The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. ( $K_{b}$ for benzene is $2.53 K k g m o l^{- 1}$ )

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answer is 58.

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Detailed Solution

Given value are:
${T^{0}}_{(b e n z e n e)} = 353.00 K; K_{b} = 2.53 K k g m o l^{- 1} T_{b (b e n z e n e)} = 354.00 K W_{s o l u t e} = 1.80 g W_{s o l v e n t} = 90 g$
The elevation in boiling point $Δ T_{b} = T_{b (s o l u t i o n)} - T_{b (s o l v e n t)}^{0}$
$= 354.11 - 353.23$
$= 0.88 K$
Molar mass of solute is given as
$M w_{s o l u t e} = \frac{K_{b} \times 1000 \times W_{s o l u t e}}{Δ T_{b} \times W_{s o l v e n t}}$
$M w_{s o l u t e} = \frac{2.53 \times 1000 \times 1.80}{0.88 \times 90} = 58.0 g m o l^{- 1}$
Hence, the molar mass of solute is $58.0 g m o l^{- 1}$