Raoult’s Law: Understanding Vapor Pressure in Solutions
The Real-Life Hook: Why Perfumes Last Longer When Mixed
Ever noticed how expensive perfumes mix different fragrances? Or why adding salt to ice makes it melt faster (the ice-cream maker’s secret)? Or why your car’s coolant is a mixture, not pure water? The answer lies in Raoult’s Law - one of the most elegant and practical laws in solution chemistry.
Daily Applications:
- Ice cream making: Salt lowers ice’s melting point (freezing point depression via vapor pressure lowering)
- Distillation: Separating alcohol from water in breweries
- Perfume industry: Mixing fragrances to control evaporation rates
- Antifreeze: Ethylene glycol lowers water’s freezing point in car radiators
Core Concept: What is Vapor Pressure?
Vapor Pressure Basics
Vapor Pressure: The pressure exerted by vapor when it’s in dynamic equilibrium with its liquid at a given temperature.
Key Points:
- Higher temperature → Higher vapor pressure
- Volatile liquids → High vapor pressure
- Non-volatile liquids → Low vapor pressure
Dynamic Equilibrium:
Liquid ⇌ Vapor
(Rate of evaporation = Rate of condensation)
Factors Affecting Vapor Pressure:
- Nature of liquid (intermolecular forces)
- Temperature (only factor that changes VP for pure liquids)
- Surface area (affects rate, not equilibrium VP)
Raoult’s Law: The Foundation
Statement
For an ideal solution:
“The partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.”
Mathematical Form
For component A:
$$\boxed{P_A = P_A^0 \times \chi_A}$$For component B:
$$\boxed{P_B = P_B^0 \times \chi_B}$$Total vapor pressure:
$$\boxed{P_{\text{total}} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B}$$Where:
- P_A = Partial vapor pressure of A in solution
- P_A^0 = Vapor pressure of pure A
- χ_A = Mole fraction of A in liquid phase
Alternative Form (For Non-volatile Solute)
When solute is non-volatile (P_B^0 = 0):
$$\boxed{P_{\text{solution}} = P_{\text{solvent}}^0 \times \chi_{\text{solvent}}}$$ $$\boxed{P_{\text{solution}} = P_{\text{solvent}}^0 \times (1 - \chi_{\text{solute}})}$$Relative Lowering of Vapor Pressure (RLVP):
$$\boxed{\frac{P^0 - P_s}{P^0} = \chi_{\text{solute}}}$$This is the first colligative property!
Memory Trick:
“Raoult Reduces Pressure Proportionally” - Vapor pressure reduces in proportion to mole fraction of solute!
Ideal Solutions: The Perfect Mixture
Definition
An ideal solution obeys Raoult’s law over the entire range of concentrations (0 to 1 mole fraction).
Conditions for Ideal Behavior
Similar Intermolecular Forces:
- A-A interactions ≈ B-B interactions ≈ A-B interactions
- ΔH_mix = 0 (no heat absorbed or released on mixing)
- ΔV_mix = 0 (no volume change on mixing)
Similar Molecular Structure:
- Similar size, shape, and polarity
- Example: Benzene-Toluene, n-Hexane-n-Heptane
Characteristics of Ideal Solutions
| Property | Ideal Solution |
|---|---|
| ΔH_mixing | 0 |
| ΔV_mixing | 0 |
| Raoult’s Law | Obeyed exactly |
| A-A, B-B, A-B forces | Equal |
| Example | Benzene + Toluene |
Examples of Nearly Ideal Solutions
- Benzene + Toluene
- n-Hexane + n-Heptane
- Ethyl bromide + Ethyl chloride
- Chlorobenzene + Bromobenzene
- CCl₄ + SiCl₄
Common Theme: Similar chemical structure and intermolecular forces!
Non-Ideal Solutions: Reality Strikes
Most real solutions deviate from Raoult’s law because A-B interactions differ from A-A and B-B interactions.
