Chemistry - Liquid Solution Question with Solution | TestHub

The aqueous solution contains $36\mathrm{g}$ glucose per ${\mathrm{dm}}^3$, so mass of solute ${\mathrm{W}}_2$ is $36\mathrm{g}$. Assuming that the density of solution is $1\mathrm{g}/{\mathrm{dm}}^3$, the mass of solvent (water) is $1000\mathrm{g}$. $$\begin{array}{rl}& \Delta {\mathrm{T}}_{\mathrm{b}}=\frac{1000{\mathrm{K}}_{\mathrm{b}}{\mathrm{W}}_2}{{\mathrm{M}}_2{\mathrm{W}}_1}\\ \Delta {\mathrm{T}}_{\mathrm{b}}& =\frac{1000\mathrm{g}{\mathrm{kg}}^{-1}\times {\mathrm{K}}_{\mathrm{b}}\times 36\mathrm{g}}{180\mathrm{g}\times 1000\mathrm{g}}\\ \Delta {\mathrm{T}}_{\mathrm{b}}& =\frac{2{\mathrm{K}}_{\mathrm{b}}}{10}\\ \Delta {\mathrm{T}}_{\mathrm{b}}& ={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^0\\ & {\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}^0+\Delta {\mathrm{T}}_{\mathrm{b}}\\ \therefore & {\mathrm{T}}_{\mathrm{b}}={\left(100+\frac{2{\mathrm{K}}_{\mathrm{b}}}{10}\right)}^{\circ }\mathrm{C}\end{array}$$