When $10 g$ of a non-volatile solute is dissolved in $100 g$ of benzene, it raises boiling point by $1 ° c$ then molecular mass of the solute is $(K_{b} for benzene = 2 .53 k - m^{- 1})$

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$243 g$

$253 g$

$233 g$

$223 g$

answer is D.

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Detailed Solution

$\begin{array}{l} Δ T_{b} = \frac{K_{b} \times w \times 1000}{m \times W} \\ m = \frac{K_{b} \times w \times 1000}{Δ T_{b} \times W} = \frac{2.53 \times 10 \times 1000}{1 \times 100} = 253 g \end{array}$

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