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 an indicator. What is the depression in freezing point of KCl solution of the given concentration? (Nearest integer). (Given : K f = 2 . 0 K kg mol - 1 ) Assume 1) 100 % ionization and 2) density of the aqueous solution as 1 g mL - 1

Step-by-step solution

At equivalence point, ⇒ mmole of KCl = mmole of AgNO 3 = 20   mmole Volume of solution = 25 ml Mass of solution ( ρ   =   1   g / mL ) = 25   gm Formula unit mass of potassium chloride = 39   +   35 . 5   gmol - 1   =   74 . 5   gmol - 1 Thus, Mass of solvent = 25 - mass of solute = 25 - 20 × 10 - 3 × 74 . 5 = 23 . 51   gm Thus, molality of KCl =  mole of  KCl  mass of solvent in  kg = 20 × 10 - 3 23 . 51 &#215