The difference between the boiling point and freezing point of an aqueous solution containing sucrose…
The difference between the boiling point and freezing point of an aqueous solution containing sucrose $\left(ight.$ molecular $\left.\mathrm{wt}=342 \mathrm{~g} \mathrm{~mole}^{-1}ight)$ in $100 \mathrm{~g}$ of water is $105^{\circ} \mathrm{C}$. If $\mathrm{K}_{\mathrm{f}}$ and $\mathrm{K}_{\mathrm{b}}$ of water are $1.86$ and $0.51 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ respectively, the weight of sucrose in the solution is about
$34.2 \mathrm{~g}$
$342 \mathrm{~g}$
$7.2 \mathrm{~g}$
$72 \mathrm{~g}$
Solution
$\left(100+\Delta \mathrm{T}_{\mathrm{b}}ight)-\left(0-\Delta \mathrm{T}_{\mathrm{f}}ight)=105$
$\Delta \mathrm{T}_{\mathrm{b}}+\Delta \mathrm{T}_{\mathrm{f}}=5$
$m\left(K_{b}+K_{f}ight)=5$
$\mathrm{m}=\frac{5}{2.37} \quad$ i.e., $\frac{5}{2.37}$ moles in $1000 \mathrm{~g}$ water
(or) $\frac{5}{2.37 \times 10}$ moles in $100 \mathrm{~g}$ water
$\therefore \quad$ Wt. of sucrose $=\frac{5}{2.37 \times 10} \times 342=72 \mathrm{~g}$