
At equivalence point,
⇒mmole of KCl= mmole of AgNO3
=20mmole
Volume of solution =25ml
Mass of solution(ρ=1g/mL)
=25gm
Formula unit mass of potassium chloride = 39+35.5gmol−1=74.5gmol−1
Thus, Mass of solvent
=25− mass of solute
=25−[20×10−3×74.5]
=23.51gm
Thus, molality of KCl= mass of solvent in kg mole of KCl
=23.51×10−320×10−3=0.85
i of KCl=2(100 ionisation )
∴ΔTf=i×Kf×m
=2×2×0.85
=3.4
≃3 K