CBSE Class 12 · 2011 · Delhi Compartment · SolutionsLong Answer

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a) Explain why a solution of chloroform and acetone shows negative deviation from Raoult's law.
b) Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 g of phenol in 1.0 kg of benzene has its freezing point lowered by 0.69 K. Calculate the fraction of phenol that has dimerised. (Given K

_{f}

for benzene = 5.1 K kg mol

^{- 1}

)

Solution

(a) In the chloroform–acetone mixture a hydrogen bond forms between the hydrogen of CHCl

_{3}

and the carbonyl oxygen of acetone. The A–B attractions become stronger than the A–A and B–B attractions, the escaping tendency of both components falls, and the vapour pressure is lower than Raoult's law predicts — negative deviation.
(b) Normal molality of phenol (

M = 94

m = \frac{20/94}{1.0} = 0.2128 mol kg^{- 1}

Calculated depression:

Δ T_{f}^{c a l c} = 5.1 \times 0.2128 = 1.085 K

van't Hoff factor:

i = \frac{0.69}{1.085} = 0.636

For dimerisation,

i = 1 - \frac{α}{2}

α = 2 (1 - 0.636)

α = 0.728 — about 73% of the phenol is dimerised

CBSE Class 12 Board · 2011 · Delhi Compartment · Subjective

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CBSE Class 12 · 2011 · Delhi Compartment · SolutionsLong Answer

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