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Question 2.32

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


Solution

Step I: Calculation of degree of dissociation
Mass of solute = 10g
Molar mass of solute
(CH3CH2CHClCOOH)=(12×4)+(1×7)+(35.5)+(16×2)=48+7+35.5+32=122.5 g mol1
Molal concentration of solution
=MassMolar massMassofsolvent (kg)=(10g)(122.5 g mol1)×(0.25 kg)=0.326m
If α is the degree of dissociation of CH3CH2CHClCOOH, then
CH3CH2CHClCOOHCH3CHClCOO+H+Initial conc. C mol1 kg00At equilibrium C(1α)CαCα
Ka=Cα×CαC(1α)=Cα2
[Considering (1α)=1 or dilute solutions]
α2=KaC or α=KaC
So, α=1.4×1030.326=42.9×104=6.55×102=0.065
Step II: Calculation of van't Hoff factor (f)
CH3CH2CHClCOOHCH3CH2CHClCOO+H+Initial no. of moles100Moles at equilibrium1ααα
Total number of moles after dissociation
=(1α)+(α)+(α)=(1+α)
Van't Hoff fraction
(i) =Total number of moles after dissociationNumber of moles before dissociation
i=(1+α)1=(1+α)
=1+0.065=1.065
Step III: Calculation of depression in freezing point (ΔTf)
ΔTf=iKfm=(1.065)×(1.86K kg mol1)×(0.326 mol kg1)
ΔTf=0.65K


Chemistry
NCERT Textbook
Standard XII

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