The freezing point of benzene decreases by 0.45∘C 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 mol−1)
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 D 94.6% Let the degree of association of acetic acid (CH3COOH) in benzene is α, then 2CH3COOH⇌(CH3COOH)2 Initialmoles10Molesatequilibrium1−αα2 ∴Totalmoles=1−α+α2=1−α2ori=1−α2 Now, depression in freezing point (ΔTf) is given as ΔTf=iKfm . . . . (i) Where, Kf - molal depression constant or cryoscopic constant. m = Molality Molality=numberofmolesofsoluteweightofsolvent(inkg)=0.260×10020 Putting the values in Eq.(i) ∴0.45=[1−α2](5.12)[100020] 1−α2=0.45×60×205.12×0.2×1000 ⇒1−α2=0.527 ⇒α2=1−0.527 ∴α=0.946 Thus, percentage of association = 94.6%