Raoult’s Law Class 12: Definition, Derivation, Deviation and Application with Examples

Understanding Raoult’s Law is essential for mastering the Solutions chapter in Class 12 Chemistry. This important law explains how the vapour pressure of a solution depends on the mole fraction of its components and forms the basis for understanding ideal and non-ideal solutions. In CBSE board exams and competitive exams like NEET, JEE Main, and other entrance tests, questions on the definition, derivation, formula, deviations, and applications of Raoult’s Law are frequently asked.

In this comprehensive guide, you will learn the definition of Raoult’s Law, its mathematical derivation, the conditions under which it is applicable, positive and negative deviations, practical applications, and solved numerical examples. The chapter is explained in simple language with clear formulas and diagrams, making it perfect for quick revision as well as detailed study. Whether you are preparing for board exams or competitive exams, these notes will strengthen your concepts and improve your problem-solving skills.

What is Raoult’s Law? (Definition)

This Law states that the partial vapor pressure of a component in a solution is directly proportional to its mole fraction in the solution.

For a solution containing a volatile solute A and solvent B, Raoult’s Law can be mathematically expressed as:

$$P_A = P_A^0 \cdot x_A$$

Where: p_A = Partial vapor pressure of component A

p⁰_A = Vapor pressure of pure component A

x_A​ = Mole fraction of component A in the solution

Similarly, for component B:

$$P_B = P_B^0 \cdot x_B$$

​The total vapor pressure of the solution is given by:

$$P_{total}=P_A+P_B=P_A^0\cdot x_A+P_B^0\cdot x_{\mathrm{B<span\; style="font-size:\; inherit;\; font-family:\; -apple-system,\; system-ui,\; BlinkMacSystemFont,\; \text{'}Segoe\; UI\text{'},\; Helvetica,\; Arial,\; sans-serif,\; \text{'}Apple\; Color\; Emoji\text{'},\; \text{'}Segoe\; UI\; Emoji\text{'},\; \text{'}Segoe\; UI\; Symbol\text{'};\; text-transform:\; none;"></span>}}$$

This law applies best to ideal solutions, where intermolecular forces remain unchanged upon mixing.

Derivation of Raoult’s Law

Step-by-Step Derivation

Consider a binary solution with two volatile liquids, A and B.

Each component exerts a partial vapor pressure, given by:

$$P_A=P_A^0\cdot x_A\text{and}{P}_B=P_B^0\cdot x_{\mathrm{B<span\; style="font-family:\; -apple-system,\; system-ui,\; BlinkMacSystemFont,\; \text{'}Segoe\; UI\text{'},\; Helvetica,\; Arial,\; sans-serif,\; \text{'}Apple\; Color\; Emoji\text{'},\; \text{'}Segoe\; UI\; Emoji\text{'},\; \text{'}Segoe\; UI\; Symbol\text{'};\; font-size:\; 17px;\; text-transform:\; none;"></span>}}$$

By Dalton’s Law of Partial Pressures, the total vapor pressure of the solution is:

$$P_{total} = P_A + P_B$$

Rearranging in terms of mole fractions:

$$P_{total} = P_A^0 \cdot x_A + P_B^0 \cdot (1 – x_A)$$

Thus, Raoult’s Law explains how the total vapor pressure depends on the composition of the solution.

Deviation from Raoult’s Law

Not all solutions follow this Law perfectly. The deviations occur due to changes in intermolecular forces upon mixing.

Positive Deviation from Raoult’s Law

• Occurs when solute-solvent interactions are weaker than those in pure components.
• The solution has higher vapor pressure than predicted.
• Example: Ethanol and Acetone – Ethanol molecules break hydrogen bonds due to acetone, leading to increased vaporization.

Negative Deviation from Raoult’s Law

• Occurs when solute-solvent interactions are stronger than those in pure components.
• The solution has lower vapor pressure than predicted.
• Example: Chloroform and Acetone – Strong hydrogen bonding reduces the escape of molecules into the vapor phase.

