Q.

Given, H2O(l)⇌H2O(g) at 373 K, ΔHo=8.31 kcal mol−1Thus, boiling point of 0.1 molal sucrose solution is

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373.52 K

373.052 K

373.06 K

374.52 K

answer is C.

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Detailed Solution

H2O(l) changes to steam H2O(g) at 373 K. Thus, this represents latent heat of vaporisation. Kb (molal elevation constant) is related to ΔHο and boiling point by equation, Kb=RT021000ΔHoHere, ΔHo is energy unit per gram of solvent.ΔHo=8.31 kcal mol−1 =8.3118kcal g−1Kb=0.002×(373)21000×8.3118 =278.25461.66=0.60o mol−1kgΔTb (sucrose solution) = molality ×Kb=0.1×0.60=0.06o∴ Boiling point of solution =373+0.06o=373.06 K

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