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Question

# The freezing point of a solution containing 0.2 g of acetic acid in 20 g of benzene is lowered by 0.45∘C.. Calculate the degree of association (β) for acetic acid. Given : (Kf)Benzene=5.12 K kg mol−1

A
0.53
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B
0.81
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C
0.25
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D
0.94
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Solution

## The correct option is D 0.94Weight of acetic acid(w)=0.2 gWeight of benzene(W)=20 g=201000 kgΔTf=0.45∘C Let mobs be observed molecular mass of acetic acid ΔTf=Kf×m m is molality mobs=1000×Kf×wW×ΔTf=1000×5.12×0.220×0.45=113.78 g mol−1 Theoretical molecular mass of acetic acid =60 g mol−1 van't Hoff factor=Theoretical molecular massObserved molecular massi=60113.78≈0.53 2CH3COOH⇌(CH3COOH)2Before associationC 0After association(C−Cβ) Cβ2 i=C−Cβ+Cβ2C0.53=1−β2 β=0.94

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