Solutions & Colligative Properties

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Expressing Concentration of Solutions

Quantitative methods to describe the amount of solute in a given solution

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Definition

A solution is a homogeneous mixture of two or more substances. The substance present in smaller amount is the solute and the substance present in larger amount is the solvent.

M

Molarity

Number of moles of solute dissolved per litre of solution.

M = n₂V (L)

mol L⁻¹

⚠ Temperature dependent (volume changes with T)

m

Molality

Number of moles of solute per kilogram of solvent.

m = n₂W₁ (kg)

mol kg⁻¹

✓ Temperature independent (mass is invariant)

χ

Mole Fraction

Ratio of moles of one component to total moles in solution.

χ_A = n_An_A + n_B

Dimensionless

χ_A + χ_B = 1 (always)

w/w%

Mass Percentage

Mass of solute per 100 g of solution.

w₂w₁+w₂ × 100

% (w/w)

v/v%

Volume Percentage

Volume of solute per 100 mL of solution.

V₂V × 100

% (v/v)

ppm

Parts per Million

Mass of solute per 10⁶ parts of solution. Used for trace constituents.

ppm = w₂w_solution × 10⁶

mg/kg or mg/L

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Key Relationships

Molarity & Molality: m = (M × 1000) / (1000ρ − M·M₂) where ρ is density (g/mL) and M₂ is molar mass of solute
For dilute aqueous solutions: molarity ≈ molality (since density ≈ 1 g/mL)
Mole fraction to molality: m = (χ₂ × 1000) / (χ₁ × M₁)

🧮 Concentration Calculator

Mass of Solute (g)

Molar Mass of Solute (g/mol)

Mass of Solvent (g)

Volume of Solution (mL)

Molar Mass of Solvent (g/mol)

Density of Solution (g/mL) (optional)

Molarity (M)

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Molality (m)

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Mole Fraction of Solute (χ₂)

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Mass % (w/w)

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Volume % (v/v)

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Vapour Pressure of Solutions & Raoult's Law

Understanding how solutes affect the vapour pressure of solutions

🌡 What is Vapour Pressure?

The vapour pressure of a liquid is the pressure exerted by its vapour when it is in dynamic equilibrium with the liquid phase at a given temperature. It is a measure of the tendency of molecules to escape into the vapour phase.

Increases with temperature (Clausius-Clapeyron equation)
When a non-volatile solute is dissolved, the vapour pressure decreases
This decrease is a colligative property — depends only on the number of solute particles

📐 Raoult's Law

"At a given temperature, the partial vapour pressure of each volatile component of a solution is directly proportional to its mole fraction in the solution."

Raoult's Law

p_A = χ_A · p°_A

Total Pressure (Binary Solution)

P_total = χ_A·p°_A + χ_B·p°_B

where p°_A and p°_B are vapour pressures of pure components A and B, and χ_A, χ_B are their mole fractions.

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Ideal Solutions — Characteristics

Obey Raoult's Law over all composition ranges
ΔH_mix = 0 — No heat is released or absorbed on mixing
ΔV_mix = 0 — Volume of solution = sum of volumes of components
Intermolecular forces: A–B = A–A = B–B (similar nature)
Examples: Benzene + Toluene, n-Hexane + n-Heptane, Ethyl bromide + Ethyl iodide, CCl₄ + SiCl₄

📊 Interactive Vapour Pressure–Composition Plot

p°A: 160 p°B: 100

Partial pressure p_A

Partial pressure p_B

Total pressure P

Ideal reference

📈 Positive Deviation from Raoult's Law

Observed VP > Predicted VP (Raoult's Law)

A–B interactions < A–A or B–B interactions
Molecules escape more easily → higher VP
ΔH_mix > 0 (endothermic mixing)
ΔV_mix > 0 (slight expansion)

Examples

Ethanol + Water, Acetone + CS₂, Ethanol + Cyclohexane, Benzene + Acetone, CCl₄ + Toluene

📉 Negative Deviation from Raoult's Law

Observed VP < Predicted VP (Raoult's Law)

A–B interactions > A–A or B–B interactions
Molecules held more tightly → lower VP
ΔH_mix < 0 (exothermic mixing)
ΔV_mix < 0 (slight contraction)

Examples

Chloroform + Acetone (H-bonding), HCl + Water, HNO₃ + Water, Phenol + Aniline, CH₃COOH + Pyridine

🌀 Azeotropes — Constant Boiling Mixtures

Azeotropes are binary mixtures having the same composition in liquid and vapour phase at a given temperature. They cannot be separated by simple distillation.

