Question

The freezing point of a solution containing $50{\text{ c}}{{\text{m}}^3}$ of ethylene glycol in 50 g of water is found to be $- {4^\circ }{\text{C}}$. Assuming ideal behavior, the density of ethylene glycol
$({{\text{K}}_{\text{f}}}{\text{ for water }} = {\text{ }}1.86{\text{ KKg mo}}{{\text{l}}^{ - 1}})$ is:
(Atomic weight of ${\text{Kr}} = 84$ )
A) $1.133{\text{ g/c}}{{\text{m}}^3}$
B) $4.433{\text{ g/c}}{{\text{m}}^3}$
C) ${\text{3}}{\text{.235 g/c}}{{\text{m}}^3}$
D) None of the above

Answer 1

For this we must use the formula for depression in freezing point. Taking d as density we will calculate mass of ethylene glycol and substitute in the formula to get the value. The concentration term used here is molality.

Formula used:
$\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}} \times {\text{m}}$ where $\Delta {{\text{T}}_{\text{f}}}$ is depression in freezing point ${{\text{K}}_{\text{f}}}$ is a constant and ${\text{m}}$ is molality of solution.
${\text{number of moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step solution:
Molality is defined as the ratio of number of moles of solute to the mass of solution in kilogram.
Our solute is ethylene glycol and solvent is water. First of all we need to calculate the number of moles of ethylene glycol.
Let the density of ethylene glycol is d, and volume is given to us as $50{\text{ c}}{{\text{m}}^3}$. So using the formula:
${\text{density }} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}$
We will get${\text{mass }} = 50 \times {\text{d}}$.
Molecular mass of ethylene glycol is $62{\text{ g}}/{\text{mol}}$ can be calculated using the molecular formula ${{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2}$ .
Hence number of moles will be:
${\text{number of moles}} = \dfrac{{{\text{50}} \times {\text{d }}}}{{{\text{62}}}}$
Mass of water is given to us that is 50 g. hence using the definition molality will be:
${\text{m}} = \dfrac{{{\text{50}} \times {\text{d}} \times 1000{\text{ }}}}{{{\text{62}} \times 50}}$
We have used 1000 to convert mass of water in Kg.
Now we will use the formula for depression in freezing point and substitute the given functions:
$\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}} \times {\text{m}}$
In the question we have been given the freezing point not the depression in freezing point. We know the freezing point of water is ${0^\circ }{\text{C}}$. So:
$\Delta {{\text{T}}_{\text{f}}} = {0^\circ }{\text{C}} - ( - {4^\circ }{\text{C)}} = {4^\circ }{\text{C}}$
$\Rightarrow 4 = 1.86 \times \dfrac{{{\text{50}} \times {\text{d}} \times 1000{\text{ }}}}{{{\text{62}} \times 50}}$
We can rearrange the equation as:
${\text{d = }}\dfrac{{4 \times {\text{62}} \times 50}}{{1.86 \times {\text{50}} \times 1000}}{\text{g}}/{\text{ml}}$
$\Rightarrow {\text{d}} = 0.133{\text{ g}}/{\text{ml}}$
This does not match any of the options and hence, the correct option is D.

Note:
The above question is actually an application of depression in freezing point, ethylene glycol and water forms an anti freeze mixture. This is used in the car and other engines to prevent freezing of water which may cause blast.