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NCERT Class XII Chemistry
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Question : 33
Total: 53
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid. Kf for water is 1.86 K kg mol–1.
Solution:  
Given, w2=19.5g,w1=500g
Kf=1.86K‌kg‌mol−1, (ΔTf)obs=1.0°C,M2=?
∴M2 (observed)=
1000Kfw2
w1ΔTf
 =
1000×1.86×19.5
500×1.0
 =72.54gmol−1
M2 (calculated) for CH2FCOOH=78gmol–1
van’t Hoff factor (i)=
(M2)cal
(M2)obs
=
78
72.54
=1.0753
Calculation of dissociation constant :
Suppose degree of dissociation at the given concentrationis α.
Then‌,CH2FCOOH→CH2FCOO–+H+
InitialC‌0‌0
At‌eqm.C(1−α)‌Cα‌Cα
Total = C(1 + α)
∴i=
C(1+α)
C
=1+α or, α=i−1=1.0753−1=0.0753
Again Ka=
[CH2FCOO−][H+]
[CH2FCOOH ]
 =
Cα×Cα
C(1−α)
=
Cα2
1−α

Taking volume of the solution as 500 mL,
C=
19.5
78
×
1
500
×1000=0.5M
∴Ka=
Cα2
1−α
=
(0.5)(0.0753)2
1−0.0753
 =3.07×10−3
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