Question

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

(Given that $\text{K}_{\text{f}}=1.86 \, \text{K kg mol}^{- \text{1}}$ and $\text{K}_{\text{b}}=0.51 \, \text{K kg mol}^{- \text{1}}$ for water)

• A. $34.2 \, g$
• B. $72.2 \, g$
• C. $342g$
• D. $460 \, g$

Answer 1

Answer: (B)

Boiling point $\left(T_{b}\right)=100+\Delta T_{b}=100+k_{b}m$
Freezing point $\left(T_{f}\right)=0-\Delta T_{f}=-k_{f}m$
$T_{b}-T_{f}=\left(100 + k_{b} m\right)-\left(\right.-k_{f}m\left.\right)$
$105=100+0.51m+1.86m$
$2.37m=5 \, orm=\frac{5}{2 .37}=2.11$
$\therefore $ Weight of sucrose to be dissolved in $100g$ water
$=\frac{2.11 \times 342}{1000}\times 100=72.2g$