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Question

# The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45oC. The degree of association of acetic acid in benzene is: [Assume acetic acid dimerizes in benzene and Kf for benzene =5.12 K kg mol−1]

A
94.5%
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B
54.9%
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C
78.2%
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D
100%
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Solution

## The correct option is A 94.5%The molar mass of acetic acid is 60 g/mol.The molality of the solution is the number of moles of solute divided by the mass of solvent (in kg). The molality of the solution will be 0.2×100060×20.0ΔTf=iKfm 0.45=i×5.12×0.2×100060×20.0i=0.5275The degree of association of acetic acid in benzene is:α=i−11n−1Here, n is the number of molecules of acetic acid that associate.α=0.5275−112−1=0.945Hence, the degree of association of acetic acid in benzene is 94.5%.

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