Question

\(25 \,mL\) of an aqueous solution of \(KCl\) was found to require \(20\, mL\) of \(1\, M\) \(AgNO _3\) 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: \(K _f=20 \,K \,kg \,mol ^{-1}\)) 
Assume 1) \(100 \%\) ionization and 
2) density of the aqueous solution as \(1 \,g \,mL ^{-1}\)

Hint:

The van ’t Hoff factor (i) accounts for the number of particles into which a so lute dissociates in solution. For ionic compounds like KCl, i is the number of ions formed

Answer 1

Correct Answer: 3

Calculation of Depression in Freezing Point:

1. Calculate the moles of AgNO₃:
   Moles of AgNO₃ = Molarity × Volume (L) = 1 × 0.020 = 0.020 mol. 
2. Determine the moles of KCl:
   From the reaction:
   AgNO₃ + KCl → AgCl + KNO₃,
   1 mole of AgNO₃ reacts with 1 mole of KCl. Therefore, the moles of KCl are:
   Moles of KCl = 0.020 mol.
3. Calculate the molality of the KCl solution:
   The volume of the solution is 25 mL, and the density is 1 g/mL, so the mass of the solution is:
   Mass of solution = 25 g.
   Since the KCl solution is dilute, the mass of the solvent is approximately:
   Mass of solvent = 25 g = 0.025 kg.
   The molality is:
   Molality = Moles of solute / Mass of solvent (kg) = 0.020 / 0.025 = 0.8 mol/kg.
4. Calculate the depression in freezing point:
   KCl dissociates completely into K⁺ and Cl⁻, so the van ’t Hoff factor (i) is 2.
   The depression in freezing point is given by:
   ∆T_f = i × K_f × Molality
   Substituting the values:
   ∆T_f = 2 × 2.0 × 0.8 = 3.2 K.
   The nearest integer is:
   ∆T_f = 3 K.

Conclusion: The depression in freezing point is 3 K.

Answer 2

The correct answer is 3

At equivalence point,
mmole of $KCl=$ mmole of $AgN{O}_3$
$=20m$ mole
Volume of solution $=25ml$
Mass of solution $=25gm$
Mass of solvent
$=25-$ mass of solute
$=25-\left[20\times 10^{-3}\times 74.5\right]$
$=23.51gm$
Molality of $KCl=\frac{\text{mole of}KCl}{\text{mass of solvent in}kg}$
$=\frac{20\times 10^{-3}}{23.51\times 10^{-3}}=0.85$
$\text{i of}KCl=2(100\%\text{ionisation})$
$\Delta T_f=i\times K_f\times m$
$=2\times 2\times 0.85$
$=3.4$
$\approx 3$