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Question

A 4% solution(w/w) of sucrose (M = 342 g mol1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol1) in water. (Given: Freezing point of pure water = 273.15 K)

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Solution

Molality of 4% solution of sucrose:
let assume that 4 grams sucrose in 96 grams of water then molality of solution
m=4×1000342×96 (1)
(Tf)sucrose(aq.)=i×m×kf
273.15271.15=1×4×1000342×96×kf (2)
the molal freezing point depression constant of the solvent which is water for both solution , from eqn. 2 the value of kf=2×342×964×1000 (3)

For 5% solution of glucose:
(Tf)glucose(aq.)=i×5×1000180×95×kf
after putting value of kf from eqn. (3)
(Tf)glucose(aq.)=4.8
273.15Tf=4.8
Tf=268.35K

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