CBSE Class 12 · 2004 · Delhi Compartment · SolutionsShort Answer

2 g of benzoic acid (C

_{6}

_{5}

COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol

^{- 1}

. What is the percentage association of the acid if it forms a dimer in solution?

Answer: ≈ 99.2% associated

Solution

Observed molality:

m_{o b s} = \frac{Δ T _{f}}{K _{f}} = \frac{1.62}{4.9} = 0.3306 mol kg^{- 1}

Observed moles in 25 g benzene =

0.3306 \times 0.025 = 8.265 \times 1 0^{- 3}

, so

M_{o b s} = \frac{2}{8.265 \times 1 0 ^{- 3}} = 242.0 g mol^{- 1}

van't Hoff factor:

i = \frac{M _{n or ma l}}{M _{o b s}} = \frac{122}{242.0} = 0.504

For dimerisation (2A

\to

_{2}

i = 1 - \frac{α}{2}

α = 2 (1 - i) = 2 (1 - 0.504)

α = 0.992 — about 99.2% associated

CBSE Class 12 Board · 2004 · Delhi Compartment · Subjective