Types of Deviations
1. Positive Deviation from Raoult’s Law
Condition: A-B interactions < A-A or B-B interactions
Characteristics:
- P_total > P_ideal (higher vapor pressure than expected)
- A and B molecules escape more easily
- ΔH_mix > 0 (endothermic mixing - heat absorbed)
- ΔV_mix > 0 (volume increases on mixing)
Graph:
P_total (actual) is ABOVE the ideal line
Molecular Explanation: When A-B attraction is weaker than A-A or B-B, molecules escape more easily from the solution surface, increasing vapor pressure.
Mathematical Form:
$$\boxed{P_A > P_A^0 \chi_A \text{ and } P_B > P_B^0 \chi_B}$$Examples:
- Ethanol + Water (most important for JEE!)
- Acetone + Carbon disulfide (CS₂)
- Ethanol + Cyclohexane
- Acetone + Ethanol
- CCl₄ + CHCl₃
Case Study: Ethanol + Water
- H-bonding in pure water is stronger than H-bonding in ethanol-water mixture
- Water molecules are “pulled apart” by ethanol
- Result: Both components evaporate more easily
- ΔH_mix = +815 J/mol (endothermic)
Memory Trick:
“Positive = Pushed apart” - Molecules push each other away, vapor pressure increases!
2. Negative Deviation from Raoult’s Law
Condition: A-B interactions > A-A or B-B interactions
Characteristics:
- P_total < P_ideal (lower vapor pressure than expected)
- A and B molecules escape less easily (stronger attraction)
- ΔH_mix < 0 (exothermic mixing - heat released)
- ΔV_mix < 0 (volume decreases on mixing)
Graph:
P_total (actual) is BELOW the ideal line
Molecular Explanation: When A-B attraction is stronger than A-A or B-B, molecules are held more tightly in solution, reducing vapor pressure.
Mathematical Form:
$$\boxed{P_A < P_A^0 \chi_A \text{ and } P_B < P_B^0 \chi_B}$$Examples:
- Chloroform + Acetone (hydrogen bonding forms!)
- Water + Nitric acid (HNO₃)
- Water + Hydrochloric acid (HCl)
- Acetic acid + Pyridine
- Chloroform + Diethyl ether
Case Study: Chloroform + Acetone
- New H-bonding forms: CHCl₃···O=C(CH₃)₂
- This H-bond is stronger than individual dipole-dipole forces
- Result: Molecules held tighter, lower vapor pressure
- ΔH_mix = -1600 J/mol (exothermic)
Reaction:
CHCl₃ + (CH₃)₂C=O → CHCl₃···O=C(CH₃)₂
(H-bonding between H of CHCl₃ and O of acetone)
Memory Trick:
“Negative = New attraction” - New interactions form, vapor pressure decreases!
Comparison Table: Positive vs Negative Deviations
| Property | Positive Deviation | Negative Deviation |
|---|---|---|
| A-B vs A-A, B-B | Weaker | Stronger |
| P_total | > P_ideal | < P_ideal |
| ΔH_mix | > 0 (endothermic) | < 0 (exothermic) |
| ΔV_mix | > 0 (increases) | < 0 (decreases) |
| Solubility | Lower | Higher |
| Example | Ethanol + Water | Chloroform + Acetone |
| Reason | Bonds break | New bonds form |
Azeotropes: When Solutions Can’t Be Separated
What are Azeotropes?
Azeotrope: A mixture that boils at constant temperature and composition. The vapor has the same composition as the liquid.
Key Feature: Cannot be separated by simple distillation!
Types of Azeotropes
1. Minimum Boiling Azeotrope (Positive Deviation)
Characteristics:
- Shows positive deviation
- Boiling point < Both pure components
- Forms at maximum vapor pressure point
Examples:
Ethanol (95.6%) + Water (4.4%) - Boils at 78.2°C
- Pure ethanol: 78.4°C
- Pure water: 100°C
- This is why you can’t get 100% ethanol by distillation!
