It is given that:

${\mathrm{w}}_1=500 \mathrm{g}$
${\mathrm{w}}_2=19.5 \mathrm{g}$
${\mathrm{K}}_{\mathrm{f}}=1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$
$∆{\mathrm{T}}_{\mathrm{f}}=1\mathrm{K}$

We know that:

${\mathrm{M}}_2=\frac{{\mathrm{K}}_{\mathrm{f}}\times {\mathrm{w}}_2\times 1000}{∆{\mathrm{T}}_{\mathrm{f}}\times {\mathrm{w}}_1}$
$=\frac{1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\times 19.5 \mathrm{g}\times 1000 \mathrm{g} {\mathrm{kg}}^{-1}}{500 \mathrm{g}\times 1\mathrm{K}}$
$=72.54 \mathrm{g} {\mathrm{mol}}^{-1}$

Therefore, observed molar mass of ${\mathrm{CH}}_2\mathrm{FCOOH}, {(\mathrm{M}}_{}$₂)obs=72.54 g mol

The calculated molar mass of CH₂FCOOH

$\left({\mathrm{M}}_2\right)\mathrm{cal}=14+19+12+16+16+1=78 \mathrm{g} {\mathrm{mol}}^{-1}$

Therefore, van’t Hoff factor, $\mathrm{i}=\frac{\left({\mathrm{M}}_2\right)\mathrm{cal} }{{(\mathrm{M}}_{}}$₂)obs is:

$=\frac{78 \mathrm{g} {\mathrm{mol}}^{-1}}{72.54 \mathrm{g} {\mathrm{mol}}^{-1}}$
$=1.0753$

Let α be the degree of dissociation of CH₂FCOOH

${\mathrm{CH}}_2\mathrm{FCOOH}\leftrightarrow {\mathrm{CH}}_2{\mathrm{FCOOH}}^{-}+{\mathrm{H}}^{+}$
$\mathrm{Initial} \mathrm{conc}. \mathrm{C} \mathrm{mol} {\mathrm{L}}^{-1} 0 0$
$\mathrm{At} \mathrm{equilibrium} \mathrm{C}(1-\mathrm{\alpha }) \mathrm{C\alpha } \mathrm{C\alpha } \mathrm{Total}=\mathrm{C}(1+\mathrm{\alpha })$

$\therefore \mathrm{i}=\frac{\mathrm{C}(1+\mathrm{\alpha })}{\mathrm{C}}$
$\Rightarrow \mathrm{i}=1+\mathrm{\alpha }$
$\Rightarrow \mathrm{\alpha }=\mathrm{i}-1$
$=1.0753-1$
$=0.0753$

Now, the value of K_a is given as:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{CH}}_2{\mathrm{FCOOH}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{CH}}_2\mathrm{FCOOH}\right]}$
$=\frac{\mathrm{C\alpha }.\mathrm{C\alpha }}{\mathrm{C}(1-\mathrm{\alpha })}$
$=\frac{{\mathrm{C\alpha }}^2}{1-\mathrm{\alpha }}$

Taking the volume of the solution as 500 mL, we have the concentration:

$\mathrm{C}=\frac{{\displaystyle \frac{19.5}{78}}}{500}\times 1000 \mathrm{M}$
$=0.5 \mathrm{M}$

Therefore, ${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{C\alpha }}^2}{1-\mathrm{\alpha }}$

$=\frac{0.5\times {(0.0753)}^2}{1-0.0753}$
$=\frac{0.5\times 0.00567}{0.9247}$
$=0.00307 \left(\mathrm{approximately}\right)$
$=3.07\times {10}^{-3}$