When 2.50 g of an unknown weak acid (ha) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of the resulting solution is -0.255 ∘c. part a calculate ka for the unknown weak acid?

When dT = Kf * molality * i
                = Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
                               = 2.5g /250g x 1 mol /85 g x1000g/kg
                               =0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution 
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and  i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
     = (0.1176 * 0.359)^2 / (1-0.359)
     = 2.8x10^-3

7is b8 is d9 is c10 is a11 is d

Answer( precipitate );