Step 1

A colligative property is a property that depends on the number number of moles of solute, not on the nature of the solute. Depression in freezing point is a colligative property and it is calculated by using the formula

$\begin{array}{rcl} ∆ T_{f} & = & K_{f} m . \\ ({T^{o}}_{f} - T_{f}) & = & K_{f} m . \end{array}$

Whereas,

K_f is the depression in freezing point constant.

m is the molality.

T^o_f is the freezing point of a pure substance.

T_f is the freezing point of the solution.

Step 2

To determine the freezing point of chloroform in CCl₄.

Given data

Moles of CHCl₃=1.15 mol.

Mass of solvent=530.0 g.

K_f=29.8 °C/m.

T^o_f = -22.9 °C

Using depression in freezing point equation.

$\begin{array}{rcl} ∆ T_{f} & = & K_{f} m . \\ ({T^{o}}_{f} - T_{f}) & = & K_{f} m . \\ (- 22.9^{o} C - T_{f}) & = & 29.8^{o} C / m \times \frac{n_{{CHCl}_{3}}}{W_{{CCl}_{4}}} \times 1000 . \\ (- 22.9^{o} C - T_{f}) & = & 29.8^{o} C / m \times \frac{1.15 mol}{530.0 g} \times 1000 . \\ (- 22.9^{o} C - T_{f}) & = & 64.7^{o} C . \\ T_{f} & = & - 64.7^{o} C - 22.9^{o} C . \\ T_{f} & = & - 87 . 6^{o} C . \end{array}$

Step 3

Answer

The freezing point of the solution is -87.6 ^oC.

What is the freezing point, in °C, of a solution made with 1.15 mol of CHCl₃ in 530.0 g of CCl₄ (Kf = 29.8 °C/m, normal freezing point, Tf = -22.9 °C)?

Public Answer