25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M 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? \underline\hspace2cm (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

Solutions PYQ — JEE Main 2023 · 1 Feb · Shift 1 · Q12 with Solution

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Correct answer: 3

The titration reaction is:

\mathrm{AgNO_3 + KCl \to AgCl\downarrow + KNO_3}

This is a $1:1$ reaction, so moles of $\mathrm{AgNO_3}$ used = moles of $\mathrm{KCl}$ present.

Given:

V_{\mathrm{AgNO_3}} = 20\,\text{mL} = 0.020\,\text{L}, \qquad M_{\mathrm{AgNO_3}} = 1\,\text{M}

So,

\text{moles of } \mathrm{AgNO_3} = M \times V = 1 \times 0.020 = 0.020\,\text{mol}

Hence, moles of $\mathrm{KCl}$ in $25$ mL solution are:

0.020\,\text{mol}

Volume of solution = $25$ mL. Given density = $1\,\text{g mL}^{-1}$ , so mass of solution is:

25\,\text{mL} \times 1\,\text{g mL}^{-1} = 25\,\text{g}

Mass of solute $\mathrm{KCl}$ :

\text{molar mass of KCl} = 39 + 35.5 = 74.5\,\text{g mol}^{-1}

\text{mass of KCl} = 0.020 \times 74.5 = 1.49\,\text{g}

Therefore, mass of solvent = mass of solution $-$ mass of solute:

25 - 1.49 = 23.51\,\text{g} = 0.02351\,\text{kg}

m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.020}{0.02351}

m \approx 0.851\,\text{mol kg}^{-1}

For $\mathrm{KCl}$ with 100% ionization:

\mathrm{KCl \to K^+ + Cl^-}

So van’t Hoff factor:

i = 2

Formula:

\Delta T_f = i K_f m

Given $K_f = 2.0\,\text{K kg mol}^{-1}$ :

\Delta T_f = 2 \times 2.0 \times 0.851

\Delta T_f \approx 3.404\,\text{K}

Nearest integer:

\boxed{3}

Stored correct answer = $3$

Our derived answer = $3$

So, the derived answer agrees with the stored correct answer.