HCl (20.2%) + Water (79.8%) - Boils at 108.6°C
Ethanol + Benzene - Boils at 67.8°C
Graph Behavior:
- Vapor pressure curve shows a maximum
- Boiling point curve shows a minimum
Memory Trick:
“Minimum BP = Maximum VP” - Positive deviation creates low boiling azeotrope!
2. Maximum Boiling Azeotrope (Negative Deviation)
Characteristics:
- Shows negative deviation
- Boiling point > Both pure components
- Forms at minimum vapor pressure point
Examples:
HNO₃ (68%) + Water (32%) - Boils at 120.5°C
- Pure HNO₃: 86°C
- Pure water: 100°C
HCl (20.2%) + Water (79.8%) - Boils at 110°C
H₂O + Formic acid (77.5%)
Graph Behavior:
- Vapor pressure curve shows a minimum
- Boiling point curve shows a maximum
Memory Trick:
“Maximum BP = Minimum VP” - Negative deviation creates high boiling azeotrope!
Interactive Demo: Calculating Vapor Pressure
Problem 1: Binary Ideal Solution
Question: Benzene (P°_benzene = 100 torr) and toluene (P°_toluene = 40 torr) form an ideal solution. If a solution contains equal moles of both, calculate: (a) Partial pressures (b) Total vapor pressure (c) Mole fraction in vapor phase
Solution:
Given:
- χ_benzene = χ_toluene = 0.5
- P°_benzene = 100 torr, P°_toluene = 40 torr
(a) Partial Pressures: Using Raoult’s law:
- P_benzene = 100 × 0.5 = 50 torr
- P_toluene = 40 × 0.5 = 20 torr
(b) Total Vapor Pressure:
- P_total = 50 + 20 = 70 torr
(c) Mole Fraction in Vapor: Using Dalton’s law:
- y_benzene = P_benzene / P_total = 50/70 = 0.714
- y_toluene = P_toluene / P_total = 20/70 = 0.286
Key Observation: Vapor is richer in the more volatile component (benzene)!
Problem 2: Non-volatile Solute
Question: 18g glucose (C₆H₁₂O₆, MW = 180) is dissolved in 90g water. Calculate the vapor pressure of solution at 25°C. (P°_water = 23.8 torr)
Solution:
Step 1: Calculate moles
- Moles of glucose = 18/180 = 0.1 mol
- Moles of water = 90/18 = 5 mol
Step 2: Calculate mole fraction
- χ_glucose = 0.1/(0.1 + 5) = 0.0196
- χ_water = 5/5.1 = 0.9804
Step 3: Apply Raoult’s law
- P_solution = P°_water × χ_water
- P_solution = 23.8 × 0.9804 = 23.33 torr
Alternative (RLVP):
- ΔP/P° = χ_solute = 0.0196
- ΔP = 23.8 × 0.0196 = 0.47 torr
- P_solution = 23.8 - 0.47 = 23.33 torr ✓
Vapor Phase Composition: Dalton’s Law Connection
Relating Liquid and Vapor Compositions
For an ideal binary solution:
Liquid phase mole fractions: χ_A, χ_B Vapor phase mole fractions: y_A, y_B
$$\boxed{y_A = \frac{P_A}{P_{\text{total}}} = \frac{P_A^0 \chi_A}{P_A^0 \chi_A + P_B^0 \chi_B}}$$Key Insight: The vapor is always richer in the more volatile component!
Mathematical Proof: If P°_A > P°_B (A is more volatile):
- y_A / y_B = (P°_A / P°_B) × (χ_A / χ_B)
- Since P°_A / P°_B > 1, we get y_A / y_B > χ_A / χ_B
- Therefore: y_A > χ_A (vapor enriched in A)
This principle is the basis of fractional distillation!
Common JEE Mistake Traps
Mistake 1: Confusing Liquid and Vapor Compositions
❌ Wrong: Assuming vapor has same composition as liquid ✅ Correct: Vapor is enriched in more volatile component
Example: For benzene-toluene with χ_benzene = 0.5 in liquid
- In vapor: y_benzene ≈ 0.71 (not 0.5!)
