Colligative Properties: Properties That Depend on Number, Not Nature
The Real-Life Hook: Why Salt Melts Ice and Antifreeze Works
Why do we throw salt on icy roads in winter? Why does your car’s radiator need antifreeze? Why do we add salt to water when cooking pasta? The answer: colligative properties - properties that depend only on the NUMBER of solute particles, not their identity!
Daily Applications:
- De-icing roads: Salt lowers freezing point of water (FP depression)
- Antifreeze in cars: Ethylene glycol prevents coolant from freezing
- Cooking: Adding salt raises boiling point of water slightly
- Reverse osmosis: Purifying seawater for drinking (osmotic pressure)
- Preserving food: High sugar/salt concentration prevents bacterial growth
What Are Colligative Properties?
Definition
Colligative Properties: Properties of solutions that depend only on the number (concentration) of solute particles, NOT on their chemical nature.
The Four Colligative Properties:
- Relative Lowering of Vapor Pressure (RLVP)
- Elevation of Boiling Point (ΔT_b)
- Depression of Freezing Point (ΔT_f)
- Osmotic Pressure (π)
Key Insight: Whether you dissolve 1 mole of glucose, urea, or sucrose in water, the effect on these properties is identical!
Memory Trick:
“ROBE” - RLVP, Osmotic pressure, Boiling point elevation, (freezing point) dEpression
1. Relative Lowering of Vapor Pressure (RLVP)
Concept
When a non-volatile solute is added to a solvent, the vapor pressure decreases.
Raoult’s Law Form:
$$\boxed{\frac{P^0 - P_s}{P^0} = \chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}}$$For dilute solutions (n_solute « n_solvent):
$$\boxed{\frac{P^0 - P_s}{P^0} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{w/M}{W/M_{\text{solvent}}}}$$Where:
- P^0 = Vapor pressure of pure solvent
- P_s = Vapor pressure of solution
- w = mass of solute, M = molar mass of solute
- W = mass of solvent, M_solvent = molar mass of solvent
Simplified for aqueous solutions:
$$\boxed{\frac{\Delta P}{P^0} = \frac{w \times 18}{M \times W}}$$(Since M_water = 18 g/mol)
Why Does This Happen?
Molecular Explanation:
- Solute particles occupy surface positions
- Fewer solvent molecules at surface → less evaporation
- Lower evaporation rate → lower vapor pressure
Important Notes: ✅ Applies only to non-volatile solutes (solute doesn’t evaporate) ✅ Valid for dilute solutions ✅ Foundation for other three colligative properties
2. Elevation of Boiling Point (ΔT_b)
Concept
Boiling Point: Temperature at which vapor pressure equals atmospheric pressure.
When vapor pressure is lowered (by adding solute), the solution must be heated to a higher temperature to make its vapor pressure equal to atmospheric pressure.
Result: Boiling point increases!
Formula
$$\boxed{\Delta T_b = T_b - T_b^0 = K_b \times m}$$Where:
- ΔT_b = Elevation in boiling point
- T_b^0 = Boiling point of pure solvent
- T_b = Boiling point of solution
- K_b = Molal elevation constant (ebullioscopic constant)
- m = molality of solution
Alternative Form:
$$\boxed{\Delta T_b = K_b \times \frac{w \times 1000}{M \times W}}$$Where:
- w = mass of solute (g)
- M = molar mass of solute (g/mol)
- W = mass of solvent (g)
K_b Values (Important for JEE)
| Solvent | K_b (K·kg/mol) | Normal BP (°C) |
|---|---|---|
| Water | 0.52 | 100 |
| Benzene | 2.53 | 80.1 |
| Chloroform | 3.63 | 61.2 |
| Ethanol | 1.22 | 78.4 |
| CCl₄ | 5.03 | 76.7 |
Memory Trick:
“Water is half, Benzene is 2.5, CCl₄ is 5” - Remember these three main K_b values!
