2 g of benzoic acid (C_6H_5COOH) dissolved i | vedclass

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(B) Given: $W_B = 2 \ g$,$K_f = 4.9 \ K \ kg \ mol^{-1}$,$W_A = 25 \ g$,$\Delta T_f = 1.62 \ K$. First,calculate the observed molar mass $(M_{obs})$ u

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$99.2$

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(B) Given: $W_B = 2 \ g$,$K_f = 4.9 \ K \ kg \ mol^{-1}$,$W_A = 25 \ g$,$\Delta T_f = 1.62 \ K$. First,calculate the observed molar mass $(M_{obs})$ using the formula: $\Delta T_f = K_f 	imes \frac{W_B}{M_{obs}} 	imes \frac{1000}{W_A}$. $M_{obs} = \frac{4.9 	imes 2 	imes 1000}{1.62 	imes 25} = 241.98 \ g \ mol^{-1}$. The theoretical molar mass of benzoic acid $(M_{theo})$ is $122 \ g \ mol^{-1}$. The van't Hoff factor $i = \frac{M_{theo}}{M_{obs}} = \frac{122}{241.98} \approx 0.504$. For the association reaction $2 C_6H_5COOH ightleftharpoons (C_6H_5COOH)_2$,the degree of association $\alpha$ is related to $i$ by $i = 1 - \alpha + \frac{\alpha}{n}$,where $n = 2$. $0.504 = 1 - \alpha + \frac{\alpha}{2} \Rightarrow 0.504 = 1 - \frac{\alpha}{2}$. $\frac{\alpha}{2} = 1 - 0.504 = 0.496 \Rightarrow \alpha = 0.992$. Thus,the percentage of benzoic acid in dimeric form is $99.2 \ \%$.

$2 \ g$ 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 for benzene is $4.9 \ K \ kg \ mol^{-1}$. The percentage of benzoic acid in dimeric form is ......... $\%$.

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