Mistake 2: Forgetting That Solute is Non-volatile
❌ Wrong: Including solute vapor pressure when it’s non-volatile ✅ Correct: P_solution = P°_solvent × χ_solvent only
Example: For glucose in water, glucose contributes ZERO to vapor pressure!
Mistake 3: Using Wrong Formula for RLVP
❌ Wrong: ΔP/P° = χ_solvent ✅ Correct: ΔP/P° = χ_solute
Memory Aid: “Lowering” is due to solute, not solvent!
Mistake 4: Sign of ΔH_mix
❌ Wrong: Positive deviation → ΔH_mix < 0 ✅ Correct: Positive deviation → ΔH_mix > 0 (endothermic)
Logic: Breaking interactions requires energy!
Practice Problems: Three-Level Mastery
Level 1: JEE Main Foundation
Q1. Pure benzene has vapor pressure 100 torr. When a non-volatile solute is added, the vapor pressure drops to 95 torr. Calculate the mole fraction of solute.
Solution
Using RLVP formula: (P° - P_s)/P° = χ_solute (100 - 95)/100 = χ_solute χ_solute = 5/100 = 0.05
Q2. At 25°C, the vapor pressures of pure A and B are 80 torr and 60 torr respectively. If χ_A = 0.4 in an ideal solution, calculate total vapor pressure.
Solution
χ_B = 1 - 0.4 = 0.6 P_A = 80 × 0.4 = 32 torr P_B = 60 × 0.6 = 36 torr P_total = 32 + 36 = 68 torr
Q3. Which of the following shows positive deviation? (a) CHCl₃ + Acetone (b) Benzene + Toluene (c) Ethanol + Water (d) HNO₃ + Water
Solution
(c) Ethanol + Water shows positive deviation
- (a) shows negative deviation (H-bonding)
- (b) is nearly ideal
- (d) shows negative deviation
Level 2: JEE Main Advanced
Q4. Two liquids A and B form an ideal solution. At 30°C, the vapor pressure of pure A is 300 torr and pure B is 800 torr. Calculate: (a) Vapor pressure when χ_A = 0.6 (b) Composition of vapor phase
Solution
(a) Vapor Pressure: χ_B = 1 - 0.6 = 0.4 P_A = 300 × 0.6 = 180 torr P_B = 800 × 0.4 = 320 torr P_total = 180 + 320 = 500 torr
(b) Vapor Composition: y_A = 180/500 = 0.36 y_B = 320/500 = 0.64
Note: B is more volatile, so vapor is richer in B!
Q5. 2 moles of a non-volatile solute is dissolved in 3 moles of solvent (P° = 600 torr). Calculate: (a) Vapor pressure of solution (b) Lowering of vapor pressure
Solution
(a) Vapor Pressure: χ_solvent = 3/(2+3) = 0.6 P_solution = 600 × 0.6 = 360 torr
(b) Lowering: ΔP = P° - P_s = 600 - 360 = 240 torr
Verification: χ_solute = 2/5 = 0.4 ΔP/P° = 0.4 ✓ ΔP = 600 × 0.4 = 240 torr ✓
Interactive Demo: Visualize Vapor Pressure Reduction
See how solute particles reduce vapor pressure at the solution surface.
Q6. A solution of two liquids A (P° = 400 torr) and B (P° = 600 torr) boils at temperature where total pressure equals atmospheric pressure (760 torr). If χ_A = 0.3, is this solution ideal or non-ideal? If non-ideal, what type of deviation?
Solution
For ideal solution: P_total (ideal) = 400 × 0.3 + 600 × 0.7 = 120 + 420 = 540 torr
Actual: P_total (actual) = 760 torr (given)
Comparison: 760 > 540
Conclusion: Shows positive deviation (actual VP > ideal VP)
Level 3: JEE Advanced Challenge
Q7. Two liquids A and B form an ideal solution at 25°C. The mole fraction of A in vapor phase is 0.6 when its mole fraction in liquid is 0.4. If P°_B = 600 torr, calculate P°_A.
