Question

The osmotic pressure of an aqueous solution at 300 K is 120 kPa. Calculate the freezing point of the solution.

Answer 1

The relation between the osmotic pressure and the depression in freezing point can be given, as follows:

$$ \Delta \mathrm{T}=\frac{\mathrm{K}_{\mathrm{f}} \pi}{\mathrm{RT} \rho_{\mathrm{H}_{2} \mathrm{O}}} $$

Here \(\pi\) is the osmotic pressure and \(\mathrm{K}_{\mathrm{f}}\) is the freezing point constant of the solution, \(\rho_{\mathrm{H}_{2} \mathrm{O}}\) is the density of water, \(\mathrm{R}\) is the universal gas constant, \(\mathrm{T}\) is the given temperature in Kelvin, and \(\Delta \mathrm{T}\) is the depression in freezing point.

Given that, \(\pi=120 \mathrm{kPa}, \mathrm{T}=300 \mathrm{~K}\)

Calculate the concentration depression in freezing point, as follows:

$$ \begin{aligned} \Delta \mathrm{T} &=\frac{\mathrm{K}_{i} \pi}{\mathrm{RT} \rho_{\mathrm{H}_{2} \mathrm{O}}} \\ &=\frac{1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} \cdot \mathrm{mol}^{-1} \times 120 \mathrm{kPa}}{8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \times 1000 \mathrm{~kg} \cdot \mathrm{m}^{-3}} \\ &=\frac{1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} \cdot \mathrm{mol}^{-1} \times 120 \times 1000 \mathrm{~J} \cdot \mathrm{m}^{-3}}{8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K} \times 1000 \mathrm{~kg} \cdot \mathrm{m}^{-3}} \\ &=0.0895^{\circ} \mathrm{C} \end{aligned} $$

The \(\Delta \mathrm{T}\) is the difference in actual freezing point and the observed freezing point. That is, \(\Delta \mathrm{T}=\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{f}}^{\prime}\)

Here, \(\mathrm{T}_{\mathrm{f}}\) is the actual freezing point and \(\mathrm{T}_{\mathrm{f}}^{\prime}\) is the observed freezing point.

As it is known, the freezing point of water is \(\mathrm{T}_{\mathrm{f}}=0^{\circ} \mathrm{C}\).

Calculate the observed freezing point, as follows:

$$ \begin{aligned} \Delta \mathrm{T} &=\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{f}}^{\prime} \\ 0.0895{ }^{\circ} \mathrm{C} &=0.0^{\circ} \mathrm{C}-\mathrm{T}_{\mathrm{f}}^{\prime} \\ \mathrm{T}_{\mathrm{f}}^{\prime} &=0.0^{\circ} \mathrm{C}-0.0895{ }^{\circ} \mathrm{C} \\ \mathrm{T}_{\mathrm{f}}^{\prime} &=-0.0895^{\circ} \mathrm{C} \end{aligned} $$

Hence, the freezing point of the solution is \(-0.0895{ }^{\circ} \mathrm{C}\).