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 \(1 \,g \,mL ^{-1}\)
\(25 \,mL\) 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 \(1 \,g \,mL ^{-1}\)
Assume 1) \(100 \%\) ionization and
2) density of the aqueous solution as \(1 \,g \,mL ^{-1}\)
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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
Updated On: Jan 15, 2025
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Correct Answer: 3
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Calculation of Depression in Freezing Point:
- Calculate the moles of AgNO3:
Moles of AgNO3 = Molarity × Volume (L) = 1 × 0.020 = 0.020 mol. - 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. - 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. - 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:
∆Tf = i × Kf × Molality
Substituting the values:
∆Tf = 2 × 2.0 × 0.8 = 3.2 K.
The nearest integer is:
∆Tf = 3 K.
Conclusion: The depression in freezing point is 3 K.
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Approach Solution -2
The correct answer is 3
At equivalence point,
mmole of mmole of
mole
Volume of solution
Mass of solution
Mass of solvent
mass of solute
Molality of
mmole of mmole of
mole
Volume of solution
Mass of solution
Mass of solvent
mass of solute
Molality of
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.