Question

If I cool a solution containing 62g of ethylene glycol in 250g water to -9.3°C, what mass of ice will separate out?

Answer 1

To answer the question we need to first understand the freezing point depression. Depression in freezing point is an example of a colligative property i.e. which depends on the concentration of solute and is independent of its identity. There is a depression in the value of the freezing point that occurs whenever we add a non-volatile solute to it.

Complete answer:
Let us first discuss the formulas to be used in the question-
The depression in freezing point is given by the formula-
$\Delta {T_f} = {K_f} \times m - - - - (1)$
Where, $\Delta {T_f}$ is the depression in freezing point which is given by the formula-
$\Delta {T_f} = {T_f}^\circ - {T_f} - - - - (2)$
Here ${T_f}^\circ$ is the freezing point of pure solvent i.e. solvent without any solute.
${T_f}$ is the freezing point of solution i.e. the solvent on adding non-volatile solute.
${K_f}$ is the cryoscopic constant or freezing point depression constant or molal depression constant that depends on the nature of the solvent.
$m$ is the molality
Now let us see how to calculate the Molality of the solution-
Molality $\left( m \right)$ is the number of moles of solute per kilogram of solvent. It is expressed as-
Number of moles/Mass of solvent in kg and;
Here number of moles=given mass/molar mass.
Given in the question we have-
${T_f} = - 9.3^\circ C = - 9.3 + 273K = 263.7K$ (Freezing point of a solution of ethylene glycol)
Here pure solvent is water whose freezing point is $0^\circ C$ or $273K$
${T_f}^\circ = 273K$ (Freezing point of pure solvent water)
Mass of solute=62g
Mass of solvent=250g (water)
Let us calculate molality $\left( m \right)$ -
Here the number of moles is given by-
$= \dfrac{{62g}}{{62g/mole}} = 1mole$
(Since the molar mass of ethylene glycol is 62g/mole)
Molality is calculated as-
$\Rightarrow m = \dfrac{{1mole}}{W} = \dfrac{1}{W}moles$
The value of cryoscopic constant for ethylene glycol; ${K_f} = 1.86Kkg/mole$
Putting these values in equation (1) we get-
$\Rightarrow \left( {273 - 263.7} \right)K = 1.86Kkg/mol \times \dfrac{1}{W}mole$
$9.3K = 1.86Kkg/mol \times \dfrac{1}{W}mole$
$\Rightarrow W = \dfrac{{1.86}}{{9.3}} = 0.2kg = 200g$
Since we started the experiment with 250g of water and the amount remaining is 200g so the rest amount will be converted into ice i.e. (250-200)g=50g.

Note:
Depression in freezing point is an important colligative property for real-life applications. The addition of salt to water decreases its freezing point. Thus when we add salt to an icy road it melts very fast at low temperatures and thus making it safe for drivers. This principle is also used while making ice-cream. In this case, the freezing point is lowered as a result of the addition of sugar as a solvent in the water of ice-cream.