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Solutions

Question
CBSEENCH12010557

Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water.
(Kf of water = 1.86 K kg mol–1)

Solution

We know,

W1 = 250 g, w2 = 60 g, mw2 = 180 g/mol, Kf = 1.86 k kg mol–1

Tf  = kf x mOrTf  = kf xw2 x 1000m.w2 x w1 (g) = 1.86 x 60 x 1000180 x 250 = 1.80 x 60018 x 25 = 1116450 Tf = 2.48 KTf = Tsolvent - Tsolution = 2.48 = 273.15 - TsolutionTsolution  = 270.67 K