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Question

The freezing point of 0.02mol fraction of acetic acid in benzene is 277.4K. Acetic acid exists partly as a dimer. Calculate the equilibrium constant for dimerization. (approximately)
(Freezing point of C6H6=278.4K and Kf of C6H6=5Km−1)

A
2.6
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B
3.2
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C
4.3
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D
5.8
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Solution

The correct option is A 3.2
Freezing point of acetic acid (mole fraction =0.02) in benzene is 277.4k. Freezing point of benzene is 278.4k. molal freezing point constant =5mg/mole
equilibrium constant for dimerisation =?
Let acetic acid =A and
Benzene =B
Assume that
α part of A forms dimer.
2AA2
1α α/2 moles after formation of dimer.
i=(1α)+α/2=1α/2(1)
mole fraction of Acetic acid =XA=0.02
mole fraction of Benzene =Xe=10.02=0.98
Molality of A in B.
m=XA×1000XB×MB=
where XA= Mole fraction of Acetic acid
XB= Mole fraction of Benzene
MB= Molecular mass of Benzene
=0.02×10000.98×78=0.262mol/kg
Since TF=KFXi×m(molality)
where
TF= depression in freezing point
KF= The freezing point constant of a solvent.
i= Vant Hoff factor
m= molality
The TF is change in freezing point
278.4277.4=TF
TF=1
TF=KFXi×molality
1=5×i×0.262
i=15×0.262
i=0.763
Substitute i in eqn (1)
0.763=1α2
A=0.7631=α2
=0.237=α2
=0.237=α2
2×0.237=α
0.474=α
A0.48
Hence the molality of A after formation of dimer =(1α)× initial molality.
(10.48)×0.262
=0.52×0.262
=0.13624
Now, the molality of A2 after formation of dimer =(α/2)×molality
=0.482×0.262
=0.24×0.262
=0.06288
The equilibrium constant.
=Keq=[A2][A1]2=0.6288(0.13624)2=3.39kg/mol

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