$$20\%$$ of acetic acid is dissociated when its 5 g is added to 500 mL of water. The depression in freezing point of such water is ______ $$\times 10^{-3}$$ °C. Atomic mass of C, H and O are 12, 1 and 16 a.m.u. respectively.
[Given: Molal depression constant and density of water are $$1.86$$ K kg mol$$^{-1}$$ and $$1$$ g cm$$^{-3}$$ respectively]

We need to find the depression in freezing point when 5 g of acetic acid (20% dissociated) is added to 500 mL of water.

Calculate the moles of acetic acid.

Molar mass of CH$$_3$$COOH = 12 + 3(1) + 12 + 16 + 16 + 1 = 60 g/mol.

$$n = \frac{5}{60} = \frac{1}{12} \text{ mol}$$

Calculate the van't Hoff factor.

Acetic acid dissociates as: CH$$_3$$COOH $$\rightleftharpoons$$ CH$$_3$$COO$$^-$$ + H$$^+$$

With degree of dissociation $$\alpha = 0.20$$:

$$i = 1 + \alpha = 1 + 0.20 = 1.2$$

Calculate the molality.

Mass of water = 500 mL $$\times$$ 1 g/cm$$^3$$ = 500 g = 0.5 kg.

$$m = \frac{1/12}{0.5} = \frac{1}{6} \text{ mol/kg}$$

Calculate the depression in freezing point.

$$\Delta T_f = i \cdot K_f \cdot m = 1.2 \times 1.86 \times \frac{1}{6}$$

$$\Delta T_f = 1.2 \times 0.31 = 0.372 \text{ °C}$$

Express in the required form.

$$\Delta T_f = 0.372 \text{ °C} = 372 \times 10^{-3} \text{ °C}$$

The correct answer is 372.