0.1 molal aqueous solution of weak acid is 30% ionized. of the Kf for water is 1.86C/m , the freezing point of the solution will be

Consider the weak acid is HA. Let "n" is the number of moles of HA that are initially present and "x" is the degree of dissociation. As 30 % dissociation occurs. So degree of dissociation is 0.3. Its dissociation can be represented as


                                                       HA   ⇌   H⁺  +  A^-
Concentration initially                      n           0         0
Concentration at equilibrium          n-0.3n     0.3n   0.3n
Total number of moles initially = n
Total number of moles at equilibrium = n-0.3n+0.3n+0.3n
= 1.3 n
Van't hoff factor, i = Number of moles at equilibrium / number of moles initially
= 1.3 n / n
= 1.3
A we knoe depression  in freezing point is given by
∆T_f = i K_f m
​= 1.3$\times$1.86$\times$0.1
= 0.242 K
∆​T_f = T_f⁰ - T_f
T_f⁰ = 273 K (freezing point of pure solvent i.e., water is 273 K)
T_f = 273 - 0.242
= 272.76 K
Freezing point of solution = 272.26 K