⬆ Minimum Boiling Azeotrope

Formed by liquids showing positive deviation. The azeotrope boils at a lower temperature than either pure component.

Example: Ethanol–Water (95.5% EtOH, bp = 78.1°C)

⬇ Maximum Boiling Azeotrope

Formed by liquids showing negative deviation. The azeotrope boils at a higher temperature than either pure component.

Example: HCl–Water (20.24% HCl, bp = 108.58°C)

🔍 Henry's Law (for Volatile Solutes)

For a volatile solute (gas dissolved in liquid), the partial pressure is proportional to its mole fraction but with Henry's constant K_H instead of p°:

Henry's Law

p = K_H · χ

K_H > p° → gas is less soluble (positive deviation for gases). Raoult's Law applies to solvent; Henry's Law applies to dilute solute.

Applications: Carbonation of drinks, scuba diving (nitrogen narcosis), blood oxygen levels

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Colligative Properties

Properties that depend on the number of solute particles, not their nature

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Definition & Significance

Colligative properties are those properties of solutions that depend only on the number of dissolved solute particles and NOT on their chemical nature or identity. They include: Relative Lowering of Vapour Pressure, Elevation of Boiling Point, Depression of Freezing Point, and Osmotic Pressure.

🔻Relative Lowering of VP

♨️Elevation of BP

❄️Depression of FP

💧Osmotic Pressure

Relative Lowering of Vapour Pressure (RLVP)

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution (p) is less than that of the pure solvent (p°). This decrease is called lowering of vapour pressure.

RLVP = Mole Fraction of Solute

p° − pp° = χ₂ = n₂n₁ + n₂

For dilute solutions (n₂ << n₁):

Dilute Solution Approximation

Δpp° ≈ n₂n₁ = w₂ / M₂w₁ / M₁

Molecular Mass from RLVP

M₂ = Δp · w₂ · M₁(p° − Δp) · w₁

Worked Example

Vapour pressure of water at 25°C = 23.8 mmHg. On dissolving 1.8 g glucose (M = 180) in 90 g water, find RLVP.

n₂ = 1.8/180 = 0.01 mol | n₁ = 90/18 = 5 mol
χ₂ = 0.01/(0.01 + 5) = 0.002 → RLVP = 0.002

Elevation of Boiling Point (ΔT_b)

The boiling point of a solution is always higher than that of the pure solvent. This is because the vapour pressure of the solution is lower, so a higher temperature is required to make the vapour pressure equal to atmospheric pressure.

Elevation of Boiling Point

ΔT_b = T_b(solution) − T_b(solvent) = K_b × m

Ebullioscopic Constant

K_b = R · T²_b · M₁1000 · ΔH_vap

Molecular Mass of Solute

M₂ = K_b × w₂ × 1000ΔT_b × w₁

K_b Values for Common Solvents

Solvent	Normal BP (°C)	K_b (K kg mol⁻¹)
Water	100.0	0.52
Benzene	80.1	2.53
Chloroform	61.2	3.63
Carbon tetrachloride	76.7	5.02
Ethanol	78.4	1.20

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Note

K_b is a property of the solvent only, not the solute. Elevation of BP is generally small — making it less accurate for molecular mass determination. K_b for water is only 0.52 K kg/mol.

Depression of Freezing Point (ΔT_f)

The freezing point of a solution is always lower than that of the pure solvent. At the freezing point, solid and liquid phases are in equilibrium. Adding a solute lowers the vapour pressure, so equilibrium occurs at a lower temperature.

Depression of Freezing Point

ΔT_f = T_f(solvent) − T_f(solution) = K_f × m

Cryoscopic Constant

K_f = R · T²_f · M₁1000 · ΔH_fus

Molecular Mass of Solute

M₂ = K_f × w₂ × 1000ΔT_f × w₁

K_f Values for Common Solvents

Solvent	Normal FP (°C)	K_f (K kg mol⁻¹)
Water	0.0	1.86
Benzene	5.5	5.12
Acetic acid	16.6	3.90
Cyclohexane	6.5	20.0
Camphor	179.8	40.0

Real-World Application: Antifreeze

Ethylene glycol (C₂H₆O₂, M = 62 g/mol) is added to automobile radiators. Adding 620 g (10 mol) to 1 kg water:
m = 10 mol/kg → ΔT_f = 1.86 × 10 = 18.6°C
FP of antifreeze solution = 0 − 18.6 = −18.6°C

🧂

Road Salting

NaCl sprinkled on icy roads dissociates into Na⁺ + Cl⁻ (i=2), effectively doubling the FP depression: ΔT_f = 2 × 1.86 × m. This melts ice at temperatures well below 0°C.

