• # question_answer A solution of urea $\left( mol.\text{ }mass\text{ }56\text{ }g\text{ }mo{{l}^{-}}^{1} \right)$ boils at 100. $18{}^\circ C$ at the atmospheric pressure. If ${{K}_{f}}\,\,and\,\,{{K}_{b}}$ for water are 1.86 and 0.512 K kg $mo{{l}^{-1}}$ respectively, the above solution will freeze at A) $0.654\,{}^\circ C$      B)        $-\,0.654\,{}^\circ C$ C) $\,6.54\,{}^\circ C$      D)        $-\,6.54\,{}^\circ C$

As $\Delta {{T}_{f}}={{K}_{f}}\,m$ $\Delta {{T}_{b}}=\,\,{{K}_{b}}.\,m$ Hence, we have $m=\frac{\Delta {{T}_{f}}}{{{K}_{f}}}\,\,=\,\,\frac{\Delta {{T}_{b}}}{{{K}_{b}}}$ $\Delta {{T}_{f}}=\Delta {{T}_{b}}\frac{{{K}_{f}}}{{{K}_{b}}}$ $\Rightarrow \,\,\,[\Delta {{T}_{b}}=100.18-100=0.18{}^\circ \,C]$ $=\,\,0.18\times \frac{1.86}{0.512}=0.654{}^\circ \,C$ As the Freezing Point of pure water is $0{}^\circ C$, $\Delta {{T}_{f}}=0-{{T}_{f}}$ $0.654=\,\,0-{{T}_{f}}.$ $\therefore \,\,{{T}_{f}}=-\,0.654$ Thus the freezing point of solution will be $-\,0.654{}^\circ C$.