The freezing point of benzene decreases by 0. | vedclass

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(D) The association reaction is: $2 \ CH_{3}COOH \rightleftharpoons (CH_{3}COOH)_{2}$Initial moles: $1 \quad 0$At equilibrium: $1 - \alpha \quad \alph

Vedclass · Accepted Answer

$94.6$

Vedclass · Answer

(D) The association reaction is: $2 \ CH_{3}COOH ightleftharpoons (CH_{3}COOH)_{2}$Initial moles: $1 \quad 0$At equilibrium: $1 - \alpha \quad \alpha / 2$Total moles at equilibrium: $1 - \alpha + \alpha / 2 = 1 - \alpha / 2$Van't Hoff factor $i = 1 - \alpha / 2$Using the formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$Molality $m = \frac{0.2 / 60}{20 / 1000} = \frac{0.2 	imes 1000}{60 	imes 20} = \frac{200}{1200} = 0.1667 \ m$$0.45 = (1 - \alpha / 2) 	imes 5.12 	imes 0.1667$$0.45 = (1 - \alpha / 2) 	imes 0.8535$$1 - \alpha / 2 = 0.5272$$\alpha / 2 = 0.4728$$\alpha = 0.9456$Percentage association $= 94.56 \% \approx 94.6 \%$

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