25mL of an aqueous solution of KCl was found to require 20mL of 1M...

At equivalence point,

$⇒$mmole of $KCl =$ mmole of ${AgNO}_{3}$

$= 20mmole$

Volume of solution $= 25ml$

Mass of solution$( ρ = 1g / mL )$

$= 25gm$

Formula unit mass of potassium chloride = $39 + 35 . 5{gmol}^{- 1} = 74 . 5{gmol}^{- 1}$

Thus, Mass of solvent

$= 25 -$ mass of solute

$= 25 - \left[20 \times {10}^{- 3} \times 74 . 5\right]$

$= 23 . 51gm$

Thus, molality of $KCl$$= \dfrac{\text{mole of}KCl}{\text{mass of solvent in}kg}$

$= \dfrac{20 \times {10}^{- 3}}{23 . 51 \times {10}^{- 3}} = 0 . 85$

i of $KCl = 2 ( 100$ ionisation )

$∴ {\Delta T}_{f} = i \times {K}_{f} \times m$

$= 2 \times 2 \times 0 . 85$

$= 3 . 4$

$≃ 3$ K