Theoretical Derivation (For Understanding)
From thermodynamics:
$$\boxed{K_b = \frac{R \times M_{\text{solvent}} \times (T_b^0)^2}{1000 \times \Delta H_{\text{vap}}}}$$Key Insights:
- K_b depends only on solvent properties
- Higher ΔH_vap → Lower K_b
- Higher molecular mass of solvent → Higher K_b
Why Does This Happen?
Graph Explanation:
Vapor Pressure vs Temperature:
Pure solvent curve reaches 760 torr at T_b^0
Solution curve is lower → reaches 760 torr at higher T_b
The horizontal gap at 760 torr = ΔT_b
Molecular Explanation: Solute particles interfere with solvent vaporization → need more energy (higher temperature) to vaporize.
3. Depression of Freezing Point (ΔT_f)
Concept
Freezing Point: Temperature at which solid and liquid phases are in equilibrium (vapor pressures equal).
When solute is added:
- Solution has lower vapor pressure
- Solid (pure solvent) has unchanged vapor pressure
- Equilibrium shifts to lower temperature
Result: Freezing point decreases!
Formula
$$\boxed{\Delta T_f = T_f^0 - T_f = K_f \times m}$$Where:
- ΔT_f = Depression in freezing point
- T_f^0 = Freezing point of pure solvent
- T_f = Freezing point of solution
- K_f = Molal depression constant (cryoscopic constant)
- m = molality of solution
Alternative Form:
$$\boxed{\Delta T_f = K_f \times \frac{w \times 1000}{M \times W}}$$K_f Values (Important for JEE)
| Solvent | K_f (K·kg/mol) | Normal FP (°C) |
|---|---|---|
| Water | 1.86 | 0 |
| Benzene | 5.12 | 5.5 |
| Chloroform | 4.68 | -63.5 |
| Acetic acid | 3.90 | 16.6 |
| Camphor | 37.7 | 179.8 |
Memory Trick:
“Water is 1.86, Benzene is 5.12, Camphor is 37.7” - Camphor has exceptionally high K_f (used for molar mass determination!)
Theoretical Derivation
$$\boxed{K_f = \frac{R \times M_{\text{solvent}} \times (T_f^0)^2}{1000 \times \Delta H_{\text{fusion}}}}$$Comparison with K_b:
- K_f usually > K_b (because ΔH_fusion < ΔH_vaporization)
- Both depend on solvent properties only
Real-Life Application: Antifreeze
Problem: Pure water freezes at 0°C, which would crack car engine blocks in winter.
Solution: Add ethylene glycol (C₂H₆O₂, MW = 62 g/mol)
Example Calculation: How much ethylene glycol per liter of water to prevent freezing down to -15°C?
Given: K_f (water) = 1.86 K·kg/mol, ΔT_f = 15 K
Solution: ΔT_f = K_f × m 15 = 1.86 × m m = 15/1.86 = 8.06 mol/kg
Mass of ethylene glycol needed per kg water: = 8.06 × 62 = 500 g per kg water
For 1L water (≈ 1 kg): Answer: 500 g or about 450 mL of ethylene glycol
This is why antifreeze solutions are roughly 50-50 water-glycol mix!
Relationship Between Colligative Properties
Unified Theory
All four colligative properties stem from lowering of vapor pressure.
Mathematical Relationships:
For dilute solutions with same molality (m):
$$\boxed{\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}}$$Example for water:
$$\frac{\Delta T_b}{\Delta T_f} = \frac{0.52}{1.86} = \frac{1}{3.58}$$Insight: Freezing point depression is about 3.5 times larger than boiling point elevation for the same molality!
Why? Because K_f > K_b (ΔH_fusion < ΔH_vaporization)
Molar Mass Determination
One of the most important JEE applications!
Method 1: From Boiling Point Elevation
$$\boxed{M = \frac{K_b \times w \times 1000}{\Delta T_b \times W}}$$Method 2: From Freezing Point Depression
$$\boxed{M = \frac{K_f \times w \times 1000}{\Delta T_f \times W}}$$Method 3: From Osmotic Pressure
$$\boxed{M = \frac{w \times R \times T}{\pi \times V}}$$Which method is best?
| Method | Advantages | Disadvantages |
|---|---|---|
| Osmotic Pressure | Most accurate, works for macromolecules | Requires special apparatus |
| Freezing Point | More sensitive (K_f > K_b) | Requires precise temperature control |
| Boiling Point | Simple apparatus | Less sensitive, volatile solutes escape |
JEE Preference: Freezing point depression is most commonly asked!
