A solution containing 3.3g of a substance in 125g of benzene (b.pt. = 80^oC) boils at 80.66^OC. If Kb for benzene is 3.28 K kg mol–1 the molecular mass of the substance will be:

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130.20

129.20

132.20

131.20

answer is D.

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Detailed Solution

$∆ T_{b} = K_{b} m ∆ T_{b} = 0.66 0.66 = K_{b} \times \frac{W_{s}}{M_{s}} \times \frac{1}{mass of solvent (kg)} 0.66 = 3.28 (\frac{3.3}{M} \times \frac{1}{0.125}) M = 131.20 g$

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