Question 2.32
Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250 g of water, Ka=1.4×10−3; Kf=1.86 K kg mol−1
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 mol−1
Molal concentration of solution
=MassMolar massMassofsolvent (kg)=(10g)(122.5 g mol−1)×(0.25 kg)=0.326m
If α is the degree of dissociation of CH3CH2CHClCOOH, then
CH3CH2CHClCOOH⇌CH3CHClCOO−+H+Initial conc. C mol−1 kg00At equilibrium C(1−α)CαCα
∴Ka=Cα×CαC(1−α)=Cα2
[Considering (1−α)=1 or dilute solutions]
α2=KaC or α=√KaC
So, α=√1.4×10−30.326=√42.9×10−4=6.55×10−2=0.065
Step II: Calculation of van't Hoff factor (f)
CH3CH2CHClCOOH⇌CH3CH2CHClCOO−+H+Initial no. of moles100Moles at equilibrium1−ααα
Total number of moles after dissociation
=(1−α)+(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 mol−1)×(0.326 mol kg−1)
ΔTf=0.65K