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")

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")

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 1 molar NaCl solution assuming NaCl to be 100% dissociated in water is: ( K_f = 1.86 K molality^-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) ):

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) ):

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)