1. Find moles of each component

Given:

• Water $=2.7\text{ kg}=2700\text{ g}$
• Acetic acid $=2.7\text{ kg}=2700\text{ g}$

Moles of water: $n_{\mathrm{H_2O}}=\frac{2700}{18}=150\text{ mol}$

Moles of acetic acid: $n_{\mathrm{CH_3COOH}}=\frac{2700}{60}=45\text{ mol}$

2. Decide which component acts as solvent

To know the freezing point of the mixture, compare the depression in freezing point from each pure component side.

• If water is solvent, acetic acid is solute.
• If acetic acid is solvent, water is solute.

The component that gives the higher freezing point for the mixture will crystallize first, and that will be the actual freezing point of the solution.

3. Assume water is the solvent

Molality of acetic acid in water: $m=\frac{45}{2.7}=16.67\,\text{mol kg}^{-1}$

Depression in freezing point of water: $\Delta T_f=K_f m=1.86\times 16.67\approx 31.0\text{ K}$

So freezing point of solution from water side: $T_f=273-31=242\text{ K}$

4. Assume acetic acid is the solvent

Molality of water in acetic acid: $m=\frac{150}{2.7}=55.56\,\text{mol kg}^{-1}$

Depression in freezing point of acetic acid: $\Delta T_f=K_f m=3.90\times 55.56\approx 216.7\text{ K}$

So freezing point from acetic acid side: $T_f=290-216.7\approx 73.3\text{ K}$

5. Determine actual freezing point

On cooling, the component with the higher equilibrium freezing temperature freezes first.

We got:

• From water side: $242\text{ K}$
• From acetic acid side: $73.3\text{ K}$

Since $242\text{ K}$ is much higher, water freezes first, so the actual freezing point of the mixture is: $242\text{ K}$

In Celsius: $242-273=-31^\circ\text{C}$

Hence, $x=31$

6. Comparison with stored answer

Stored correct answer: $31$

Our derived answer matches it.