\(\Delta T_b\,=\,K_b\,.\,m\)

Hence , We have \(m\, = \frac{{\Delta {T_f}}}{{{K_f}}}\, = \,\frac{{\Delta {T_b}}}{{{K_b}}}\)

\(\Delta {T_f}\, = \,\Delta {T_b}\frac{{{K_f}}}{{{K_b}}}\)

\( \Rightarrow \,[\,\Delta {T_b}\, = \,100.18\, - \,100\, = \,0.18\,{\,^o}C]\)

\( = \,0.18\, \times \,\frac{{1.86}}{{0.512}}\, = 0.654{\,^o}C\)

As the Freezing Point of pure water is \(0\,^oC\)

\(\Delta {T_f}\, = \,0\, - \,{T_f}\)

\(0.654\, = \,0\, - \,{T_f}\)

\(\therefore {T_f}\, = \,\, - \,0.654\)

Thus the freezing point of solution will be \( - \,0.654\,{\,^o}C\)