Raoult’s Law Class 12: 10 Long-Type Numericals with Solutions

Numerical 1: Vapour Pressure of an Ideal Solution

Question:

A solution is prepared by mixing 2 moles of benzene (vapour pressure = 100 mmHg) and 3 moles of toluene (vapour pressure = 40 mmHg) at the same temperature. Calculate:

1. Mole fraction of each component
2. Partial vapour pressure of each component
3. Total vapour pressure of the solution

Solution

Given

• $n_{\text{benzene}}=2$
• $n_{\text{toluene}}=3$
• $P^0_{\text{benzene}}=100$ mmHg
• $P^0_{\text{toluene}}=40$ mmHg

Step 1: Mole Fractions

Total moles$$2+3=5$$$$X_{\text{benzene}}=\frac{2}{5}=0.4$$$$X_{\text{toluene}}=\frac{3}{5}=0.6$$

Step 2: Partial Vapour Pressures

$$P_A=P_A^0X_A$$$$=100\times0.4=40\text{ mmHg}$$ $$P_B=40\times0.6=24\text{ mmHg}$$

Step 3: Total Vapour Pressure

$$P_{\text{total}}=40+24=64\text{ mmHg}$$

Answer

• Mole fraction of benzene = 0.4
• Mole fraction of toluene = 0.6
• Partial pressures = 40 mmHg and 24 mmHg
• Total vapour pressure = 64 mmHg

Numerical 2: Finding Vapour Pressure

Question

The vapour pressure of pure acetone is 300 mmHg. A solution contains 0.70 mole fraction of acetone.

Find its vapour pressure.

Solution

$$P=P^0X$$$$=300\times0.70$$$$=210\text{ mmHg}$$

Answer

210 mmHg

Numerical 3: Finding Mole Fraction

Question

The vapour pressure of pure ethanol is 80 mmHg. The solution has vapour pressure 48 mmHg.

Find the mole fraction of ethanol.

Solution

$$X=\frac{P}{P^0}$$$$=\frac{48}{80}$$$$=0.60$$

Answer

Mole fraction = 0.60

Numerical 4: Vapour Pressure Lowering

Question

Pure water has vapour pressure 23.8 mmHg. A solution has vapour pressure 21.4 mmHg.

Calculate:

1. Lowering in vapour pressure
2. Relative lowering

Solution

Lowering$$23.8-21.4=2.4\text{ mmHg}$$

Relative lowering$$\frac{2.4}{23.8}=0.1008$$

Answer

Lowering = 2.4 mmHg

Relative lowering = 0.101

Numerical 5: Mole Fraction of Solute

Question

The relative lowering of vapour pressure is 0.08.

Find the mole fraction of the solute.

Solution

According to Raoult’s law,$$\frac{P^0-P}{P^0}=X_{\text{solute}}$$

Therefore,$$X_{\text{solute}}=0.08$$

Answer

0.08

Numerical 5: Mole Fraction of Solute

Question

The relative lowering of vapour pressure is 0.08.

Find the mole fraction of the solute.

Solution

According to Raoult’s law,$$\frac{P^0-P}{P^0}=X_{\text{solute}}$$

Therefore,$$X_{\text{solute}}=0.08$$

Answer

0.08

Numerical 7: Binary Solution

Question

A solution contains 4 moles of A and 6 moles of B.

The vapour pressures are

• A = 120 mmHg
• B = 60 mmHg

Find the total vapour pressure.

Solution

$$X_A=\frac{4}{10}=0.4$$$$X_B=0.6$$

Partial pressure of A$$120\times0.4=48$$

Partial pressure of B$$60\times0.6=36$$

Total$$48+36=84\text{ mmHg}$$

Answer

84 mmHg

Numerical 8: Unknown Vapour Pressure

Question

The mole fraction of solvent is 0.85.

Its partial vapour pressure is 102 mmHg.

Calculate the vapour pressure of the pure solvent.

Solution

$$P=P^0X$$$$102=P^0\times0.85$$ $$P^0=\frac{102}{0.85}$$​ $$=120\text{ mmHg}$$

Answer

120 mmHg

Numerical 9: Vapour Pressure of Binary Solution

Question

A solution contains

• 1 mole chloroform
• 2 moles carbon tetrachloride

Pure vapour pressures

• Chloroform = 200 mmHg
• Carbon tetrachloride = 90 mmHg

Calculate total vapour pressure.

Solution

Total moles$$1+2=3$$$$X_{\text{CHCl}_3}=\frac13=0.333$$$$X_{\text{CCl}_4}=\frac23=0.667$$

Partial pressures$$200\times0.333=66.7$$$$90\times0.667=60.0$$

Total$$66.7+60=126.7\text{ mmHg}$$

Answer

126.7 mmHg

Numerical 10: Relative Lowering of Vapour Pressure

Question

2 g of a non-volatile solute is dissolved in 98 g water.

The relative lowering of vapour pressure is 0.01.

Calculate the molar mass of the solute.

Solution

Relative lowering$$\frac{n_{\text{solute}}}{n_{\text{solvent}}} =0.01$$

Moles of water$$=\frac{98}{18} =5.444$$

Moles of solute$$=0.01\times5.444 =0.05444$$

Molar mass$$=\frac{2}{0.05444} =36.7\text{ g/mol}$$

Answer

Molar mass of the solute = 36.7 g/mol