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When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by 1^oC then molecular mass of the solute is: K_b for benzene = 2.53 k-m^-1)

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223 g

233 g

243 g

253 g

answer is D.

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Detailed Solution

${ΔT}_{b} = \frac{K_{b} \times W \times 1000}{m \times W} m = \frac{K_{b} \times W \times 1000}{{ΔT}_{b} \times W} = \frac{2.53 \times 10 \times 1000}{1 \times 100} = 253 g$

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