SolutionHard

Question

The boiling point of pure benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. The value of Kb for benzene is 2.53 K kg mol−1.

Options

A.85 g mol−1
B.57.5 g mol−1
C.23 g mol−1
D.38.4 g mol−1

Solution

$\Delta T_{b} = K_{b}.m \Rightarrow (354.11 - 353.23) = 2.53 \times \frac{1.8/M}{90} \times 100$

$\therefore M = 57.5$

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