Colligative Properties | B.Sc. 1st Year Chemistry

1. Introduction to Colligative Properties

Colligative properties are properties of dilute solutions that depend mainly on the number of solute particles present in the solution and not on the chemical nature of the solute particles.

The word "colligative" is related to the idea of particles being considered collectively. For example, one mole of glucose and one mole of another non-electrolyte solute produce the same number of solute molecules when present in the same amount, provided there is no association or dissociation.

Main colligative properties

Relative lowering of vapour pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

These properties are especially useful for determining the molar mass of a dissolved substance.

Basic idea: Colligative properties depend on the number of dissolved particles, not simply on the identity of the solute.

2. Solutions and Dilute Solutions

A solution is a homogeneous mixture of two or more substances.

The substance present in a larger amount is generally called the solvent, while the substance dissolved in it is called the solute.

Solution = Solvent + Solute

Example

In a salt-water solution:

Water = solvent

NaCl = solute

Colligative-property equations are commonly discussed for dilute solutions, especially solutions containing a non-volatile solute in a volatile solvent.

Volatile and non-volatile substances

A volatile substance has appreciable vapour pressure and can readily enter the vapour phase.

A non-volatile solute has negligible vapour pressure compared with the solvent.

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solvent decreases.

3. Vapour Pressure

Vapour pressure is the pressure exerted by the vapour of a liquid when the liquid and vapour are in dynamic equilibrium at a given temperature.

Consider a pure solvent. At a particular temperature it has a definite vapour pressure, represented by:

P°₁ = vapour pressure of pure solvent

When a non-volatile solute is added, some surface positions are occupied by solute particles. Fewer solvent molecules escape into the vapour phase.

P₁ < P°₁

Therefore, the vapour pressure of the solvent decreases.

Why does vapour pressure decrease?

In a solution containing a non-volatile solute, the fraction of solvent molecules at the surface is lower than in the pure solvent. Consequently, fewer solvent molecules escape into the vapour phase.

Pure solvent → more solvent molecules escape → higher vapour pressure.
Solution → fewer solvent molecules escape → lower vapour pressure.

4. Raoult's Law

For a solution containing a volatile component, the partial vapour pressure of a component is proportional to its mole fraction in the solution.

For a volatile component A:

P_A = X_A P°_A

Where:

P_A = partial vapour pressure of A in solution
X_A = mole fraction of A
P°_A = vapour pressure of pure A

For a non-volatile solute

If the solute is non-volatile, the vapour pressure of the solution is due almost entirely to the solvent.

P = X₁ P°₁

Therefore:

P°₁ − P = P°₁ − X₁P°₁

P°₁ − P = P°₁(1 − X₁)

Since:

X₁ + X₂ = 1

Therefore:

P°₁ − P = P°₁X₂

Relative lowering of vapour pressure

(P°₁ − P)/P°₁ = X₂

For a dilute solution:

X₂ ≈ n₂/n₁

(P°₁ − P)/P°₁ ≈ n₂/n₁

For dilute solution: Relative lowering of vapour pressure is approximately equal to the mole fraction of the solute.

5. Relative Lowering of Vapour Pressure

Relative lowering of vapour pressure is defined as the decrease in vapour pressure divided by the vapour pressure of the pure solvent.

Relative lowering = (P°₁ − P₁)/P°₁

According to Raoult's law:

(P°₁ − P₁)/P°₁ = X₂

For a dilute solution:

X₂ = n₂/(n₁ + n₂)

Since n₂ is very small compared with n₁:

X₂ ≈ n₂/n₁

If:

n₁ = w₁/M₁

n₂ = w₂/M₂

Then:

(P°₁ − P₁)/P°₁ = (w₂/M₂)/(w₁/M₁)

(P°₁ − P₁)/P°₁ = w₂M₁/(w₁M₂)

Determination of molar mass

Rearranging:

M₂ = w₂M₁P°₁ / [w₁(P°₁ − P₁)]

This equation can be used to determine the molar mass of a non-volatile solute.

6. Elevation of Boiling Point

The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external pressure.

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lower than that of the pure solvent. Therefore, the solution must be heated to a higher temperature to reach the external pressure.

