Q. The difference between the boiling point and freezing point of an aqueous solution containing sucrose (mol wt. = 342 g$mol^{- 1}$) in 100 g of water is $105.0^{\circ}C.$ If $K_{f}and\ K_{b}\$of water are 1.86 and 0.51 K kg $mol^{- 1}$ respectively, the weight of sucrose in the solution is about
A) 34.2 g
B) 342 g
C) 7.2 g
D) 72 g
Explanation
(d)
Given, $T_{b} - T_{f} = {105.0}^{\circ}C$
$\Rightarrow$ $(100 = \bigtriangleup T_{b})\ –(0 - \bigtriangleup T_{f}) = \ {105.0}^{\circ}C\ \$
$\bigtriangleup T_{b} + \bigtriangleup T_{f} = 5\ \$
$\bigtriangleup T_{b} + \bigtriangleup T_{f}(k_{b} + k_{f}) \times m\ \ \ \ \ \ \ \ \ \ \$ (m =molality)
$\Rightarrow$ 5= $(1.86 \times 0.51) \times \frac{w \times 1000}{342 \times 100}$
$\therefore\ \ \ \ \ \ \ \ \ w = \frac{1710}{23.7} = 72\ g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.