If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $$(\Delta T_f)$$, when $$0.01$$ mol of sodium sulphate is dissolved in $$1$$ kg of water, is ($$K_f = 1.86$$ K kg mol$$^{-1}$$)

The freezing point depression of a solution is given by the formula $$\Delta T_f = i\,K_f\,m$$ where

$$i$$ = van’t Hoff factor (total particles produced per formula unit),
$$K_f$$ = cryoscopic constant of the solvent, and
$$m$$ = molality of the solution (moles of solute per kilogram of solvent).

1. Calculation of the van’t Hoff factor.
Sodium sulphate dissociates as $$Na_2SO_4 \rightarrow 2\,Na^+ + SO_4^{2-}$$.
Total ions formed per formula unit = $$2 + 1 = 3$$, so $$i = 3$$.

2. Molality of the solution.
Given $$0.01$$ mol of $$Na_2SO_4$$ dissolved in $$1$$ kg of water:
$$m = 0.01 \text{ mol kg}^{-1}$$.

3. Substitute into the formula.
$$\Delta T_f = 3 \times 1.86 \,\text{K kg mol}^{-1} \times 0.01 \,\text{mol kg}^{-1}$$.

4. Perform the multiplication:
First, $$1.86 \times 0.01 = 0.0186$$.
Then, $$3 \times 0.0186 = 0.0558$$.

Hence the change in freezing point is $$\Delta T_f = 0.0558 \text{ K}$$.

Option B which is: $$0.0558 \text{ K}$$.