Question

The amount (in grams) of sucrose ( $molwt.=342\text{ g}$ ) 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. $34.2g$
• B. $72.2g$
• C. $342g$
• D. $460g$

Answer 1

Answer: (B)

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