75.2 g of C_6H_5OH (phenol) is dissolved in | vedclass

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(B) The molar mass of phenol $(C_6H_5OH)$ is $M = 94 \ g \ mol^{-1}$.Given mass of phenol $w = 75.2 \ g$,so moles $n = \frac{75.2}{94} = 0.8 \ mol$.As

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$75$

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(B) The molar mass of phenol $(C_6H_5OH)$ is $M = 94 \ g \ mol^{-1}$.Given mass of phenol $w = 75.2 \ g$,so moles $n = \frac{75.2}{94} = 0.8 \ mol$.Assuming $1 \ kg$ of solvent,molality $m = 0.8 \ mol \ kg^{-1}$.The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$.Given $\Delta T_f = 7 \ K$ and $K_f = 14 \ K \ kg \ mol^{-1}$,we have $7 = i 	imes 14 	imes 0.8$.$i = \frac{7}{14 	imes 0.8} = \frac{1}{1.6} = 0.625$.For dimerization,$i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.$0.625 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.375 \implies \alpha = 0.75$.Thus,the percentage of phenol that dimerises is $75\%$.

$75.2 \ g$ of $C_6H_5OH$ (phenol) is dissolved in a solvent of $K_f = 14 \ K \ kg \ mol^{-1}$. If the depression in freezing point is $7 \ K$,then find the percentage of phenol that dimerises. (in $\%$)

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