{M}_{2}}$ = Molar mass of the solute = $329$
${{w}_{1}}$ = Mass of the solvent (water) = $100 g$
Now putting all these in the formula:
$\Rightarrow \Delta {{T}_{f}}=4\ \text{x 1}\text{.86 x }\dfrac{0.1}{329}\text{ x }\dfrac{1000}{100}$
$\Rightarrow \Delta {{T}_{f}}=2.3\text{ x 1}{{\text{0}}^{-2}}$
The $\Delta {{T}_{f}}$ is calculated as the difference of freezing point of water to the solution.
$\Rightarrow \Delta {{T}_{f}}={{T}_{f}}^{\circ }-{{T}_{f}}$
So ${{T}_{f}}$ will be: ${{T}_{f}}={{T}_{f}}^{\circ }-\Delta {{T}_{f}}$
$\Rightarrow {{T}_{f}}$= $0-2.3\text{ x 1}{{\text{0}}^{-2}}=-2.3\text{ x 1}{{\text{0}}^{-2}}^{\circ }C$

So, the correct answer is an option (A) $-2.3\text{ x 1}{{\text{0}}^{-2}}$.

Note: The mass of the solvent should always be taken in grams, if it is given in litres, then convert it into grams. ${{K}_{f}}$ is also called the cryoscopic constant of the solvent. Depression in freezing point is a colligative property because it depends on the number of moles of the solute.