Hence, the boiling point increases.

T_b(solution) > T_b(pure solvent)

Elevation in boiling point

ΔT_b = T_b − T°_b

For a dilute solution:

ΔT_b = K_b m

Where:

ΔT_b = elevation of boiling point
K_b = molal elevation constant or ebullioscopic constant
m = molality of solution

Molality

m = moles of solute / kg of solvent

If w₂ grams of solute having molar mass M₂ are dissolved in w₁ grams of solvent:

m = (w₂/M₂) / (w₁/1000)

m = 1000w₂/(M₂w₁)

Therefore:

ΔT_b = K_b × 1000w₂/(M₂w₁)

Molar mass from elevation of boiling point

M₂ = K_b × 1000w₂/(ΔT_bw₁)

Memory: Boiling point goes up → ΔT_b = K_bm.

7. Depression of Freezing Point

The freezing point of a liquid is the temperature at which the liquid and solid phases are in equilibrium.

When a non-volatile solute is dissolved in a solvent, the freezing point of the solution becomes lower than the freezing point of the pure solvent.

T_f(solution) < T°_f(pure solvent)

Depression in freezing point

ΔT_f = T°_f − T_f

For a dilute solution:

ΔT_f = K_f m

Where:

ΔT_f = depression of freezing point
K_f = molal depression constant or cryoscopic constant
m = molality

Using:

m = 1000w₂/(M₂w₁)

We get:

ΔT_f = K_f × 1000w₂/(M₂w₁)

Molar mass from depression of freezing point

M₂ = K_f × 1000w₂/(ΔT_fw₁)

Practical example

Salt is spread on icy roads in cold regions because dissolving salt in the thin water layer lowers the freezing point of water.

Memory: Freezing point goes down → ΔT_f = K_fm.

8. Osmotic Pressure

Osmosis is the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration toward a region of higher solute concentration.

A semipermeable membrane allows solvent molecules to pass but restricts the passage of solute particles.

Osmotic pressure

Osmotic pressure is the minimum external pressure that must be applied to a solution to stop osmosis.

It is represented by π.

van't Hoff equation

πV = nRT

Since:

n = w/M

Therefore:

πV = wRT/M

Rearranging:

M = wRT/(πV)

If concentration C is expressed in mol L⁻¹:

π = CRT

Why osmotic pressure is useful

Osmotic pressure can be measured at relatively low temperatures and is particularly useful for determining the molar mass of large molecules such as polymers and biological molecules.

Memory: Osmotic pressure uses πV = nRT, similar in form to the ideal gas equation PV = nRT.

9. Reverse Osmosis

In normal osmosis, solvent passes through a semipermeable membrane toward the solution with higher solute concentration.

In reverse osmosis, an external pressure greater than the osmotic pressure is applied to the concentrated solution. This forces solvent molecules in the opposite direction.

Applied pressure > Osmotic pressure

Use of reverse osmosis

Reverse osmosis is widely used for water purification and desalination.

For example, dissolved salts can be removed from water by forcing water through a suitable semipermeable membrane.

Remember: Osmosis occurs naturally due to concentration difference. Reverse osmosis requires applied pressure.

10. Determination of Molar Mass from Elevation of Boiling Point

The relationship is:

ΔT_b = K_bm

For:

m = 1000w₂/(M₂w₁)

Substitution gives:

ΔT_b = K_b × 1000w₂/(M₂w₁)

Therefore:

M₂ = K_b × 1000w₂/(ΔT_bw₁)

Terms used

w₂ = mass of solute in grams
M₂ = molar mass of solute
w₁ = mass of solvent in grams
K_b = boiling point constant
ΔT_b = observed elevation of boiling point

Common mistake: Molality uses kilograms of solvent. If the solvent mass is given in grams, the factor 1000 must be included.

11. Determination of Molar Mass from Depression of Freezing Point

Start with:

ΔT_f = K_fm

Since:

m = 1000w₂/(M₂w₁)

Therefore:

ΔT_f = K_f × 1000w₂/(M₂w₁)

Rearranging:

M₂ = K_f × 1000w₂/(ΔT_fw₁)

Why freezing point measurements are useful

The change in freezing point can be measured accurately for dilute solutions, making cryoscopic methods useful for determining molar mass.

