Chemistry - Liquid Solution Question with Solution | TestHub

$$\begin{array}{rl}& {\mathrm{W}}_2=6\mathrm{g},{\mathrm{W}}_1=100\mathrm{g},{\mathrm{K}}_{\mathrm{b}}=0.52\mathrm{Kg}{\mathrm{mol}}^{-1},{\mathrm{T}}_{\mathrm{b}}100.52^{\circ }\mathrm{C}\\ & \therefore \Delta {\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\circ }=(100.52+273)-(100+273)\\ & =0.52\mathrm{k}\end{array}$$ Now, ${\mathrm{M}}_2=\frac{1000\cdot {\mathrm{K}}_{\mathrm{b}}\cdot {\mathrm{W}}_2}{\Delta {\mathrm{T}}_{\mathrm{b}}\cdot {\mathrm{W}}_1}=\frac{1000\mathrm{g}{\mathrm{kg}}^{-1}\times 0.52\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}\times 6\mathrm{g}}{0.52\mathrm{K}\times 100\mathrm{g}}$ $$\therefore {\mathrm{M}}_2=60\mathrm{g}{\mathrm{mol}}^{-1}$$