Camphor is Special: With K_f = 37.7, even small temperature changes correspond to significant molar masses → excellent for molar mass determination!
Interactive Demo: Complete Calculation
Problem: Multi-Part Colligative Property
Question: 6.0 g of urea (NH₂CONH₂) is dissolved in 90 g water. Calculate: (a) RLVP at 25°C (P°_water = 23.8 torr) (b) Boiling point of solution (c) Freezing point of solution
Given Data:
- Molar mass of urea = 60 g/mol
- K_b (water) = 0.52 K·kg/mol
- K_f (water) = 1.86 K·kg/mol
Solution:
Step 1: Calculate molality Moles of urea = 6/60 = 0.1 mol Mass of water = 90 g = 0.09 kg m = 0.1/0.09 = 1.11 mol/kg
(a) Relative Lowering of Vapor Pressure:
Method 1 (Exact): Moles of water = 90/18 = 5 mol χ_urea = 0.1/(0.1 + 5) = 0.0196
ΔP/P° = 0.0196 ΔP = 23.8 × 0.0196 = 0.466 torr
P_solution = 23.8 - 0.466 = 23.33 torr
Method 2 (Approximate - dilute solution): Using: ΔP/P° ≈ n_solute/n_solvent ΔP/P° = 0.1/5 = 0.02 (close to 0.0196!)
(b) Boiling Point Elevation:
ΔT_b = K_b × m ΔT_b = 0.52 × 1.11 = 0.577 K = 0.577°C
Boiling point = 100 + 0.577 = 100.577°C
(c) Freezing Point Depression:
ΔT_f = K_f × m ΔT_f = 1.86 × 1.11 = 2.06 K = 2.06°C
Freezing point = 0 - 2.06 = -2.06°C
Verification: ΔT_f / ΔT_b = 2.06/0.577 = 3.57 ≈ K_f/K_b = 1.86/0.52 = 3.58 ✓
Key Observations:
- FP depression (2.06°C) is much larger than BP elevation (0.577°C)
- Ratio is approximately K_f/K_b
- All three properties calculated from same molality!
Common JEE Mistake Traps
Mistake 1: Confusing ΔT_f and T_f
❌ Wrong: ΔT_f = T_f (taking solution’s freezing point as depression) ✅ Correct: ΔT_f = T_f^0 - T_f (depression is the difference)
Example: If solution freezes at -2°C:
- ΔT_f = 0 - (-2) = 2°C (correct)
- NOT ΔT_f = -2°C (wrong!)
Mistake 2: Using Molarity Instead of Molality
❌ Wrong: ΔT_b = K_b × M (using molarity) ✅ Correct: ΔT_b = K_b × m (using molality)
Why? Colligative property formulas use molality, not molarity!
Mistake 3: Wrong Units for w and W
❌ Wrong: Using w in kg and W in g ✅ Correct: Both in grams, OR both in kg - be consistent!
Standard Form: M = (K_f × w × 1000) / (ΔT_f × W) where w and W are in grams!
Mistake 4: Forgetting Factor of 1000
❌ Wrong: M = (K_f × w) / (ΔT_f × W) ✅ Correct: M = (K_f × w × 1000) / (ΔT_f × W)
Memory Aid: The 1000 converts g to kg to match molality units!
Mistake 5: Not Accounting for Ionization
❌ Wrong: NaCl gives same effect as glucose (both 1 mole) ✅ Correct: NaCl ionizes → 2 particles → double the effect!
Solution: Use van’t Hoff factor (covered in next chapter!)
