Given, $w_{2}=19.5\ \text{g},\ w_{1}=500\ \text{g}$ $K_{f}=1.86\ \text{K kg mol}^{-1}$, $(\Delta T_f)_{\text{obs}}=1.0^{\circ}\text{C},\ M_2=?$ $\therefore M_2$ (observed)$= \frac{1000 K_f w_2}{w_1 \Delta T_f}$ $= \frac{1000 \times 1.86 \times 19.5}{500 \times 1.0}$ $=72.54\ \text{g mol}^{-1}$$M_2$ (calculated) for $\mathrm{CH_2FCOOH} = 78\ \text{g mol}^{-1}$ van’t Hoff factor $(i)= \frac{(M_2)_{\text{cal}}}{(M_2)_{\text{obs}}}= \frac{78}{72.54}=1.0753$ Calculation of dissociation constant : Suppose degree of dissociation at the given concentrationis α.Total = C(1 + α)$\therefore i= \frac{C(1+\alpha)}{C}=1+\alpha$ or, $\alpha=i-1=1.0753-1=0.0753$Again $K_{a}= \frac{[\mathrm{CH_2FCOO}^{-}][\mathrm{H}^{+}]}{[\mathrm{CH_2FCOOH}]}$ $= \frac{C \alpha \times C \alpha}{C(1-\alpha)} = \frac{C \alpha^2}{1-\alpha}$Taking volume of the solution as 500 mL,$C= \frac{19.5}{78} \times \frac{1}{500} \times 1000 = 0.5\ \text{M}$$\therefore K_{a}= \frac{C \alpha^2}{1-\alpha}= \frac{(0.5)(0.0753)^2}{1-0.0753}$ $= 3.07 \times 10^{-3}$