The elevation in boiling point of a solution of $$13.44\,g$$  of $$CuC{l_2}$$  in $$1\,kg$$  of water using the following information will be ( Molecular weight of $$CuC{l_2} = 134.4\,g$$    and $${K_b} = 0.52\,K\,kg\,mo{l^{ - 1}}$$     )

Solution :
$$\eqalign{ & \left( i \right)\,\,i = \frac{{{\text{No}}{\text{. of particles after ionisation}}}}{{{\text{No}}{\text{. of particles before ionisation}}}} \cr & \left( {ii} \right)\,\Delta {T_b} = i \times {K_b} \times m \cr & CuC{l_2} \to C{u^{2 + }} + 2C{l^{ - 1}} \cr & \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & \left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & i = \frac{{1 + 2\alpha }}{1};\, = 1 + 2\alpha \cr & {\text{Assuming}}\,100\% \,{\text{ionization}} \cr & {\text{So,}}\,\,i = 1 + 2 = 3 \cr & \Delta {T_b} = 3 \times 0.52 \times 0.1 \cr & = 0.156 \approx 0.16 \cr & \left( {m = \frac{{13.44}}{{134.4}} = 0.1} \right) \cr} $$