12. Determination of Molar Mass from Osmotic Pressure

Starting from:

πV = nRT

Since:

n = w/M

Therefore:

πV = wRT/M

Rearranging:

M = wRT/(πV)

Importance for large molecules

Osmotic pressure is particularly useful for determining the molar mass of macromolecules because other colligative changes may be extremely small for high-molar-mass substances.

Examples include:

Proteins
Polysaccharides
Polymers

Exam point: Osmotic pressure is preferred for large biomolecules because it can be measured using relatively dilute solutions without requiring large temperature changes.

13. van't Hoff Factor

The colligative properties depend on the number of solute particles. If the solute associates or dissociates in solution, the actual number of particles becomes different from the number expected from the original formula.

To account for this, van't Hoff introduced a factor represented by i.

i = observed colligative property / calculated colligative property

Therefore:

i = observed value / normal value

Modified equations

ΔT_b = iK_bm

ΔT_f = iK_fm

π = iCRT

Relative lowering of vapour pressure = iX₂

Interpretation of i

i = 1 → no association or dissociation

i > 1 → dissociation

i < 1 → association

Particle count: Dissociation increases the number of particles, while association decreases the number of particles.

14. Association of Solute Particles

Association occurs when two or more solute particles combine to form larger particles.

Example: Dimerization

2A ⇌ A₂

Before association, two particles are present. After complete association, they become one particle.

Therefore, the number of particles decreases.

Number of particles ↓

i < 1

Degree of association

Let α be the degree of association for:

2A ⇌ A₂

If one mole of A is initially present:

Initial particles = 1

If α is the fraction associated:

A remaining = 1 − α

Amount of A₂ formed:

A₂ = α/2

Total particles:

Total = (1 − α) + α/2

Total = 1 − α/2

Therefore:

i = 1 − α/2

Hence:

α = 2(1 − i)

15. Dissociation of Solute Particles

Dissociation occurs when a compound separates into two or more particles in solution.

Example

NaCl → Na⁺ + Cl⁻

One particle of NaCl produces two ions when completely dissociated.

Number of particles ↑

i > 1

Degree of dissociation

For:

AB ⇌ A⁺ + B⁻

Suppose one mole of AB is initially present and α is the degree of dissociation.

AB remaining = 1 − α

A⁺ formed = α

B⁻ formed = α

Total particles:

Total = (1 − α) + α + α

Total = 1 + α

Therefore:

i = 1 + α

α = i − 1

For a completely dissociated 1:1 electrolyte:

i ≈ 2

For:

CaCl₂ → Ca²⁺ + 2Cl⁻

Complete dissociation gives three particles:

i ≈ 3

16. Abnormal Molar Mass

The molar mass calculated from colligative properties may differ from the actual molar mass when association or dissociation occurs.

This is called abnormal molar mass.

Relationship with van't Hoff factor

i = Normal molar mass / Observed molar mass

Therefore:

Observed molar mass = Normal molar mass / i

Rearranging:

i = M_normal/M_observed

For association

Since i is less than 1:

M_observed > M_normal

For dissociation

Since i is greater than 1:

M_observed < M_normal

Easy way to remember:
Association → fewer particles → i < 1 → observed molar mass becomes larger.
Dissociation → more particles → i > 1 → observed molar mass becomes smaller.

17. Comparison of Colligative Properties

Property	Main equation	Effect on solvent
Relative lowering of vapour pressure	(P° − P)/P° = X₂	Vapour pressure decreases
Elevation of boiling point	ΔT_b = K_bm	Boiling point increases
Depression of freezing point	ΔT_f = K_fm	Freezing point decreases
Osmotic pressure	π = CRT	Pressure needed to stop osmosis

With van't Hoff factor

ΔT_b = iK_bm

ΔT_f = iK_fm

π = iCRT

Relative lowering = iX₂

Four-property memory:
Vapour pressure ↓
Boiling point ↑
Freezing point ↓
Osmotic pressure = pressure required to stop osmosis

18. Worked Numerical Problems

Numerical 1: Relative lowering of vapour pressure

Calculate the relative lowering of vapour pressure when 1 mole of a non-volatile solute is dissolved in 99 moles of solvent.

