Considering acetic acid dissociates in water, | vedclass

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Question

(A) Mass of $CH_3COOH = V 	imes d = 5 \ mL 	imes 1.2 \ g \ mL^{-1} = 6 \ g$.Moles of $CH_3COOH = \frac{6 \ g}{60 \ g \ mol^{-1}} = 0.1 \ mol$.Molali

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) Mass of $CH_3COOH = V 	imes d = 5 \ mL 	imes 1.2 \ g \ mL^{-1} = 6 \ g$.Moles of $CH_3COOH = \frac{6 \ g}{60 \ g \ mol^{-1}} = 0.1 \ mol$.Molality $(m) = \frac{0.1 \ mol}{1 \ kg \ 	ext{water}} = 0.1 \ m$.For the dissociation $CH_3COOH ightleftharpoons CH_3COO^{-} + H^{+}$,$K_{a} = \frac{C\alpha^{2}}{1-\alpha}$.Since $\alpha$ is small,$1-\alpha \approx 1$,so $K_{a} \approx C\alpha^{2}$.$\alpha = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{6.25 	imes 10^{-5}}{0.1}} = \sqrt{6.25 	imes 10^{-4}} = 25 	imes 10^{-3} = 0.025$.van't Hoff factor $(i) = 1 + \alpha(n-1) = 1 + 0.025(2-1) = 1.025$.Freezing point depression $\Delta T_{f} = i 	imes K_{f} 	imes m = 1.025 	imes 1.86 	imes 0.1 = 0.19065 \ {}^{\circ}C$.Given $\Delta T_{f} = x 	imes 10^{-2} \ {}^{\circ}C$,so $x \approx 19$.

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