1.2 , mL of acetic acid is dissolved in water | vedclass

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(B) $1$. Calculate the mass of acetic acid: $Mass = Density 	imes Volume = 1.02 \, g \, mL^{-1} 	imes 1.2 \, mL = 1.224 \, g$.$2$. Calculate the mol

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$5$

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(B) $1$. Calculate the mass of acetic acid: $Mass = Density 	imes Volume = 1.02 \, g \, mL^{-1} 	imes 1.2 \, mL = 1.224 \, g$.$2$. Calculate the moles of acetic acid: $Moles = \frac{1.224 \, g}{60 \, g \, mol^{-1}} = 0.0204 \, mol$.$3$. Calculate molality $(m)$: Since the solution is dilute,we assume the density of the solution is approximately $1 \, g \, mL^{-1}$,so $2 \, L$ of solution is $2 \, kg$ of water. $m = \frac{0.0204 \, mol}{2 \, kg} = 0.0102 \, mol \, kg^{-1}$.$4$. Use the depression in freezing point formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$.$5$. For acetic acid dissociation $(CH_{3}COOH ightleftharpoons CH_{3}COO^{-} + H^{+})$,$i = 1 + \alpha$.$6$. Substitute values: $0.0198 = (1 + \alpha) 	imes 1.85 	imes 0.0102$.$7$. Solve for $\alpha$: $1 + \alpha = \frac{0.0198}{1.85 	imes 0.0102} \approx 1.04928$.$8$. $\alpha = 0.04928$,which is $4.928 \% \approx 5 \%$. The nearest integer is $5$.

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