75.2 g of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ (phenol) is dissolved in a solvent of $\mathrm{K}_{\mathrm{f}}=14$. If the depression in freezing point is 7 K then find the % of phenol that dimerises.

JEE Advanced · 2006 · 09 Apr 2006 · Solutions

75.2 g of

C_{6} H_{5} OH

(phenol) is dissolved in a solvent of

K_{f} = 14

. If the depression in freezing point is 7 K then find the % of phenol that dimerises.

Answer: 75

Solution

Phenol undergoes dimerization:

2 C_{6} H_{5} OH ⇌ (C_{6} H_{5} OH)_{2}

Molality of phenol:

m = \frac{75.2}{94 \times 1} = 0.8

mol/kg (assuming 1 kg solvent)

For dimerization with degree of association

α

Initial moles: 1
At equilibrium: $(1 - α)$ monomers and $\frac{α}{2}$ dimers
Total particles: $1 - α + \frac{α}{2} = 1 - \frac{α}{2}$
Van't Hoff factor: $i = 1 - \frac{α}{2}$

Using

Δ T_{f} = i \cdot K_{f} \cdot m

7 = (1 - \frac{α}{2}) \times 14 \times 0.8

7 = 11.2 (1 - \frac{α}{2})

1 - \frac{α}{2} = \frac{7}{11.2} = 0.625

\frac{α}{2} = 0.375

α = 0.75 = 75%

JEE Advanced · 2006 · 09 Apr 2006 · Numerical answer

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