25 mL of an aqueous solution of KCl was found to require 20 mL of 1M AgNO$$_3$$ solution when titrated using K$$_2$$CrO$$_4$$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration?
(Nearest integer).
(Given: K$$_f$$ = 2.0 K kg mol$$^{-1}$$)
Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL$$^{-1}$$

We have 25 mL of KCl solution titrated with 20 mL of 1M $$AgNO_3$$ solution. Given: $$K_f = 2.0$$ K kg $$mol^{-1}$$, 100% ionization, density of solution = 1 g/mL.

The reaction is: $$KCl + AgNO_3 \rightarrow AgCl\downarrow + KNO_3$$

Moles of $$AgNO_3$$ used = 0.020 $$\times$$ 1 = 0.02 mol. Since the ratio is 1:1, moles of KCl = 0.02 mol.

Now, mass of solution = 25 $$\times$$ 1 = 25 g, and mass of KCl = 0.02 $$\times$$ 74.5 = 1.49 g. So the mass of solvent (water) = 25 - 1.49 = 23.51 g = 0.02351 kg.

The molality is:

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

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

So the depression in freezing point is:

$$\Delta T_f = i \times K_f \times m = 2 \times 2.0 \times 0.8507 = 3.40 \text{ K}$$

Rounding to the nearest integer, the answer is $$3$$ K.