Practice Problems: Three-Level Mastery
Level 1: JEE Main Foundation
Q1. Calculate the boiling point of a solution containing 10 g urea (MW = 60) in 500 g water. (K_b = 0.52 K·kg/mol)
Solution
Step 1: Calculate molality m = (10/60) / 0.5 = 0.167/0.5 = 0.333 mol/kg
Step 2: Calculate ΔT_b ΔT_b = 0.52 × 0.333 = 0.173 K
Step 3: Calculate T_b T_b = 100 + 0.173 = 100.173°C
Q2. A solution of glucose freezes at -0.93°C. Calculate the molality. (K_f = 1.86 K·kg/mol)
Solution
ΔT_f = 0 - (-0.93) = 0.93°C
Using: ΔT_f = K_f × m 0.93 = 1.86 × m m = 0.93/1.86 = 0.5 mol/kg
Q3. The vapor pressure of pure water at 25°C is 23.76 torr. What is the vapor pressure when 12 g glucose (MW = 180) is dissolved in 100 g water?
Solution
Moles: Glucose = 12/180 = 0.0667 mol Water = 100/18 = 5.56 mol
Mole fraction: χ_water = 5.56/(5.56 + 0.0667) = 5.56/5.627 = 0.988
Vapor pressure: P_s = 23.76 × 0.988 = 23.47 torr
Level 2: JEE Main Advanced
Q4. A solution containing 12.5 g of non-electrolyte in 175 g water freezes at -0.372°C. Calculate the molar mass of solute. (K_f = 1.86 K·kg/mol)
Solution
Given:
- w = 12.5 g
- W = 175 g
- ΔT_f = 0 - (-0.372) = 0.372°C
Using formula: M = (K_f × w × 1000) / (ΔT_f × W) M = (1.86 × 12.5 × 1000) / (0.372 × 175) M = 23250 / 65.1 M = 357 g/mol
Q5. 2 g of benzoic acid dissolved in 25 g benzene shows a depression in freezing point of 1.62 K. Calculate the molar mass of benzoic acid. (K_f for benzene = 5.12 K·kg/mol)
Solution
Given:
- w = 2 g
- W = 25 g
- ΔT_f = 1.62 K
Calculation: M = (K_f × w × 1000) / (ΔT_f × W) M = (5.12 × 2 × 1000) / (1.62 × 25) M = 10240 / 40.5 M = 253 g/mol
Note: Theoretical molar mass of benzoic acid (C₆H₅COOH) = 122 g/mol. The observed value is approximately double! This suggests association (dimerization) of benzoic acid in benzene. We’ll study this in van’t Hoff factor chapter!
Q6. The boiling point of a solution of 0.1 g of a substance in 16 g ether (K_b = 2.16 K·kg/mol) is 0.100°C higher than pure ether. Calculate molar mass.
Solution
Given:
- w = 0.1 g
- W = 16 g
- ΔT_b = 0.100 K
Calculation: M = (K_b × w × 1000) / (ΔT_b × W) M = (2.16 × 0.1 × 1000) / (0.100 × 16) M = 216 / 1.6 M = 135 g/mol
Level 3: JEE Advanced Challenge
Q7. (JEE Advanced Type) Calculate the freezing point of a solution containing 8.1 g HBr in 100 g water assuming 90% ionization. (K_f = 1.86 K·kg/mol, MW of HBr = 81)
Solution
Step 1: Calculate moles Moles of HBr = 8.1/81 = 0.1 mol
Step 2: Account for ionization HBr → H⁺ + Br⁻
If 90% ionizes:
- HBr remaining = 0.1 × 0.1 = 0.01 mol
- H⁺ formed = 0.1 × 0.9 = 0.09 mol
- Br⁻ formed = 0.1 × 0.9 = 0.09 mol
Total particles = 0.01 + 0.09 + 0.09 = 0.19 mol
Step 3: Calculate molality m = 0.19/0.1 = 1.9 mol/kg
Step 4: Calculate ΔT_f ΔT_f = 1.86 × 1.9 = 3.534 K
Step 5: Calculate freezing point T_f = 0 - 3.534 = -3.534°C
Alternative (using van’t Hoff factor): i = 1 + α(n-1) = 1 + 0.9(2-1) = 1.9 ΔT_f = i × K_f × m_initial ΔT_f = 1.9 × 1.86 × 1 = 3.534 K ✓
Q8. (JEE Advanced 2019 Type) Two solutions are prepared:
- Solution A: 10 g urea (MW = 60) in 100 g water
- Solution B: 10 g glucose (MW = 180) in 100 g water
