5 g of ${\mathrm{Na}}_2{\mathrm{SO}}_4$ was dissolved in x g of $H_{2}O$. The change in freezing point was found to be 3.82°C. If $N{a}_{2}S{O}_4$ is 81.5% ionised, the value of x ($K_f\text{ for water }=1.86°{\mathrm{Ckgmol}}^{-1}$) is approximately: $(\hspace{0.17em}\text{ molar mass of }\mathrm{S}=32{\mathrm{gmol}}^{-1} and that of Na=36g mol m^{-1})$

$$\begin{gathered} \text{ Molar mass of }{\mathrm{Na}}_2{\mathrm{SO}}_4=2\left(23\right)+32+4\left(16\right)=142 \mathrm{g}/\mathrm{mol} \\ 5{\mathrm{gNa}}_2{\mathrm{SO}}_4=\frac{5 \mathrm{g}}{142 \mathrm{g}/\mathrm{mol}}=0.0352 \mathrm{mol} \\ \text{ Mass of water }=\mathrm{xg}=\frac{\mathrm{x}}{1000} \mathrm{kg}=0.001\mathrm{xkg} \\ \text{ Molality }\mathrm{m}=\frac{0.0352{\mathrm{molNa}}_2{\mathrm{SO}}_4}{0.001x}=\frac{35.2}{\mathrm{x}}\mathrm{molkg} \\ {\mathrm{Na}}_2{\mathrm{SO}}_4\text{ is }81.5\%\text{ ionised. } \\ \text{degree of dissociation }\mathrm{\alpha }=\frac{81.5}{100}=0.815 \\ \text{Vant Hoff factor }\mathrm{i}=[1+(\mathrm{n}-1\left)\mathrm{\alpha }\right] \\ \mathrm{i}=[1+(3-1)\times 0.815]=2.63 \\ \text{The decrease in freezing point }{\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m} \\ \begin{array}{l}3.82°\mathrm{C}=2.63\times 1.86{}^{\circ }\mathrm{CKg}/\mathrm{mol}\times \frac{35.2}{\mathrm{x}}\mathrm{mol}/\mathrm{kg}\\ \mathrm{x}=45 \mathrm{g}\end{array} \end{gathered}$$

Hence, the correct option is option (A) 45g.