25 mL of an aqueous solution of KCl was found to require 20 mL of 1M $$AgNO_3$$ solution when titrated using $$K_2CrO_4$$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? (Nearest integer).
(Given: $$K_f = 2.0 \text{ K kg mol}^{-1}$$)
Assume 1) 100% ionization and 2) density of the aqueous solution as $$1 \text{ g mL}^{-1}$$

We have a 25 mL KCl solution that requires 20 mL of 1 M $$AgNO_3$$; we need to find the depression in freezing point.

To begin, we find the concentration of KCl by considering the titration reaction: $$KCl + AgNO_3 \rightarrow AgCl + KNO_3$$.

Moles of $$AgNO_3$$ = $$20 \times 10^{-3} \times 1 = 0.02$$ mol.

Since the molar ratio is 1:1, moles of $$KCl = 0.02$$ mol in 25 mL of solution.

Next, we calculate molality.

Mass of solution = $$25 \text{ mL} \times 1 \text{ g/mL} = 25$$ g (given density = 1 g/mL).

Mass of KCl = $$0.02 \times 74.5 = 1.49$$ g.

Mass of solvent (water) = $$25 - 1.49 = 23.51$$ g $$= 0.02351$$ kg.

Molality $$m = \frac{0.02}{0.02351} = 0.8507$$ mol/kg.

We then apply the freezing point depression formula.

For KCl with 100% ionization: $$KCl \rightarrow K^+ + Cl^-$$, so the van’t Hoff factor $$i = 2$$.

$$\Delta T_f = i \cdot K_f \cdot m = 2 \times 2.0 \times 0.8507 = 3.40$$.

Rounding to the nearest integer gives $$\Delta T_f \approx 3$$.

The answer is 3.