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Question
CBSEENCH12010145

18 g of glucose, C6H12O6 (Molar Mass = 180 g mol-1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil? 

 (Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K)

Solution

w1 = weight of solvent (H2O) = 1 kg

w2 = weight of solute (C6H12O6) = 18 gm

M2 = Molar mass of solute (C6 H12O6) = 180 g mol-1

Kb = 0.52 K Kg mol-1

straight T subscript straight b superscript 0 space equals 373.15 straight K

increment straight T subscript straight b space equals fraction numerator straight K subscript straight b space straight x space 1000 space straight x space straight w subscript 2 over denominator straight M subscript 2 space straight x space straight w subscript 1 end fraction space equals space fraction numerator 0.52 space straight x space 1000 space straight x space 18 over denominator 180 space straight x space 1000 end fraction equals 0.052 straight K

Therefore comma

increment straight T subscript straight b equals space straight T subscript straight b minus space straight T subscript straight b superscript 0 space
0.052 space straight K space equals space straight T subscript straight b minus 373.15

straight T subscript straight b equals 373.202 space straight K