At equivalence point, mmole of KCl= mmole of AgNO3=20 mmole Volume of solution =25ml Mass of solution =25gm Mass of solvent =25 - mass of solute =25−[20×10−3×74.5] =23.51gm Molality of KCl=
mole of KCl
mass of solvent in kg
=
20×10−3
23.51×10−3
=0.85 i of KCl=2(100% ionisation ) ∆Tf=i×Kf×m =2×2×0.85 =3.4 ≃3