It is given that,

w₁ = 500 g

w₂ = 19.5 g

K_f = 1.86 K kg mol^-1

ΔT_f = 1K

We know that

M₂ = \(\frac{K_f \times w_2 \times 1000}{ΔT_f \times w_1}\)

= \(\frac{1.86 K \ kg mol^{-1} \times 19.5g \times 1000 g kg^{-1}}{500g \times 1K}\)

= 72.54 gmol^-1

Therefore, observed molar mass of CH₂FCOOH, (M₂)_obs = 72.54 g mol

The calculated molar mass of CH₂FCOOH is

(M₂)_cal = 14 + 19 + 12 + 16 + 16 + 1

= 78 gmol^-1

Therefore, van't Hoff factor, i

= \(\frac{(M_2)_{cal}}{(M_2)_{obs}}\)

= \(\frac{78g \ mol^{-1}}{72.54 g \ mol^{-1}}\)

= 1.0753

Let α be the degree of dissociation of CH₂FCOOH

CH₂FCOOH ↔ CH₂FCOO^- + H⁺

Initial conc.    C mol L^-1   0            0

At equilibrium  C(1 - α)   Cα         Cα

Total = C(1 + α)

∴ i = \(\frac{C(1 - \alpha)}{C}\)

⇒ i = 1 + α

⇒ α = i - 1

= 1.0753 - 1

= 0.0753

Now, the value of K_a is given as

K_a = \(\frac{[CH_2FCOO^-][H^+]}{[CH_2FCOOH]}\)

= \(\frac{C \alpha . C \alpha}{C(1 - \alpha)}\)

= \(\frac{C \alpha^2}{1 - \alpha}\)

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

C = \(\frac{\frac{19.5}{78}}{500}\)x 1000 M

= 0.5 M

Therefore, K_a = \(\frac{0.5 \times (0.0753)^2}{1-0.0753}\)

= \(\frac{0.5 \times 0.00567}{0.9247}\)

= 0.00307 (approximately)

= 3.07 x 10^-3