Calculate the ratio of: (a) Their boiling points (b) Their freezing points (c) Their vapor pressures at 25°C
Solution
For Solution A (Urea): m_A = (10/60) / 0.1 = 1.667 mol/kg n_urea = 0.167 mol, n_water = 5.56 mol χ_urea = 0.167/5.727 = 0.0291
For Solution B (Glucose): m_B = (10/180) / 0.1 = 0.556 mol/kg n_glucose = 0.0556 mol, n_water = 5.56 mol χ_glucose = 0.0556/5.616 = 0.0099
(a) Boiling Points: ΔT_b(A) = 0.52 × 1.667 = 0.867 K → T_b(A) = 100.867°C ΔT_b(B) = 0.52 × 0.556 = 0.289 K → T_b(B) = 100.289°C
Ratio = 100.867/100.289 ≈ 1.0058 (very close to 1)
For JEE approximation: Ratio ≈ 1 + (ΔT_b(A) - ΔT_b(B))/100 = 1 + (0.867 - 0.289)/100 = 1.0058
(b) Freezing Points: ΔT_f(A) = 1.86 × 1.667 = 3.10 K → T_f(A) = -3.10°C ΔT_f(B) = 1.86 × 0.556 = 1.03 K → T_f(B) = -1.03°C
Ratio = -3.10/-1.03 = 3.01
Note: Freezing points are negative, so lower number = lower temperature!
(c) Vapor Pressures: P_A = P° × (1 - 0.0291) = 0.9709 P° P_B = P° × (1 - 0.0099) = 0.9901 P°
Ratio = 0.9709/0.9901 = 0.981
Summary: Solution A (urea) has higher concentration → greater effect on all colligative properties!
Q9. (Challenging) A solution is prepared by dissolving 1.8 g glucose (MW = 180) and 3.42 g sucrose (MW = 342) in 100 g water. Calculate: (a) Total lowering of vapor pressure at 25°C (P° = 23.8 torr) (b) Boiling point (c) Freezing point
Solution
Step 1: Calculate total moles of solute Moles of glucose = 1.8/180 = 0.01 mol Moles of sucrose = 3.42/342 = 0.01 mol Total moles = 0.02 mol
Step 2: Calculate molality m = 0.02/0.1 = 0.2 mol/kg
(a) RLVP: Moles of water = 100/18 = 5.56 mol χ_solute = 0.02/(0.02 + 5.56) = 0.00358
ΔP = 23.8 × 0.00358 = 0.085 torr P_solution = 23.8 - 0.085 = 23.715 torr
(b) Boiling Point: ΔT_b = 0.52 × 0.2 = 0.104 K T_b = 100 + 0.104 = 100.104°C
(c) Freezing Point: ΔT_f = 1.86 × 0.2 = 0.372 K T_f = 0 - 0.372 = -0.372°C
Key Point: Two different solutes, but we just add their moles! Colligative properties depend on TOTAL number of particles, not their type!
Interactive Demo: Visualize Colligative Properties
See how solute particles affect boiling point, freezing point, and osmotic pressure.
Q10. (JEE Advanced Level) 0.90 g of a non-electrolyte dissolved in 90 g water gave a boiling point elevation of 0.052 K. The same mass of solute in 60 g benzene gave a boiling point elevation of 0.125 K. Calculate: (a) Molar mass from water data (b) Molar mass from benzene data (c) Are they consistent?
Solution
(a) From water data: K_b (water) = 0.52 K·kg/mol w = 0.90 g, W = 90 g, ΔT_b = 0.052 K
M = (0.52 × 0.90 × 1000) / (0.052 × 90) M = 468 / 4.68 M = 100 g/mol
(b) From benzene data: K_b (benzene) = 2.53 K·kg/mol w = 0.90 g, W = 60 g, ΔT_b = 0.125 K
M = (2.53 × 0.90 × 1000) / (0.125 × 60) M = 2277 / 7.5 M = 303.6 g/mol
(c) Consistency: NOT consistent! Water gives ~100 g/mol, benzene gives ~304 g/mol
Reason: The solute likely associates (forms dimers or trimers) in benzene but remains as single molecules in water. This is common for substances that can hydrogen bond with water but form aggregates in non-polar solvents.
