Mass of water = 500 $$\times$$ 0.997 = 498.5 g and that of acetic acid = 3 $$\times$$ 10^$$-$$3 kg = 3 g.

No. of moles of acetic acid in solution = $${3 \over {60}} = 0.05$$ mol [$$\because$$ Molar mass of acetic acid = 60 g mol^$$-$$1]

Molality (m) of the solution = $${{0.05} \over {498.5}} \times 1000 = 0.1$$ mol kg^$$-$$1

Given : $$\alpha$$ = 0.23. Now, 1 molecule of CH₃COOH on dissociation in aqueous solution produces 2 ions,

CH₃COOH(aq) $$\to$$ CH₃COO^$$-$$(aq) + H⁺(aq)

So, n = 2

$$\therefore$$ $$0.23 = {{i - 1} \over {2 - 1}}$$ or, i = 1.23

For a solution of an electrolyte, the freezing point depression of the solution is given by, $$\Delta$$T_f = i $$\times$$ k_f $$\times$$ m

$$\therefore$$ $$\Delta$$T_f = 1.23 $$\times$$ 1.86 $$\times$$ 0.1 = 0.228 K

Hence, freezing point depression of solution = 0.228 K