$75.2 g$ of 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.

75

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80

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70

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100

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Solution

The correct option is A $75$ %
$2 C_{6} H_{5} O H ⇌ (C_{6} H_{5} O H)_{2}$
Initially $C$ $0$
At equil $C (1 - α)$ $C α / 2$

Total number of moles $= C - C α + (C α / 2)$
$= C - \frac{C α}{2} = \frac{2 C - C α}{2} = \frac{C (2 - α)}{2}$

$C = \frac{W e i g h t}{M o l e c u l a r w e i g h t} = \frac{75.2}{94} = 0.8$

$Δ T_{f} = K_{f} \times m$

$\Rightarrow 7 = 14 \times 0.8 (\frac{2 - α}{2})$ , or $1 = 0.8 (2 - α)$ $α = 0.75 = 75$

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