25 mL of an aqueous solution of KCl was found to require 20 mL of `1 M AgNO_(3)` solution when titrated using `K_(2)CrO_(4)` as indicator . Depression in freezing point of KCl solution with `100%` ionization will be :
(`K_(f) = (10)/(9)` k /molal)

25 mL of an aqueous solution of KCl was found to requires 20 mL of 1M AgNO_(3) solution when titrated using a K_(2)CrO_(4) as indicator. Depression in freezing point of KCl solution with 100% ionisation will be : (K_(f) = 2.0 mol^(-1) kg "and molarity = molality")

The freezing point of a 0.05 molal solution of a non-electrolyte in water is [K_(f)=1.86K//m]

Calculate depression of freezing point for 0.56 molal aq. Solution of KCl. (Given : K_f(H_(2)O) = 1.8 kg mol^(-1) ).

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

Equal volume of 1.0 M KCl and 1.0 M AgNO_(3) are mixed . The depression of freezing point of the resulting solution will be : ( K_(f) (H_(2)O) = 1.86K kg mol^(-1) , Assume : 1M = 1m)

KBr is 80% ionized in solution. The freezing point of 0.4 molal solution of KBr is : K_f (H_2O) = 1.86 (K kg)/("mole")