Question

The amount (in grams) of sucrose $\left(\right. \text{mol} \text{.wt} \text{.} = 342 \, \text{g} \left.\right)$ 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 $k_{ƒ}=1.86 \, Kkgmol^{- 1}$ and $k_{b}=0.52 \, Kkgmol^{- 1}$ for water)

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

Answer 1

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 $=0.211 \mathrm{~mol} \times 342 \mathrm{~g} / \mathrm{mole}=72.2 \mathrm{~g}$