Question

$ 10\,g $ of a non-volatile solute when dissolved in $ 100\,g $ of benzene raises its boiling point by $ 1^{\circ} $ . What is the molecular mass of the solute? ( $ K_{b} $ for benzene $ =2.53\,km^{-1} $ )

• A. $ 352 $
• B. $ 253 $
• C. $ 275 $
• D. $ 310 $

Answer 1

Answer: (B)

$w_{1}=100\, g, w_{2}=10 \,g$,
$\Delta T_{b}=1^{\circ}, K_{b}=2.53\, Km ^{-1}$
$M_{2}=\frac{1000\, K_{b} \times w_{2}}{w_{1} \times \Delta T_{b}}$
$=\frac{1000 \times 2.53 \times 10}{100 \times 1}$
$ = 253\,g$