$19.5 \ g$ of $CH_2FCOOH$ is dissolved in $500 \ g$ of water. The depression in the freezing point of water observed is $1.0 \ ^\circ C$. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

(N/A) It is given that:
$w_1 = 500 \ g$
$w_2 = 19.5 \ g$
$K_f = 1.86 \ K \ kg \ mol^{-1}$
$\Delta T_f = 1 \ K$
We know that:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
$= \frac{1.86 \ K \ kg \ mol^{-1} \times 19.5 \ g \times 1000 \ g \ kg^{-1}}{500 \ g \times 1 \ K}$
$= 72.54 \ g \ mol^{-1}$
Therefore,observed molar mass of $CH_2FCOOH$,$(M_2)_{obs} = 72.54 \ g \ mol^{-1}$
The calculated molar mass of $CH_2FCOOH$ is $(M_2)_{cal} = 12 + 2 + 19 + 12 + 16 + 16 + 1 = 78 \ g \ mol^{-1}$
Therefore,van't Hoff factor,$i = \frac{(M_2)_{cal}}{(M_2)_{obs}} = \frac{78}{72.54} = 1.0753$
For the dissociation reaction: $CH_2FCOOH \rightleftharpoons CH_2FCOO^{-} + H^{+}$
At equilibrium: $C(1-\alpha)$,$C\alpha$,$C\alpha$
Total moles $= C(1+\alpha)$
$\therefore i = 1 + \alpha \Rightarrow \alpha = i - 1 = 0.0753$
Concentration $C = \frac{19.5 \ g}{78 \ g \ mol^{-1}} \times \frac{1000 \ g}{500 \ g} \approx 0.5 \ M$
$K_a = \frac{C\alpha^2}{1 - \alpha} = \frac{0.5 \times (0.0753)^2}{1 - 0.0753} = \frac{0.5 \times 0.00567}{0.9247} \approx 3.07 \times 10^{-3}$