${\text{2KI}}_{\left(\mathrm{aq}\right)}+{\mathrm{HgI}}_{2\left(\mathrm{s}\right)}\to {\mathrm{K}}_2{\mathrm{HgI}}_{4\left(\mathrm{aq}\right)}\mathrm{\hspace{0.17em}}\mathrm{or}\mathrm{\hspace{0.17em}}2{\mathrm{K}}^{+}+2{\mathrm{I}}^{-}+{\mathrm{HgI}}_{2\left(\mathrm{s}\right)}\to 2{\mathrm{K}}^{+}+{\left[{\mathrm{HgI}}_4\right]}^{2-}$

As a result of complex ion formation, the number of mole particles in the solution decreases in the ratio 4 : 3. So ${\mathrm{\Delta T}}_{\mathrm{f}}$ decreases until all the ${\mathrm{I}}^{-}$ ions are consumed with 0.05 moles of ${\mathrm{HgI}}_2$. The excess of ${\mathrm{HgI}}_2$ remain insoluble. Hence after 0.05 moles of ${\mathrm{HgI}}_2$, the freezing point of solution remains constant.