19.5 text{ g} of fluoroacetic acid (molar mas | vedclass

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(D) $1$. Calculate the number of moles of fluoroacetic acid: $	ext{Moles} = \frac{19.5 	ext{ g}}{78 	ext{ g mol}^{-1}} = 0.25 	ext{ mol}$.$2$. Cal

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$3 	imes 10^{-3}$

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(D) $1$. Calculate the number of moles of fluoroacetic acid: $	ext{Moles} = \frac{19.5 	ext{ g}}{78 	ext{ g mol}^{-1}} = 0.25 	ext{ mol}$.$2$. Calculate the molality $(m)$: $m = \frac{0.25 	ext{ mol}}{0.5 	ext{ kg}} = 0.5 	ext{ m}$.$3$. Use the freezing point depression formula: $\Delta T_f = i \cdot K_f \cdot m$.$4$. Given $\Delta T_f = 1 	ext{ K}$, $K_f = 1.86 	ext{ K kg mol}^{-1}$, and $m = 0.5 	ext{ m}$, we have: $1 = i 	imes 1.86 	imes 0.5$.$5$. Solving for the van't Hoff factor $(i)$: $i = \frac{1}{0.93} \approx 1.075$.$6$. Since $i = 1 + \alpha$ for a weak acid, $\alpha = i - 1 = 1.075 - 1 = 0.075$.$7$. Calculate the dissociation constant $(K_a)$: $K_a = C \alpha^2 = 0.5 	imes (0.075)^2 = 0.5 	imes 0.005625 = 0.0028125 \approx 3 	imes 10^{-3}$.

Similar Questions

$A$ $0.1 \ m$ aqueous solution of a weak acid $HX$ is $30\%$ ionized. If $K_f$ for water is $1.86 \ ^{\circ}C/m,$ the freezing point of the solution will be ......... $^{\circ}C$.

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Consider the following solutions at $25\,^oC$:
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Select the correct statement.

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