🌡️ Colligative Properties: A Complete Guide

Unlocking the Secrets of Solutions: A Deep Dive into Colligative Properties

Have you ever tossed salt on an icy driveway on a frigid winter morning and watched the ice melt before your eyes? Or perhaps you’ve wondered why you add a pinch of salt to a pot of boiling water for pasta?

These everyday actions are perfect examples of some of the most fascinating concepts in chemistry: colligative properties.

Don't let the fancy name intimidate you. "Colligative" comes from the Latin word colligatus, meaning "bound together." In the world of chemistry, these properties are "bound together" by a single, simple idea: they depend only on the number of solute particles in a solution, not on what kind of particles they are.

Let's break down what that means and explore the four main colligative properties that quietly govern everything from our cooking to our body's functions.

The Main Character: The Solute Particle

Imagine you have two separate glasses of water. To the first, you add a spoonful of table sugar (sucrose). To the second, you add a spoonful of salt (NaCl). You stir until both are dissolved.

According to the principle of colligative properties, if you were to measure the boiling point of these two solutions, they would not be the same. Why? Because of the number of particles.

Sugar (Sucrose) is a molecular compound. When it dissolves, it stays as whole sugar molecules. One spoonful of sugar dissolves into many, many individual sugar molecules.
Salt (NaCl) is an ionic compound. When it dissolves, it dissociates, or breaks apart, into sodium ions (Na⁺) and chloride ions (Cl⁻). One spoonful of salt dissolves into roughly twice as many particles as the sugar!

This is the golden rule of colligative properties: More particles = Bigger effect.

Now, let's meet the fantastic four.

Definition Properties that depend only on the number of solute particles in a solution, not on what kind of particles they are.

Why Are Colligative Properties Important?

Colligative properties play a vital role in:

Automotive antifreeze systems
Food preparation (boiling & freezing processes)
Medical treatments (IV fluids & osmosis)
Determining molar mass in laboratories

Types of Colligative Properties

There are four main colligative properties:

1. Relative lowering of vapour pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

1. Relative lowering of vapour pressure

Definition

It is defined as the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent.

\text{Relative lowering of vapour pressure} = \frac{P^0 - P}{P^0}

Where:

$P^0$ = Vapour pressure of pure solvent
$P$ = Vapour pressure of solution

Using Raoult’s Law

According to Raoult's Law:

P = X_{\text{solvent}} \cdot P^0

So,

\frac{P^0 - P}{P^0} = X_{\text{solute}}

👉 Conclusion:
Relative lowering of vapour pressure = Mole fraction of solute

For Dilute Solutions

If the solution is dilute:

X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}

Since solute is very small:

X_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}

In Terms of Mass

\frac{P^0 - P}{P^0} = \frac{w_2 / M_2}{w_1 / M_1}

Where:

$w_1, w_2$ = mass of solvent and solute
$M_1, M_2$ = molar masses

Key Points

RLVP is a colligative property (depends only on number of particles, not nature).
Applicable mainly for non-volatile solutes.
Helps in determining molar mass of solute.

Example

If vapour pressure of pure solvent = 100 mmHg
and solution = 90 mmHg

\text{RLVP} = \frac{100 - 90}{100} = 0.1

👉 So, mole fraction of solute = 0.1

Why Vapour Pressure Decreases?

Solute particles occupy surface → fewer solvent molecules escape
Hence vapour pressure decreases

Applications

Determination of molar mass
Used in boiling point elevation & freezing point depression
Important in chemical industries

Important Observation (From PYQs)

👉 CBSE mostly asks RLVP in these forms:

Numericals (very important ⭐)
Derivation (2–3 marks)
Assertion–Reason / Case study
Concept-based short answers

CBSE Previous Year Questions (RLVP)

1. MCQ (CBSE 2023 Board)

The colligative property used for determination of molar mass of proteins is:
(a) Relative lowering of vapour pressure
(b) Elevation in boiling point
(c) Depression in freezing point
(d) Osmotic pressure

2. Case Study Based Question (CBSE Pattern)

A sugar solution shows lowering in vapour pressure = 0.061 mm Hg.
Vapour pressure of pure water = 17.5 mm Hg.

(i) Calculate relative lowering of vapour pressure
(a) 0.00348
(b) 0.061
(c) 0.122
(d) 1.75

(ii) Vapour pressure of solution will be:
(a) 17.5
(b) 0.61
(c) 17.439
(d) 0.00348

(iii) Mole fraction of solute is:
(a) 0.00348
(b) 0.9965
(c) 0.061
(d) 1.75

3. 2 Marks Question (Very Common PYQ Type)

Show that relative lowering of vapour pressure is a colligative property.

4. 3 Marks Numerical (Board Pattern)

846 g of water is mixed with 30 g of a non-volatile solute (M = 60 g/mol).
Vapour pressure of water = 23.8 mm Hg.
Calculate vapour pressure of the solution.

5. 2 Marks Concept Question

State Raoult's Law for a solution containing a non-volatile solute.

6. 1–2 Marks Theory Question

Why does the vapour pressure of a solvent decrease when a non-volatile solute is added?

7. 3 Marks Derivation Question

Derive the relation:

\frac{P^0 - P}{P^0} = X_{\text{solute}}

8. Assertion–Reason (CBSE Pattern)

Assertion (A): Relative lowering of vapour pressure depends only on number of solute particles.
Reason (R): It is a colligative property.

9. Numerical (PYQ Type)

2 g of a solute is dissolved in 100 g of water.
Relative lowering of vapour pressure = 0.01.
Calculate molar mass of solute.

10. Case-Based / Conceptual

Define colligative properties. Name the four colligative properties and explain RLVP as one of them.

numericals for practice

Q1. The vapour pressure of pure benzene at a certain temperature is 100 mmHg. A solution is prepared by dissolving 5 g of a non-volatile solute in 95 g benzene. The vapour pressure of the solution is 95 mmHg. Calculate the molar mass of the solute.

Q2. A solution contains 2 g of a non-volatile solute dissolved in 50 g of water. The vapour pressure of the solution is 99.5% of pure water. Calculate the molar mass of the solute.

Q3. 10 g of a solute is dissolved in 90 g of water. The relative lowering of vapour pressure is found to be 0.02. Calculate the molar mass of the solute.

Q4. The vapour pressure of pure acetone is 400 mmHg. When 2 g of a non-volatile solute is dissolved in 100 g acetone, the vapour pressure becomes 396 mmHg. Calculate the molar mass of solute.

Q5. A solution of glucose in water shows a relative lowering of vapour pressure equal to 0.01. Calculate the number of moles of glucose present if 10 moles of water are used.

Q6. 6 g of urea is dissolved in 90 g of water. Calculate the relative lowering of vapour pressure of the solution.

Q7. The vapour pressure of pure solvent is 80 mmHg and that of solution is 76 mmHg. Calculate the mole fraction of solute.

Q8.A solution contains 1 mole of solute and 9 moles of solvent. Calculate the relative lowering of vapour pressure.

Q9. 5 g of an unknown non-volatile solute is dissolved in 50 g of benzene. The vapour pressure of benzene decreases from 100 mmHg to 98 mmHg. Calculate the molar mass of the solute.

Q10.The relative lowering of vapour pressure of a solution containing 2 g solute in 100 g solvent is 0.005. Calculate the molar mass of solute. (Molar mass of solvent = 18 g/mol)

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Wednesday, March 18, 2026