Solution
Given:
- χ_A (liquid) = 0.4, χ_B = 0.6
- y_A (vapor) = 0.6, y_B = 0.4
- P°_B = 600 torr
Using vapor composition formula: y_A / y_B = (P°_A / P°_B) × (χ_A / χ_B)
0.6/0.4 = (P°_A / 600) × (0.4/0.6)
1.5 = (P°_A / 600) × (2/3)
1.5 × 3/2 = P°_A / 600
2.25 = P°_A / 600
P°_A = 1350 torr
Verification: P_A = 1350 × 0.4 = 540 torr P_B = 600 × 0.6 = 360 torr P_total = 900 torr y_A = 540/900 = 0.6 ✓
Q8. An aqueous solution of urea contains 6g urea (MW = 60) per 90g water. Calculate: (a) Vapor pressure of solution at 25°C (P°_water = 23.8 torr) (b) Lowering of vapor pressure (c) Relative lowering
Solution
Step 1: Calculate moles Moles of urea = 6/60 = 0.1 mol Moles of water = 90/18 = 5 mol Total moles = 5.1 mol
Step 2: Mole fractions χ_urea = 0.1/5.1 = 0.0196 χ_water = 5/5.1 = 0.9804
(a) Vapor Pressure: P_solution = 23.8 × 0.9804 = 23.33 torr
(b) Lowering: ΔP = 23.8 - 23.33 = 0.47 torr
(c) Relative Lowering: ΔP/P° = 0.47/23.8 = 0.0196 (equals χ_solute ✓)
Q9. (JEE Advanced 2018 Type) Benzene and toluene form nearly ideal solution. At 80°C, P°_benzene = 1000 torr, P°_toluene = 400 torr. Starting with an equimolar liquid mixture, calculate: (a) Composition after 50% by moles vaporizes (b) Total pressure when 50% vaporizes
Solution
Initial: 1 mol benzene + 1 mol toluene (total = 2 mol)
After 50% vaporizes:
- Liquid remaining = 1 mol
- Vapor formed = 1 mol
Step 1: Vapor composition (initially) At start: χ_benzene = χ_toluene = 0.5 P_benzene = 1000 × 0.5 = 500 torr P_toluene = 400 × 0.5 = 200 torr P_total = 700 torr
y_benzene = 500/700 = 5/7 y_toluene = 200/700 = 2/7
Step 2: Moles in vapor (first infinitesimal amount) Since benzene is more volatile, vapor is richer in benzene.
Step 3: After 50% vaporization (approximation) Let’s say x mol benzene and y mol toluene remain in liquid. x + y = 1 (remaining liquid)
In vapor: (1-x) benzene and (1-y) toluene (1-x) + (1-y) = 1 2 - x - y = 1 x + y = 1 ✓
Average composition: Approximately, more benzene vaporizes. Let x ≈ 0.417 mol benzene remains y ≈ 0.583 mol toluene remains
(a) Liquid composition after 50% vaporization: χ_benzene ≈ 0.417/1 = 0.417 χ_toluene ≈ 0.583/1 = 0.583
(b) Pressure: P_benzene = 1000 × 0.417 = 417 torr P_toluene = 400 × 0.583 = 233 torr P_total ≈ 650 torr
Note: This is an approximate solution. Exact solution requires integration!
Q10. (JEE Advanced Level) A solution shows positive deviation with maximum vapor pressure of 800 torr at χ_A = 0.5. The vapor pressures of pure A and B are 600 torr and 500 torr. Calculate the % deviation from ideality.
Solution
For ideal solution at χ_A = 0.5: P_ideal = 600 × 0.5 + 500 × 0.5 = 300 + 250 = 550 torr
Actual: P_actual = 800 torr (given)
Deviation: Deviation = P_actual - P_ideal = 800 - 550 = 250 torr
% Deviation: % = (250/550) × 100 = 45.5% positive deviation
Interpretation: This is a large positive deviation indicating very weak A-B interactions!