Actual behavior:
- In water: Single molecules (MW ≈ 100)
- In benzene: Trimers (3 molecules stick together) → MW appears 3× larger
This demonstrates why colligative properties can reveal molecular associations!
Summary Table: Colligative Properties
| Property | Formula | K values (Water) | Depends on |
|---|---|---|---|
| RLVP | ΔP/P° = χ_solute | - | Mole fraction |
| BP Elevation | ΔT_b = K_b × m | 0.52 K·kg/mol | Molality |
| FP Depression | ΔT_f = K_f × m | 1.86 K·kg/mol | Molality |
| Osmotic Pressure | π = CRT | - | Molarity |
Common Features: ✅ All depend on number of solute particles ✅ All independent of nature of solute ✅ All apply to dilute solutions ✅ All stem from vapor pressure lowering
Cross-Connections to Other Chapters
🔗 Link to Raoult’s Law
- RLVP is direct application of Raoult’s law
- Vapor pressure lowering is the foundation
- See: Raoult’s Law
🔗 Link to van’t Hoff Factor
- Electrolytes ionize → more particles
- Need to multiply by van’t Hoff factor (i)
- See: van’t Hoff Factor
🔗 Link to Thermodynamics
- ΔG = 0 at phase transitions (freezing, boiling)
- Entropy of mixing always increases
- See: Gibbs Free Energy
🔗 Link to Chemical Kinetics
- Boiling/freezing involves equilibrium rates
- Arrhenius equation relates to temperature
- See: Temperature Effect on Rates
🔗 Link to Equilibrium
- Phase equilibria at freezing/boiling points
- Le Chatelier’s principle applies
- See: Phase Equilibrium
Formula Sheet: Quick Reference
RLVP
$$\boxed{\frac{P^0 - P_s}{P^0} = \chi_{\text{solute}}}$$Boiling Point Elevation
$$\boxed{\Delta T_b = K_b \times m = K_b \times \frac{w \times 1000}{M \times W}}$$Freezing Point Depression
$$\boxed{\Delta T_f = K_f \times m = K_f \times \frac{w \times 1000}{M \times W}}$$Molar Mass Determination
$$\boxed{M = \frac{K_f \times w \times 1000}{\Delta T_f \times W}}$$Important K Values
Water:
- K_b = 0.52 K·kg/mol
- K_f = 1.86 K·kg/mol
Benzene:
- K_b = 2.53 K·kg/mol
- K_f = 5.12 K·kg/mol
Camphor:
- K_f = 37.7 K·kg/mol (highest!)
Final JEE Strategy
✅ Highest Weightage Topics:
- Molar mass determination (70% probability)
- Freezing point depression (60% probability)
- Mixed solute problems (50% probability)
- Conceptual questions on K_f, K_b (40% probability)
✅ Must-Remember Values:
- K_f (water) = 1.86, K_b (water) = 0.52
- K_f (benzene) = 5.12, K_b (benzene) = 2.53
- K_f (camphor) = 37.7
✅ Common Question Types:
- Calculate molar mass from ΔT_f or ΔT_b
- Calculate freezing/boiling point of solution
- Compare colligative effects of different solutions
- Applications: antifreeze, de-icing, etc.
✅ Calculation Checklist:
- Units: w and W in grams? ✓
- Don’t forget the 1000 factor? ✓
- ΔT_f = T_f^0 - T_f (not T_f itself)? ✓
- Using molality, not molarity? ✓
- For electrolytes, use van’t Hoff factor? ✓
Previous Topic: Raoult’s Law Next Topic: Osmotic Pressure - the fourth and most practically important colligative property!
Last Updated: June 2025 | For JEE Main & Advanced 2026