When 2.44 g of benzoic acid (C_6H_5COOH) is | vedclass

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(B) Given: $W_B = 2.44 \ g$,$K_f = 5.0 \ K \ kg \ mol^{-1}$,$W_A = 25 \ g$,$\Delta T_f = 2.2 \ K$. First,calculate the observed molar mass $(M_{obs})$

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Given: $W_B = 2.44 \ g$,$K_f = 5.0 \ K \ kg \ mol^{-1}$,$W_A = 25 \ g$,$\Delta T_f = 2.2 \ K$. First,calculate the observed molar mass $(M_{obs})$ using the formula: $\Delta T_f = K_f 	imes \frac{W_B}{M_{obs}} 	imes \frac{1000}{W_A}$. $M_{obs} = \frac{5.0 	imes 2.44 	imes 1000}{2.2 	imes 25} = 221.8 \ g \ mol^{-1}$. The theoretical molar mass of benzoic acid $(C_6H_5COOH)$ is $122 \ g \ mol^{-1}$. For association: $2 C_6H_5COOH ightleftharpoons (C_6H_5COOH)_2$. Van't Hoff factor $i = \frac{M_{theoretical}}{M_{observed}} = \frac{122}{221.8} \approx 0.55$. Also,$i = 1 + (\frac{1}{n} - 1)x$,where $n=2$ for dimer and $x$ is the degree of association. $0.55 = 1 + (\frac{1}{2} - 1)x \implies 0.55 = 1 - 0.5x$. $0.5x = 0.45 \implies x = 0.90$. Percentage association = $90\%$.

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