2 g of benzoic acid (M_w = 122 g/mol) disso | vedclass

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(B) Step $1$: Calculate the theoretical depression in freezing point $(\Delta T_{cal})$ assuming no association.$\Delta T_{cal} = K_f 	imes m = K_f \

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$99.2$

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(B) Step $1$: Calculate the theoretical depression in freezing point $(\Delta T_{cal})$ assuming no association.$\Delta T_{cal} = K_f 	imes m = K_f 	imes \frac{w 	imes 1000}{M_w 	imes W} = 4.9 	imes \frac{2 	imes 1000}{122 	imes 25} = 3.213 \ K$.Step $2$: Calculate the van't Hoff factor $(i)$.$i = \frac{\Delta T_{exp}}{\Delta T_{cal}} = \frac{1.62}{3.213} \approx 0.5042$.Step $3$: Relate $i$ to the degree of association $(\alpha)$ for dimerization $(n=2)$.$i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$.$0.5042 = 1 - \frac{\alpha}{2} \implies \frac{\alpha}{2} = 0.4958 \implies \alpha = 0.9916$.Percentage association = $99.16 \% \approx 99.2 \%$.

$2 \ g$ of benzoic acid $(M_w = 122 \ g/mol)$ dissolved in $25 \ g$ of $C_6H_6$ shows a depression in freezing point equal to $1.62 \ K$. Given $K_f (C_6H_6) = 4.9 \ K \ kg \ mol^{-1}$. If the acid forms a dimer in the solution,then the percentage association of the acid is $......... \%$.

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