2,g of benzoic acid (C_6H_5COOH) dissolved in | vedclass

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(A) Given: Mass of solute $(w_2)$ = $2\,g$,Mass of solvent $(w_1)$ = $25\,g$,$\Delta T_f = 1.62\,K$,$K_f = 4.9\,K\,kg\,mol^{-1}$,Molar mass of benzoic

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$99.2$

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(A) Given: Mass of solute $(w_2)$ = $2\,g$,Mass of solvent $(w_1)$ = $25\,g$,$\Delta T_f = 1.62\,K$,$K_f = 4.9\,K\,kg\,mol^{-1}$,Molar mass of benzoic acid $(M_2)$ = $122\,g\,mol^{-1}$.We know,$\Delta T_f = i 	imes K_f 	imes m$.$1.62 = i 	imes 4.9 	imes \frac{2 	imes 1000}{122 	imes 25}$.$1.62 = i 	imes 3.213$.$i = \frac{1.62}{3.213} = 0.5042$.Benzoic acid exists as a dimer $(n=2)$ in benzene: $i = 1 - \alpha + \frac{\alpha}{n} = 1 - \frac{\alpha}{2}$.$0.5042 = 1 - \frac{\alpha}{2}$.$\frac{\alpha}{2} = 0.4958$.$\alpha = 0.9916$.Percentage degree of association = $99.16\% \approx 99.2\%$.

$2\,g$ of benzoic acid $(C_6H_5COOH)$ dissolved in $25\,g$ of benzene shows a depression in freezing point equal to $1.62\,K$. The molal depression constant $(K_f)$ for benzene is $4.9\,K\,kg\,mol^{-1}$. What is the percentage degree of association of the acid?

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$2 \ g$ of benzoic acid $(C_6H_5COOH)$ dissolved in $25 \ g$ of benzene shows a depression in freezing point equal to $1.62 \ K$. The molal depression constant for benzene is $4.9 \ K \ kg \ mol^{-1}$. The percentage of benzoic acid in dimeric form is ......... $\%$.

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