Osmotic Pressure (π)

Osmosis is the spontaneous flow of solvent molecules from a dilute solution (or pure solvent) into a concentrated solution through a semi-permeable membrane. The osmotic pressure is the excess pressure that must be applied to the solution to prevent osmosis.

Van't Hoff Equation

π = C·R·T = n₂V·R·T

where C = concentration (mol/L), R = 0.0821 L·atm·K⁻¹·mol⁻¹, T = temperature in Kelvin

Molecular Mass from Osmometry

M₂ = w₂ · R · Tπ · V

🔵 Isotonic Solutions

Same osmotic pressure. RBCs remain normal. Normal saline = 0.9% NaCl (isotonic with blood, π ≈ 7.6 atm at 37°C).

🔴 Hypo & Hypertonic

Hypotonic: π_soln < π_cell → Water enters cell → hemolysis.
Hypertonic: π_soln > π_cell → Water leaves cell → crenation.

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Why Osmometry is Best for Large Molecules

For macromolecules like proteins (M ≈ 10⁴–10⁶ g/mol), ΔT_b and ΔT_f are too small to measure accurately. But osmotic pressure can be very large — for a 1 mg/mL solution, π ≈ several mmHg — easily measurable. Hence osmometry is the preferred method for biological macromolecules.

Application: Reverse Osmosis

If external pressure > osmotic pressure, solvent flows from concentrated → dilute (reverse osmosis). This principle is used in water purification plants and desalination. Applied pressure must exceed π of seawater (~30 atm).

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Determination of Molecular Mass

Using colligative properties to find unknown molar masses

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Choosing the Right Method

Each colligative property can give M₂ but accuracy varies: RLVP is simple but needs precise VP measurements; ΔT_b gives small changes (less precise); ΔT_f is more accurate (larger K_f); Osmometry is best for macromolecules. Always prefer FP depression for small molecules and osmometry for large ones.

📗 Worked Example 1 — Using FP Depression

Problem: 1.5 g of an unknown compound dissolved in 75 g benzene (K_f = 5.12 K kg/mol) depresses the freezing point by 0.512 K. Find the molar mass.

1

Write the formula: M₂ = (K_f × w₂ × 1000) / (ΔT_f × w₁)

2

Substitute values: M₂ = (5.12 × 1.5 × 1000) / (0.512 × 75)

3

M₂ = 7680 / 38.4 = 200 g/mol

📗 Worked Example 2 — Using Osmotic Pressure

Problem: 200 cm³ of an aqueous solution containing 1.26 g of haemoglobin exerts an osmotic pressure of 3.6 × 10⁻³ atm at 27°C. Find M of haemoglobin.

1

T = 27 + 273 = 300 K, V = 0.200 L, R = 0.0821 L·atm/mol·K

2

π·V = n₂·R·T → n₂ = πV/RT = (3.6×10⁻³ × 0.200) / (0.0821 × 300)

3

n₂ = 7.2 × 10⁻⁴ / 24.63 = 2.924 × 10⁻⁵ mol

4

M₂ = w₂/n₂ = 1.26 / 2.924 × 10⁻⁵ = ≈ 43,100 g/mol

📗 Worked Example 3 — Using RLVP

Problem: VP of water at 20°C = 17.5 mmHg. VP of a solution of 6 g urea (M=60) in 90 g water = ?

1

n₂ = 6/60 = 0.1 mol; n₁ = 90/18 = 5 mol

2

χ₂ = 0.1/(0.1+5) = 0.1/5.1 = 0.0196

3

RLVP = Δp/p° = χ₂ = 0.0196
Δp = 0.0196 × 17.5 = 0.343 mmHg

4

p = 17.5 − 0.343 = 17.157 mmHg

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Van't Hoff Factor & Abnormal Molar Mass

Accounting for dissociation and association in electrolyte solutions

Definition of Van't Hoff Factor (i)

When electrolytes dissociate or molecules associate in solution, the observed colligative property differs from the calculated value. Van't Hoff introduced factor i to account for this:

