Phenol associates in benzene to a certain ext | vedclass

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Question

(C) The molar mass of phenol $(C_6H_5OH)$ is $94 \ g \ mol^{-1}$.First,calculate the molality $(m)$ of the solution:$m = \frac{	ext{mass of solute (g

Vedclass · Accepted Answer

$73.3$

Vedclass · Answer

(C) The molar mass of phenol $(C_6H_5OH)$ is $94 \ g \ mol^{-1}$.First,calculate the molality $(m)$ of the solution:$m = \frac{	ext{mass of solute (g)}}{	ext{molar mass} 	imes 	ext{mass of solvent (kg)}} = \frac{20 \ g}{94 \ g \ mol^{-1} 	imes 1 \ kg} = 0.2128 \ mol \ kg^{-1}$.Using the freezing point depression formula: $\Delta T_f = i 	imes K_f 	imes m$.$0.69 = i 	imes 5.12 	imes 0.2128$.$i = \frac{0.69}{1.0895} \approx 0.633$.For the association of phenol into a dimer $(2C_6H_5OH ightleftharpoons (C_6H_5OH)_2)$,the van't Hoff factor is $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.$0.633 = 1 - \frac{\alpha}{2}$.$\frac{\alpha}{2} = 1 - 0.633 = 0.367$.$\alpha = 0.734$,which is $73.4 \%$. The closest option is $73.3 \%$.

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