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The freezing point of benzene decreases by 0.45C when 0.2g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
(Kf for benzene = 5.12K kg mol1)

A
64.6%
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B
80.4%
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C
74.6%
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D
94.6%
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Solution

The correct option is B 94.6%
Let the degree of association of acetic acid (CH3COOH) in benzene is α, then
2CH3COOH(CH3COOH)2
Initial moles10Moles at equilibrium1αα2
Total moles=1α+α2=1α2 or i=1α2
Now, depression in freezing point (ΔTf) is given as
ΔTf=i Kfm . . . . (i)
Where, Kf - molal depression constant or cryoscopic constant.
m = Molality
Molality=number of moles of soluteweight of solvent (in kg)=0.260×10020
Putting the values in Eq.(i)
0.45=[1α2] (5.12)[100020]
1α2=0.45×60×205.12×0.2×1000
1α2=0.527
α2=10.527
α=0.946
Thus, percentage of association = 94.6%

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