The freezing pint of benzene decreases by ${0.45}^{o} C$ when $0.2 g$ of acetic acid is added to $20 g$ of benzene. If acetic acid is associates to form a dimer in benzene, find the percentage association of acetic acid in benzene. $(K_{f} f o r b e n z e n e = 5.12 k g m o l^{- 1})$

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Solution

$I n b e n z e n e 2 {C H}_{3} C O O H ⇌ ({C H}_{3} C O O H)_{2} i = 1 + (\frac{1}{2} - 1) i - 1 - \frac{α}{2}$

Here $α$ is degree of association

$Δ T_{f} = i K_{f} m 0.45 = (1 - \frac{α}{2}) (5.12) \frac{(\frac{0.2}{60})}{\frac{20}{1000}} 1 - \frac{α}{2} = 0.527 α = 0.945$

percentage degree of association=94.5

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