Question

A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boilling point of ${80.31}^0$C. Determine the molar mass of this compound.
(Boilling point of pure benzene=${80.10}^0$C and $K_b$ for benzene=${2.53}^0$C kg $mo{l}^{-}1$).

Answer 1

$\begin{array}{c}\Delta T_b=K_p\times molality\\ \Rightarrow 80.31-80.10=2.53\times \frac{\frac{1.25}{M}\times 100}{99}\\ \Rightarrow 0.21=2.56\times \frac{1250}{99M}\\ \Rightarrow M=\frac{2.53\times 1250}{99\times 0.21}=152gmmo{l}^{-1}\end{array}$

Hence the molar mass of this compound is $152gmmo{l}^{-1}$