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NCERT Class XII Chemistry
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Question : 32
Total: 53
Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. (Ka=1.4×10–3,Kf=1.86K‌kg‌mol–1)
Solution:  
Number of moles of CH3CH2CHClCOOH =
10
122.5
 mole =8.16×10−2 mole
∴ Molality of the solution (m) =
8.16×10−2‌mol
250
×1000kg−1 =0.3264m
If α is the degree of dissociation of CH3CH2CHClCOOH, then
‌CH3CH2CHClCOOH⇌CH3CH2CHClCOO–+H+
Initial‌conc.C‌mol‌L–1‌0‌0
At‌equilibriumC(1–α)‌Cα‌Cα
∴Ka=
Cα×Cα
C(1−α)
=Cα2 or, α=√
Ka
C
 =√
1.4×10−3
0.3264
=0.065
To calculate van’t Hoff factor:
‌CH3CH2CHClCOOH⇌CH3CH2CHClCOO–+H+
Initial‌conc.1‌0‌0
At‌equilibrium1–α‌α‌α
Total moles = 1 + a
∴ i = 1 + 0.065 = 1.065
ΔTf=iKfm = (1.065)(1.86)(0.3264) = 0.65°
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