$ = \,\frac{{\Delta {T_f}}}{{{K_f}}}\, = \,\frac{{3.82}}{{1.86}}\, = \,2.054\,mol/1000\,g$ solvent

Molality (theoretical)

$ = \,\frac{{mole\,\,of\,solute}}{{wt.\,of\,solventing\,(g)}} \times 1000$

$ = \,\frac{{5\,g\,/\,142\,g\,/\,mole}}{x} \times 1000$

                                            $N{a_2}S{O_4}\, \to \,2N{a^ + }\, + \,SO_4^{2 - }$

Moles before dissociation              $1$              $0$              $0$

Moles after dissociation              $1-x$              $2x$              $x$

Von't Hoff Factor $(i) = \,\frac{{Moles\,after\,\,dissociation}}{{Moles\,before\,\,dissociation}}$

                               $ = \frac{{(1 - x)\, + \,2x\, + \,x}}{1}$

$Na_2SO_4$ is ionised $81.5\%$ means $x\,=\,0.815$

                             $ = \frac{{(1 - 0.815)\, + \,2 \times 0.815\, + \,0.815}}{1}$

                              $=\,2.63$

$i\, = \,\frac{{Observed\,\,molarrity}}{{Calculated\,\,molarity}}$

$ \Rightarrow \,2.63\, = \,\frac{{2.054}}{{\frac{{0.0352}}{x} \times 1000}}\, = \,45.07\,g$