CBSE Class 12 · 2015 · Delhi · SolutionsShort Answer

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3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of the solute (associated or dissociated). (Given: molar mass of benzoic acid = 122 g mol

^{- 1}

, K

_{f}

for benzene = 4.9 K kg mol

^{- 1}

)

Answer: i ≈ 0.51 — the acid is associated (dimerised) in benzene.

Solution

Observed molality:

m_{o b s} = \frac{1.62}{4.9} = 0.3306 mol kg^{- 1}

Observed moles =

0.3306 \times 0.049 = 0.0162

mol, so

M_{o b s} = \frac{3.9}{0.0162} = 240.7 g mol^{- 1}

van't Hoff factor:

i = \frac{M _{n or ma l}}{M _{o b s}} = \frac{122}{240.7}

i = 0.51

Since

i < 1

, benzoic acid is associated — it exists largely as dimers (hydrogen-bonded pairs) in benzene.

CBSE Class 12 Board · 2015 · Delhi · Subjective

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