19.5 g of fluoro acetic acid (molar mass = 78 g mol ^ -1 ) is...

The molality of the solution is given by:

$m = \dfrac{\text{Mass of solute}}{\text{Molar mass of solute} \times \text{Mass of solvent in kg}}$

$m = \dfrac{19.5}{78 \times 0.5} = 0.5 \text{ mol kg}^{-1}$

The depression in freezing point is given by:

$\Delta T_f = i \times K_f \times m$

Substituting the given values:

$1 = i \times 1.86 \times 0.5$

$1 = i \times 0.93$

$i = \dfrac{1}{0.93} = \dfrac{100}{93}$

For the dissociation of a weak acid $HA \rightleftharpoons H^+ + A^-$, the van 't Hoff factor is $i = 1 + \alpha$, where $\alpha$ is the degree of dissociation.

$1 + \alpha = \dfrac{100}{93}$

$\alpha = \dfrac{100}{93} - 1 = \dfrac{7}{93}$

The acid dissociation constant $K_a$ is given by:

$K_a = \dfrac{C \alpha^2}{1 - \alpha}$

Since molarity and molality have the same values, $C = 0.5 \text{ M}$.

$K_a = \dfrac{0.5 \times \left(\dfrac{7}{93}\right)^2}{1 - \dfrac{7}{93}}$

$K_a = \dfrac{0.5 \times \dfrac{49}{93^2}}{\dfrac{86}{93}} = \dfrac{0.5 \times 49}{93 \times 86}$

$K_a = \dfrac{24.5}{7998} \approx 3.06 \times 10^{-3}$

The closest value is $3 \times 10^{-3}$.

Answer: $3 \times 10^{-3}$