Chemistry - Liquid Solution Question with Solution | TestHub

$$\begin{array}{rl}\Delta {\mathrm{T}}_f& ={\mathrm{K}}_{f}m\dots (1)\\ \Delta {\mathrm{T}}_b& ={\mathrm{K}}_{b}m\dots (2)\\ \Rightarrow \frac{\Delta {\mathrm{T}}_f}{\Delta {\mathrm{T}}_b}& =\frac{{\mathrm{K}}_f}{{\mathrm{K}}_b}\dots (3)\end{array}$$ $$\begin{array}{rl}& \text{ b.p. of water }=100^{\circ }\mathrm{C};{\mathrm{K}}_f=1.86\mathrm{kg}\\ & {\mathrm{mol}}^{-1}\\ & \text{ b.p. of urea in water }=100.18^{\circ }\mathrm{C};{\mathrm{K}}_b\\ & =0.512\mathrm{kg}{\mathrm{mol}}^{-}\\ & \Rightarrow \Delta {\mathrm{T}}_b=0.18\\ & \text{ f.p. of water }=0^{\circ }\mathrm{C}\\ & f.p.\text{ of urea in water }=-{\mathrm{T}}^{\circ }\mathrm{C}\\ & \Rightarrow \Delta {\mathrm{T}}_f=\mathrm{T}\\ & \Rightarrow \text{ from eq. }(3)\\ & \frac{\mathrm{T}}{0.18}=\frac{1.86}{0.512}\\ & \Rightarrow \mathrm{T}=0.6539\\ & \Rightarrow f.p.\text{ of urea in water }=-0.654^{\circ }\mathrm{C}\end{array}$$