Given:

n₂ = 1 mol

n₁ = 99 mol

Formula:

Relative lowering = X₂

Mole fraction of solute:

X₂ = n₂/(n₁ + n₂)

X₂ = 1/(99 + 1)

X₂ = 1/100 = 0.01

Therefore:

Relative lowering = 0.01

Or:

Relative lowering = 1%

Numerical 2: Elevation of boiling point

Calculate the elevation in boiling point when 0.20 mol of a non-electrolyte is dissolved in 2 kg of solvent. Take K_b = 0.52 K kg mol⁻¹.

Formula:

ΔT_b = K_bm

First calculate molality:

m = 0.20/2

m = 0.10 mol kg⁻¹

Substitution:

ΔT_b = 0.52 × 0.10

ΔT_b = 0.052 K

Therefore, the boiling point rises by 0.052 K.

Numerical 3: Depression of freezing point

Calculate the depression in freezing point for a solution having molality 0.50 mol kg⁻¹. Take K_f = 1.86 K kg mol⁻¹.

Formula:

ΔT_f = K_fm

Substitution:

ΔT_f = 1.86 × 0.50

ΔT_f = 0.93 K

Therefore, the freezing point decreases by 0.93 K.

Numerical 4: Molar mass from elevation of boiling point

2.0 g of a non-volatile solute is dissolved in 100 g of water. The elevation in boiling point is 0.52 K. If K_b for water is 0.52 K kg mol⁻¹, calculate the molar mass of the solute.

Given:

w₂ = 2.0 g

w₁ = 100 g

ΔT_b = 0.52 K

K_b = 0.52 K kg mol⁻¹

Formula:

M₂ = K_b × 1000w₂/(ΔT_bw₁)

Substitution:

M₂ = (0.52 × 1000 × 2.0)/(0.52 × 100)

M₂ = 2000/100

M₂ = 20 g mol⁻¹

Therefore, the molar mass is 20 g mol⁻¹.

Numerical 5: Molar mass from osmotic pressure

A solution contains 2.0 g of a non-electrolyte solute in 1.0 L solution at 300 K. Its osmotic pressure is 0.246 atm. Calculate the molar mass.

Formula:

M = wRT/(πV)

Using:

w = 2.0 g

R = 0.0821 L atm mol⁻¹ K⁻¹

T = 300 K

π = 0.246 atm

V = 1.0 L

Substitution:

M = (2.0 × 0.0821 × 300)/(0.246 × 1.0)

M ≈ 200.2 g mol⁻¹

Therefore, the molar mass is approximately 200 g mol⁻¹.

Numerical 6: van't Hoff factor

The calculated freezing-point depression of a solution is 0.50 K, while the experimentally observed depression is 0.75 K. Calculate the van't Hoff factor.

Formula:

i = observed value / calculated value

Substitution:

i = 0.75/0.50

i = 1.5

Since:

i > 1

The result indicates that the solute produces additional particles, suggesting dissociation.

Numerical 7: Degree of dissociation

A compound AB dissociates according to:

AB ⇌ A⁺ + B⁻

If the van't Hoff factor is 1.4, calculate the degree of dissociation.

For a 1:1 electrolyte:

i = 1 + α

Therefore:

α = i − 1

α = 1.4 − 1

α = 0.4

Therefore:

Degree of dissociation = 40%

Numerical 8: Degree of association

A solute associates according to:

2A ⇌ A₂

If i = 0.80, calculate the degree of association.

For dimerization:

i = 1 − α/2

Therefore:

0.80 = 1 − α/2

α/2 = 0.20

α = 0.40

Degree of association = 40%

19. Common Mistakes

1. Confusing boiling point and freezing point changes

Boiling point ↑

Freezing point ↓

2. Using molarity instead of molality

The usual equations for ΔT_b and ΔT_f use molality:

ΔT_b = K_bm

ΔT_f = K_fm

3. Forgetting the 1000 factor

When solvent mass is given in grams:

m = 1000w₂/(M₂w₁)

4. Confusing solute and solvent mass

w₂ is commonly used for solute mass and w₁ for solvent mass in these equations. Always check the definitions before substitution.

