CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

Question : 29

Total: 35

When 19.5 g of F− CH2− COOH (Molar mass =78 g mol−1 ), is dissolved in 500 g of water, the depression in freezing point is observed to be 1°C.
Calculate the degree of dissociation of F− CH2− COOH.
[Given : Kf for water =1.86 K kg mol−1 ]

Solution: 👈: Video Solution

Molecular mass of 78 g∕ mol
(F−CH2COOH)
No. of moles of fluoroacetic acid is

19.5

=0.25
Molality is the number of moles of solute in 1 kg of solvent
Molality =

0.25

500

1000

=0.50 m
ΔTf=Kf×m=1.86×0.50 =0.93 K
i=

1.0

0.93

=1.0753
CH2FCOOH→CH3FCOO−+H+
C(1−α)‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌Cα‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌Cα
Total number of moles =C(1−α)+Cα+Cα =C(1+α)
i=

C(1+α)

=1+α=1.0753
α=0.0753
(ii) (α=0.50×0.0753=0.03765
C(1−α)=0.50(1−0.0753)=0.462

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