5 g of Na$$_2$$SO$$_4$$ was dissolved in $$x$$ g of H$$_2$$O. The change in freezing point was found to be 3.82°C. If Na$$_2$$SO$$_4$$ is 81.5% ionised, the value of $$x$$ is approximately: ($$K_f$$ for water = 1.86°C kg mol$$^{-1}$$) (molar mass of S = 32 g mol$$^{-1}$$ and that of Na = 23 g mol$$^{-1}$$)

We are told that 5 g of sodium sulphate has been dissolved and the freezing-point depression produced is 3.82 °C. Our aim is to find the mass $$x$$ of water that was used.

First we need the molar mass of the solute. Sodium sulphate has the formula $$Na_2SO_4$$, so

$$M_{Na_2SO_4}=2(M_{Na})+M_S+4(M_O)=2(23)+32+4(16)=46+32+64=142\ \text{g mol}^{-1}.$$

Now, the number of moles present in the given 5 g is

$$n=\frac{5\ \text{g}}{142\ \text{g mol}^{-1}}=0.0352\ \text{mol}.$$

The depression of the freezing point for solutions is governed by the equation

$$\Delta T_f = i\,K_f\,m,$$

where $$\Delta T_f$$ is the observed fall in freezing point, $$K_f$$ is the cryoscopic constant of the solvent, $$m$$ is the molality of the solution in mol kg$$^{-1}$$, and $$i$$ is the van’t Hoff factor.

Sodium sulphate dissociates as $$Na_2SO_4 \rightarrow 2\,Na^+ + SO_4^{2-}$$ giving a total of $$z=3$$ ions. If $$\alpha$$ is the degree of ionisation, then

$$i=1+(z-1)\alpha.$$ Substituting $$z=3$$ and $$\alpha=0.815$$ (since 81.5 % = 0.815), we get

$$i=1+(3-1)(0.815)=1+2(0.815)=1+1.63=2.63.$$

The numerical data are $$\Delta T_f = 3.82\ ^\circ\text{C}$$ and $$K_f = 1.86\ ^\circ\text{C kg mol}^{-1}.$$ Putting these into the freezing-point equation,

$$3.82 = 2.63 \times 1.86 \times m.$$

Solving for the molality $$m$$,

$$m = \frac{3.82}{2.63 \times 1.86} = \frac{3.82}{4.8918} = 0.782\ \text{mol kg}^{-1}\;(\text{approximately}).$$

Molality is defined as $$m = \dfrac{n}{\text{mass of solvent in kg}}.$$ Therefore,

$$\text{mass of solvent (kg)} = \frac{n}{m} = \frac{0.0352\ \text{mol}}{0.782\ \text{mol kg}^{-1}} = 0.0450\ \text{kg}.$$

Converting to grams, $$0.0450\ \text{kg} = 45.0\ \text{g}.$$

Thus the required mass of water is about 45 g.

Hence, the correct answer is Option D.