25   mL of an aqueous solution of KCl was found to require 20   mL of 1 M   AgNO 3 solution…

At equivalence point,

$\Rightarrow$mmole of $\mathrm{KCl}=$ mmole of ${\mathrm{AgNO}}_3$

$=20 \mathrm{mmole}$

Volume of solution $=25\mathrm{ml}$

Mass of solution$(\mathrm{\rho } = 1 \mathrm{g}/\mathrm{mL})$

$=25 \mathrm{gm}$

Formula unit mass of potassium chloride = $39 + 35.5 {\mathrm{gmol}}^{-1} = 74.5 {\mathrm{gmol}}^{-1}$

Thus, Mass of solvent

$=25-$ mass of solute

$=25-\left[20\times {10}^{-3}\times 74.5\right]$

$=23.51 \mathrm{gm}$

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

$=\frac{20\times {10}^{-3}}{23.51\times {10}^{-3}}=0.85$

i of $\mathrm{KCl}=2(100\%$ ionisation )

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

$=2\times 2\times 0.85$

$=3.4$

$\simeq 3$ K