Question

The amount of ice that will separate on cooling a solution containing $50{\text{g}}$ ethylene glycol in $200{\text{g}}$ water to $- {9.3^{\text{o}}}{\text{C}}$ is:
A.$38.71{\text{g}}$
B.$38.71\,{\text{mg}}$
C.${\text{42g}}$
D.$42{\text{mg}}$

Answer 1

To answer this question, you should recall the concept of lowering of the freezing point of a solvent on the addition of a non-volatile solute as well as basic points about colligative properties. By substituting the appropriate values in the formula (given below) we can calculate the amount of ice that will separate.

Formula used:
$\Delta {\text{T}} = \dfrac{{1000 \times {{\text{K}}_{\text{f}}} \times {\text{w}}}}{{{{W \times m}}}}$
where $\Delta {\text{T}} =$Lowering in freezing point, ${{\text{K}}_{\text{f}}} =$ Cryoscopic constant, ${\text{w}} =$the mass of solute, ${\text{W}} =$ of solvent, ${\text{m}} =$ molar mass of solute

Complete step by step answer:
We know that the term Freezing point depression refers to the lowering of the freezing point of solvents upon the addition of solutes. Colligative properties depend on the presence of dissolved particles and their number but not on the identity of the dissolved particles.
The depression in the freezing point of a solution can be described by the following formula:
$\Delta {{\text{T}}_{\text{F}}}\; = \;{{\text{K}}_{\text{F}}}\; \times \;{\text{b}}\; \times \,{\text{i}}$
where $\Delta {{\text{T}}_{\text{F}}}$ is the freezing-point depression, ${{\text{K}}_{\text{F}}}\;$is cryoscopic constant, ${\text{b}}$ is the molality and ${\text{i}}$ is the van 't Hoff factor.
Upon the addition of a solute which is non-volatile, the vapour pressure of the solution is found to be lower than the vapour pressure of the pure solvent. This causes the solid and the solution to reach equilibrium at lower temperatures. This temperature change can be
$\Delta {\text{T}} = \dfrac{{1000 \times {{\text{K}}_{\text{f}}} \times {\text{w}}}}{{{{W \times m}}}}$
where $\Delta {\text{T}} =$Lowering in freezing point, ${{\text{K}}_{\text{f}}} =$Cryoscopic constant, ${\text{w}} =$the mass of solute, ${\text{W}} =$Mass of solvent, ${\text{m}} =$ molar mass of solute
Substituting the values given in the question:
$\Rightarrow 9.3 = \dfrac{{1000 \times 1.86 \times 50}}{{62 \times {\text{W }}}}$.
After solving we get the value:
${\text{W}} = 161.29$
$\therefore$Ice separated $= 200 - 161.29 = 38.71{\text{g}}$.

Hence, the correct option is A.

Note:
You should know the difference between Association and Dissociation.
In Association the observed molar mass is greater than the predicted value and the value of the Van’t Hoff factor is less than one. The values of the colligative properties are lower than expected.
Example: reduced boiling point and freezing point.
While in Dissociation the observed value of molar mass is lesser than the normal value, the value of i is greater than one. Higher values of colligative properties are observed.