\(\therefore \,\frac{{\Delta {T_f}\,for\,\,KCl}}{{\Delta {T_f}\,\,for\,\,BaC{l_2}}}\, = \,\frac{{i\,\,for\,\,KCl}}{{i\,\,for\,\,BaC{l_2}}}\)
\(\therefore \,\frac{{\Delta {T_f}\,for\,\,KCl}}{{\Delta {T_f}\,\,for\,\,BaC{l_2}}}\, = \frac{2}{3}\,\) \((\because \,\Delta \,{T_f}\,for\,KCl\,\, = \,2)\)
\(\therefore \,\,\Delta {T_f}\,\,for\,\,BaC{l_2}\, = \,\frac{{3 \times 2}}{2}\, = \,3\)
\(\therefore \) Freezing point of \(BaCl_2\,=\,-\,3\,^oC\)