Question

The amount (in grams) of sucrose $(\text{mol}\text{.wt}\text{.}=342\text{g})$ that should be dissolved in $100\text{g}$ water in order to produce a solution with a $105.0^{\text{o}}\text{C}$ difference between the boiling point and freezing point is

(Given that $k_{ƒ}=1.86Kkgmo{l}^{-1}$ and $k_b=0.52Kkgmo{l}^{-1}$ for water)

Report your answer by rounding it up to the nearest whole number.

Answer 1

Answer: 72

Let the molality of the soluiton is $\mathrm{m}$.
$\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}\text{ and }\Delta {\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{f}}\mathrm{b}\times \mathrm{m}$
$\Delta {\mathrm{T}}_{\mathrm{f}}+\Delta {\mathrm{T}}_{\mathrm{b}}=\left({\mathrm{K}}_{\mathrm{f}}+{\mathrm{K}}_{\mathrm{b}}\right)\times \mathrm{m}=(1.86+0.51)\times \mathrm{m}$
$=2.37\times \mathrm{m}$
Hence, $\mathrm{m}=\frac{5}{2.37}=2.11$ and $\frac{2.11\times 100}{1000}=0.211$ moles of sucrose must be dissolved in $100\mathrm{gram}$ of water. Required mass of sucrose $=0.211\mathrm{mol}\times 342\mathrm{g}/\mathrm{mole}=72.2\mathrm{g}$