Pure benzene freezes at 5.3,^oC. A solution o | vedclass

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(B) Given: $T_f^o = 5.3\,^oC$,$T_f = 4.47\,^oC$,$\Delta T_f = 5.3 - 4.47 = 0.83\, K$. Mass of solute $(w_2) = 0.223\, g$,Mass of solvent $(w_1) = 4.4\

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exists as a dimer in benzene

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(B) Given: $T_f^o = 5.3\,^oC$,$T_f = 4.47\,^oC$,$\Delta T_f = 5.3 - 4.47 = 0.83\, K$. Mass of solute $(w_2) = 0.223\, g$,Mass of solvent $(w_1) = 4.4\, g = 0.0044\, kg$. Molar mass of $C_6H_5CH_2COOH$ $(M_2) = 136\, g/mol$. Using the formula: $\Delta T_f = i 	imes K_f 	imes m = i 	imes K_f 	imes \frac{w_2 	imes 1000}{M_2 	imes w_1}$. $0.83 = i 	imes 5.12 	imes \frac{0.223 	imes 1000}{136 	imes 4.4}$. $0.83 = i 	imes 5.12 	imes 0.3728$. $i = \frac{0.83}{1.908} \approx 0.435$. Since the van't Hoff factor $i $0.435 = 1 - \alpha/2 \implies \alpha/2 = 0.565 \implies \alpha = 1.13$. Since the value of $i$ is approximately $0.5$,it indicates that phenylacetic acid exists as a dimer in benzene.

Pure benzene freezes at $5.3\,^oC$. $A$ solution of $0.223\, g$ of phenylacetic acid $(C_6H_5CH_2COOH)$ in $4.4\, g$ of benzene $(K_f = 5.12\, K\, kg\, mol^{-1})$ freezes at $4.47\,^oC$. From this,it can be concluded that phenylacetic acid ........

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