5. Forgetting van't Hoff factor for electrolytes

For dissociating or associating substances:

ΔT_f = iK_fm

and similarly for the other colligative properties.

6. Thinking i > 1 means association

It is the opposite:

i > 1 → dissociation

i < 1 → association

7. Using the wrong gas constant in osmotic-pressure calculations

If pressure is in atm and volume is in L, a convenient value is:

R = 0.0821 L atm mol⁻¹ K⁻¹

If SI units are used, use the corresponding SI value consistently.

8. Forgetting that osmotic pressure is a pressure

It is not a temperature change like ΔT_b or ΔT_f.

20. TU Examination Preparation

Important 2-Mark Questions

Define colligative properties.
State Raoult's law.
Define relative lowering of vapour pressure.
What is elevation of boiling point?
What is depression of freezing point?
Define osmotic pressure.
What is reverse osmosis?
Define van't Hoff factor.
What is abnormal molar mass?
Differentiate association and dissociation.

Important 5-Mark Questions

Explain Raoult's law for a solution containing a non-volatile solute.
Derive the expression for relative lowering of vapour pressure.
Explain elevation of boiling point and derive ΔT_b = K_bm.
Explain depression of freezing point and derive ΔT_f = K_fm.
Explain osmotic pressure and derive π = CRT.
Explain reverse osmosis and its applications.
Explain van't Hoff factor and its significance.

Important 10-Mark Questions

Discuss all four colligative properties and explain their use in determination of molar mass.
Derive the relation for elevation of boiling point and explain the determination of molar mass using the method.
Derive the relation for depression of freezing point and explain the determination of molar mass.
Explain osmotic pressure, van't Hoff equation and the determination of molar mass by osmotic pressure.
Explain abnormal molar mass due to association and dissociation with van't Hoff factor.
Derive expressions for van't Hoff factor and degree of association/dissociation.

For derivations: Start from the basic law, define each symbol, show the substitution step-by-step and finish with the final equation.

21. Quick Revision

Colligative properties depend on the number of solute particles.
The four main properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.
Raoult's law for a non-volatile solute: P = X₁P°₁.
Relative lowering: (P°₁ − P₁)/P°₁ = X₂.
Elevation of boiling point: ΔT_b = K_bm.
Depression of freezing point: ΔT_f = K_fm.
Osmotic pressure: πV = nRT.
For dilute solution: π = CRT.
Molar mass from osmotic pressure: M = wRT/(πV).
van't Hoff factor: i = observed colligative property / calculated colligative property.
i > 1 indicates dissociation.
i < 1 indicates association.
For association: M_observed > M_normal.
For dissociation: M_observed < M_normal.

Formula memory:

Relative lowering: (P° − P)/P° = X₂

Boiling point: ΔT_b = K_bm

Freezing point: ΔT_f = K_fm

Osmotic pressure: π = CRT

With van't Hoff factor: ΔT_b = iK_bm
ΔT_f = iK_fm
π = iCRT

22. Quick Check

1. What do colligative properties depend on?

They depend mainly on the number of solute particles present in the solution.

2. What happens to the boiling point when a non-volatile solute is added?

The boiling point increases.

3. What happens to the freezing point?

The freezing point decreases.

4. Write the equation for elevation of boiling point.

ΔT_b = K_bm for a dilute solution of a non-electrolyte.

5. Write the equation for depression of freezing point.

ΔT_f = K_fm.

6. What is osmotic pressure?

Osmotic pressure is the minimum external pressure required to stop osmosis.

7. What does i > 1 indicate?

It generally indicates dissociation because the number of particles increases.

8. What does i < 1 indicate?

It indicates association because the number of particles decreases.

9. Which colligative property is especially useful for determining the molar mass of proteins?

Osmotic pressure is especially useful for large molecules such as proteins and polymers.

10. What is the main use of van't Hoff factor?

It corrects colligative-property calculations when the solute undergoes association or dissociation.