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Question

Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water.
[Ka=1.4×103,Kf=1.86 K kg mol1]

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Solution

The molar mass of 2-chloro butanoic acid is 122.5 g/mol.

Number of moles =10122.5=0.0816 mol

Molality of solution =0.0816×1000250=0.3265 m

Let α be the degree of dissociation and c be the initial concentration. The concentration after dissociation is as shown.
CH3CH2CHClCOOHc(1α)CH3CH2CHClCOOcα+Hcα+

The equilibrium constant expression is

K=cα×cαc(1α)=cα2

α=Kc=1.4×1030.3265=0.065

Calculation of vant Hoff factor:

CH3CH2CHClCOOH(1α)CH3CH2CHClCOOα+αH+

i=1α+α+α1=1+α=1+0.065=1.065

The depression in the freezing point ΔTf=iKfm=1.065×1.86×0.3265=0.647o

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