Cross-Connections to Other Chapters
🔗 Link to Thermodynamics
- ΔH_mix relates to enthalpy changes
- ΔS_mix always positive for ideal solutions (entropy increases)
- Gibbs free energy: ΔG_mix = ΔH_mix - TΔS_mix
- See: Thermodynamics of Mixing
🔗 Link to Chemical Kinetics
- Evaporation rate depends on vapor pressure
- Higher VP → Faster evaporation
- See: Rate Laws
🔗 Link to Phase Diagrams
- Azeotropes appear as maxima/minima in phase diagrams
- Lever rule applies to liquid-vapor equilibria
- See: Phase Equilibria
🔗 Link to Colligative Properties
- RLVP is the foundation of all colligative properties
- Links to boiling point elevation, freezing point depression
- See: Colligative Properties
🔗 Link to Equilibrium
- Vapor-liquid equilibrium is a physical equilibrium
- K_p connects to partial pressures
- See: Chemical Equilibrium
Advanced Topic: Henry’s Law (For Gases)
Statement
For gases dissolving in liquids (low concentration):
$$\boxed{P_{\text{gas}} = K_H \times \chi_{\text{gas}}}$$Where K_H = Henry’s law constant (different for each gas-solvent pair)
Comparison with Raoult’s Law:
- Raoult’s law: P = P° × χ (for liquids)
- Henry’s law: P = K_H × χ (for gases)
Applications:
- Carbonated drinks (CO₂ solubility)
- Scuba diving (nitrogen narcosis at high pressure)
- Oxygen solubility in blood
Example: K_H for CO₂ in water at 25°C = 1.67 × 10⁸ torr
Note: At high pressures, even gases can show deviations!
Formula Sheet: Quick Reference
Raoult’s Law (Ideal Solutions)
$$\boxed{P_A = P_A^0 \times \chi_A}$$ $$\boxed{P_{\text{total}} = P_A^0 \chi_A + P_B^0 \chi_B}$$For Non-volatile Solute
$$\boxed{P_{\text{solution}} = P_{\text{solvent}}^0 \times \chi_{\text{solvent}}}$$ $$\boxed{\frac{P^0 - P_s}{P^0} = \chi_{\text{solute}}}$$Vapor Composition
$$\boxed{y_A = \frac{P_A}{P_{\text{total}}} = \frac{P_A^0 \chi_A}{P_{\text{total}}}}$$Deviations
| Property | Positive | Negative |
|---|---|---|
| A-B force | Weaker | Stronger |
| P_total | > P_ideal | < P_ideal |
| ΔH_mix | > 0 | < 0 |
| Azeotrope | Min BP | Max BP |
Final JEE Strategy
✅ High-Weightage Topics:
- RLVP calculations (non-volatile solute) - 90% chance in JEE Main
- Ideal solution vapor pressure - 70% chance
- Positive vs negative deviation - 60% chance
- Azeotropes concept - 40% chance in JEE Advanced
✅ Must-Remember Examples:
- Ideal: Benzene + Toluene
- Positive: Ethanol + Water
- Negative: CHCl₃ + Acetone
- Min BP azeotrope: 95.6% Ethanol + Water
- Max BP azeotrope: 68% HNO₃ + Water
✅ Common Question Types:
- Calculate vapor pressure given mole fraction
- Find composition in vapor phase
- Identify type of deviation
- RLVP with non-volatile solute
- Conceptual questions on azeotropes
✅ Quick Checks:
- χ_solute + χ_solvent = 1? ✓
- Vapor richer in more volatile component? ✓
- ΔP/P° = χ_solute (not χ_solvent)? ✓
- Positive deviation → endothermic? ✓
Previous Topic: Concentration Methods Next Topic: Colligative Properties - where we’ll see how RLVP leads to BP elevation and FP depression!
Last Updated: June 2025 | For JEE Main & Advanced 2026