Question

When 0.9g of a nonelectrolyte was dissolved in 87.9g of benzene, the boiling point of benzene was raised by 0.25 degree celsius. The molar mass of nonelectrolyte is 103g/mol. Calculate the molal elevation constant of benzene.

Answer 1

$$\begin{gathered} Accordingtotheelevationofboilingpoint \\ ∆T_b=K_b\times m \\ =K_b\times \frac{W_2}{W_1\times M_2}\times 1000 \\ 0.25=K_b\times \frac{0.9}{87.9\times 103}\times 1000 \\ {K}_b=\frac{87.9\times 103\times 0.25}{0.9\times 1000}=2.51Kg{}^{0}C/mol \end{gathered}$$