If a solution prepared by adding 6.1 g of be | vedclass

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Question

(B) $1$. Calculate molality $(m)$: $m = \frac{	ext{mass of solute}}{	ext{molar mass of solute} 	imes 	ext{mass of solvent in kg}} = \frac{6.1 \ g}

Vedclass · Accepted Answer

$95$

Vedclass · Answer

(B) $1$. Calculate molality $(m)$: $m = \frac{	ext{mass of solute}}{	ext{molar mass of solute} 	imes 	ext{mass of solvent in kg}} = \frac{6.1 \ g}{122 \ g/mol 	imes 0.5 \ kg} = 0.1 \ m$. $2$. Calculate depression in freezing point $(\Delta T_f)$: $\Delta T_f = T_f^o - T_f = 5.50 \ ^oC - (-0.290 \ ^oC) = 5.79 \ ^oC$. $3$. Calculate van't Hoff factor $(i)$: $\Delta T_f = i 	imes K_f 	imes m \implies 5.79 = i 	imes 5.12 	imes 0.1 \implies i = \frac{5.79}{0.512} \approx 1.13$. $4$. For association of benzoic acid: $2C_6H_5COOH ightleftharpoons (C_6H_5COOH)_2$. $5$. The degree of association $(\alpha)$ is given by $i = 1 + \alpha(\frac{1}{n} - 1)$,where $n=2$. $6$. $1.13 = 1 + \alpha(\frac{1}{2} - 1) \implies 0.13 = -0.5\alpha \implies \alpha = 0.26$. $7$. Note: The provided solution $95 \%$ is inconsistent with the data provided in the prompt. Based on standard calculation,the result is $26 \%$.

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