Chemistry Fun & Learn

📌 Introduction

Raoult’s Law is a fundamental concept in physical chemistry that explains how the vapour pressure of a solution depends on the composition of its components. It plays a crucial role in understanding solutions, distillation, and colligative properties.

If you're a student, researcher, or chemistry enthusiast, this guide will walk you through Raoult’s Law in a simple, step-by-step way.

🔍 What is Raoult’s Law?

Raoult’s Law states that:

👉 The vapour pressure of a component in a solution is directly proportional to its mole fraction.

It is named after the French chemist François-Marie Raoult, who discovered this relationship.

🧪 Mathematical Expression of Raoult’s Law

For a solution containing a volatile component:

P = X \times P^0

Where:

$P$ = Vapour pressure of the component in solution
$XX= Mole fraction of the component$
$P^0$ = Vapour pressure of the pure component

🧩 For Binary Solutions (Two Liquids)

If a solution has two volatile liquids A and B:

P_{total} = P_A + P_B

P_A = X_A \cdot P_A^0,\quad P_B = X_B \cdot P_B^0

⚙️ Explanation

Step 1: Pure Liquid

Each pure liquid has its own vapour pressure.

Step 2: Mixing Liquids

When two liquids are mixed:

Molecules of one component reduce the number of molecules of the other at the surface.

Step 3: Vapour Pressure Decreases

Vapour pressure of each component becomes lower than its pure value.

Step 4: Equilibrium Established

A new vapour pressure is established based on composition.

📉 Vapour Pressure Lowering

Raoult’s Law explains why vapour pressure decreases when a non-volatile solute is added:

\Delta P = P^0 - P

Relative lowering:

\frac{\Delta P}{P^0} = X_{solute}

🧠 Key Assumptions of Raoult’s Law

Raoult’s Law works best for ideal solutions, which follow these conditions:

Intermolecular forces are similar (A–A ≈ B–B ≈ A–B)
No heat is absorbed or released (ΔH = 0)
No volume change on mixing

⚠️ Limitations of Raoult’s Law

Real solutions often deviate:

🔺 Positive Deviation

Weaker intermolecular forces
Higher vapour pressure than expected

🔻 Negative Deviation

Stronger intermolecular forces
Lower vapour pressure than expected

🔥 Real-Life Applications of Raoult’s Law

1. Distillation Process

Used in separation of liquids based on boiling points.

2. Colligative Properties

Helps in studying:

Boiling point elevation
Freezing point depression

3. Industrial Chemistry

Used in chemical manufacturing and solvent systems.

4. Pharmaceuticals

Helps in drug formulation and stability.

🧪 Example Problem

Q: A solution contains 1 mole of ethanol and 1 mole of water. Calculate mole fraction of ethanol.

X=1/(1+1)

X=0.5

Q. The vapour pressure of pure benzene is 100 mmHg. A solution is prepared by mixing 2 moles of benzene with 3 moles of toluene. Calculate the partial vapour pressure of benzene.

✅ Solution:

$X_{benzene} = \frac{2}{2+3} = \frac{2}{5} = 0.4$

Using Raoult’s Law:

$P = X \cdot P^0 = 0.4 \times 100 = 40 \, mmHg$

👉 Answer: 40 mmHg

Numerical for practice

1.Vapour pressures of pure liquids A and B are 80 mmHg and 120 mmHg respectively. If mole fractions are 0.6 and 0.4, find total vapour pressure.

2. The vapour pressure of pure water is 30 mmHg. After adding a non-volatile solute, it becomes 27 mmHg. Calculate mole fraction of solute.

3. A solution has vapour pressure 45 mmHg. Mole fraction of solvent is 0.9. Find vapour pressure of pure solvent.

4.18 g of glucose (non-volatile) is dissolved in 180 g of water. Calculate mole fraction of solute.