Molar mass of CH₃CH₂CHClCOOH

= 15+14+13+35.5+12+16+16+1

= 122.5 g mol^-1

∴ No. of moles present in 10 g of CH₃CH₂CHClCOOH

= \(\frac{10 \ g}{122.5 \ gmol^{-1}}\)

= 0.0816 mol

It is given that 10 g of CH₃CH₂CHClCOOH is added to 250 g of water

∴ Molality of the solution = \(\frac{0.0186}{250}\)x 1000

= 0.3264 mol kg^-1

Let αbe the degree of dissociation of CH₃CH₂CHClCOOH.

CH₃CH₂CHClCOOH undergoes dissociation according to the following equation:

CH₃CH₂CHClCOOH ↔ CH₃CH₂CHClCOO^- + H⁺

Initial conc.    C mol L^-1   0            0

At equilibrium  C(1 - α)   Cα         Cα

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

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

Since αis very small with respect to 1, 1 - α

Now, K_a = \(\frac{C \alpha^2}{1}\)

⇒ K_a = Cα²

⇒ α = \(\sqrt{\frac{K_a}{C}}\)

= \(\sqrt{\frac{1.4 \times 10^{-3}}{0.3264}}\) (∴ K_a = 1.4 x 10^-3)

= 0.0655

Again, 

CH₃CH₂CHClCOOH ↔ CH₃CH₂CHClCOO^- + H⁺

Initial moles    1   0            0

At equilibrium  1 - α   α         Cα

Total moles of equilibrium = 1 - α+ α+ α

= 1 + α

∴ i = \(\frac{1 + \alpha}{1}\)

= 1 + α

= 1+0.0655

= 1.0655

Hence, the depression in the freezing point of water is given as:

ΔT_f = i.K_f m

= 1.0655 x 1.86 K kg mol^-1 x 0.3264 mol kg^-1

= 0.65 K