Question

Ethylene glycol (molar mass=$62gmo{l^{ - 1}}$) is a common automobile antifreeze. Calculate the freezing point of a solution containing $12.4g$ of this substance in $100g$ of water. Would it be advisable to keep this substance in the car radiator during summer? Given $kf$ for water=$1.86kkg/mol$ kb for water=$0.512kkg/mol$.

Answer 1

We have to know that depression in any solution is called a colligative property. We can find the freezing point of a solution using the molality, freezing point depression constant/boiling point elevation constant and freezing point depression/boiling point elevation constant. We have to calculate the molality of the solution by converting the grams of solute to moles of solute using molar mass and the grams of water is converted into kilograms of water.

Complete answer:
The freezing point of a solution is lesser when compared to the freezing point of the pure solvent. The reduction of the freezing point is related to the decrease of the vapor pressure of the solvent. This freezing-point depression is directly related to the concentration of solute particles. Molality is the unit of concentration used in freezing-point depression.
We can write the formula as,
$\Delta {T_f} = m{k_f}$
Where, $\Delta {T_f}$ is depression in freezing point
We can represent the molality as $m$.
We can represent the freezing point depression constant as ${k_f}$.
Let us now calculate the molality of the solution.
First, we have to convert the grams to moles of ethylene glycol using molar mass.
$Moles = \dfrac{{Grams}}{{Molarmass}}$
Now we have to substitute the known values we get,
$Moles = \dfrac{{12.4g}}{{62g/mol}}$
On simplification,
$Moles = 0.2mol$
The moles of ethylene glycol are $0.2mol$.
Grams of water are converted into kilograms of water. The kilograms of water is $0.1kg$.
Let us now calculate the molality of the solution as,
$m = \dfrac{{Moles Of Solute}}{{Ki\log rams solvent}}$
Now we have to substitute the known values we get,
$m = \dfrac{{0.2mol}}{{0.1kg}}$
On simplification,
$m = 2m$
We have calculated the molality of the solution as $0.2m$.
Let us calculate the difference in freezing point. We know that the value of molality of the solution and freezing point depression are constant.
$\Delta {T_f} = m{k_f}$
Now we have to substitute the known values we get,
$\Delta {T_f} = \left( {2\,m} \right)\left( {1.86\,\dfrac{K}{m}} \right)$
On simplification,
$\Delta {T_f} = 3.72K$
From this we can calculate the freezing point of the solution as,
${T_f} = 273 - 3.72\,$
${T_f} = 269.28K$
We have calculated the freezing point of the solution as $269.28K$.
Let us calculate the difference in boiling point. We know that the value of the molality of the solution and boiling point depression are constant.
$\Delta {T_b} = m{k_b}$
Now we have to substitute the known values we get,
$\Delta {T_f} = \left( {2\,m} \right)\left( {0.512\,\dfrac{K}{m}} \right)$
On simplification,
$\Delta {T_b} = 1.024K$
From this we can calculate the boiling point of the solution as,
${T_b} = 373 + 1.024$
${T_b} = 374.02K$
We have calculated the boiling point of the solution as $374.024K$.
There would be an increase in boiling point when there is an increase in the percentage of ethylene glycol. When ethylene glycol is used, there could be a decrease in freezing point and increase in boiling point. It behaves as an anti-freezing agent and so could be kept in car radiators during the summer season.

Note:
We have to remember while calculating the molality of the solution, we have to convert the grams of solute to moles of solute using molar mass and also the unit of solvent should be in kilograms. For strong electrolytes like ionic compounds, we have to use van’t Hoff factor for calculating the change in freezing point/change in boiling point. For weak electrolytes, the van't Hoff factor would be one.