Boiling point of benzene is 353.24 K. When 1.8 g of non volatile solute was dissolved in 90g of benzene the boiling point raised to 354.11K. Molar mass of the solute is ( $k_{b}$ of benzene = 2.523 K kg $m o l^{- 1}$ )

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$72 g {mol}^{- 1}$

$36 g {mol}^{- 1}$

$58 g {mol}^{- 1}$

$84 g {mol}^{- 1}$

answer is C.

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Detailed Solution

$Δ T_{b} = 0.88 = k_{b} m$
$0.87 = \frac{2.523 \times 1000 \times 1.8}{x \times 90}$
$\Rightarrow x = \frac{2.523 \times 1000 \times 1.8}{90 \times 0.87} = 58 g m o l^{- 1}$

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