Van't Hoff Factor

i = Observed value of colligative propertyCalculated value (if no dissociation/association)

Equivalent Definitions

i = Total particles after dissociationOriginal formula units dissolved

🔸 Dissociation (Electrolytes) → i > 1

Electrolyte	Dissociation	i (complete)
NaCl	Na⁺ + Cl⁻	2
KCl	K⁺ + Cl⁻	2
MgCl₂	Mg²⁺ + 2Cl⁻	3
Na₂SO₄	2Na⁺ + SO₄²⁻	3
AlCl₃	Al³⁺ + 3Cl⁻	4

🔹 Association → i < 1

When molecules associate (form aggregates), number of particles decreases → i < 1

Example: Acetic acid in benzene dimerizes:

2CH₃COOH ⇌ (CH₃COOH)₂

For complete dimerization: i = 0.5

Other examples: Benzoic acid in benzene, HF in many solvents

Modified Colligative Property Formulas

RLVP

Δpp° = i · χ₂

Elevation of BP

ΔT_b = i · K_b · m

Depression of FP

ΔT_f = i · K_f · m

Osmotic Pressure

π = i · C · R · T

Degree of Dissociation & Association

For Electrolyte Dissociation (1 → n ions):

Let α = degree of dissociation. For 1 mole of electrolyte:

Initial: 1 formula unit
At eq: (1−α) + nα = 1 + α(n−1)

α = i − 1n − 1

For Association (n molecules → 1 complex):

Let β = degree of association. For 1 mole of solute:

Initial: 1 mol
At eq: (1−β) + β/n = 1 − β(1−1/n)

β = n(1 − i)n − 1

Abnormal Molar Mass

Due to dissociation or association, the observed (apparent) molar mass differs from the true molar mass:

Apparent Molar Mass

M_apparent = M_truei

• Dissociation: i > 1 → M_apparent < M_true

• Association: i < 1 → M_apparent > M_true

This is why electrolyte solutions show larger-than-expected colligative properties.

🧮 Van't Hoff Factor Calculator

Observed ΔT_f (K)

Molality m (mol/kg)

K_f of Solvent (K kg/mol)

Number of ions (n)

Significance of Van't Hoff Factor

Explains why electrolyte solutions deviate from ideal colligative behaviour
Helps determine degree of dissociation or association
Allows calculation of true molar mass from abnormal molar mass
Essential for designing IV fluids (isotonic solutions need correct osmotic pressure)
Used in pharmaceutical formulations to determine tonicity

📝

Practice Quiz

Test your understanding of Solutions & Colligative Properties

🎯 MCQ — 12 Questions

Question 1 of 12

QUESTION 01

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Formula Cheat Sheet

All key formulas and constants at a glance

Concentration Methods

MolarityM = n₂/V(L)

Molalitym = n₂/W₁(kg)

Mole fractionχ₂ = n₂/(n₁+n₂)

Mass %(w₂/w) × 100

Volume %(V₂/V) × 100

ppm(w₂/w) × 10⁶

Raoult's Law

Partial VP (component A)pA = χA × p°A

Total VP (ideal)P = χA·p°A + χB·p°B

Henry's Lawp = KH × χ

Ideal: ΔHmix= 0

Ideal: ΔVmix= 0

Colligative Properties

RLVPΔp/p° = χ₂

Elevation of BPΔTb = Kb × m

Depression of FPΔTf = Kf × m

Osmotic pressureπ = CRT

With Van't Hoff iProp × i

Molar Mass Formulas

From RLVPM₂=Δp·w₂·M₁/((p°-Δp)·w₁)

From ΔTbM₂=Kb·w₂·1000/(ΔTb·w₁)

From ΔTfM₂=Kf·w₂·1000/(ΔTf·w₁)

From πM₂=w₂RT/(π·V)

Van't Hoff Factor

i (definition)i = obs/calcd

Degree of dissociationα=(i−1)/(n−1)

Degree of associationβ=n(1−i)/(n−1)

Apparent molar massM_app = M_true/i

Dissociation: i vs 1i > 1

Association: i vs 1i < 1

Important Constants

Kf (water)1.86 K kg/mol

Kb (water)0.52 K kg/mol

Kf (benzene)5.12 K kg/mol

Kb (benzene)2.53 K kg/mol

Kf (camphor)40.0 K kg/mol

R0.0821 L·atm/mol·K

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Quick Memory Tricks

BORD: Boiling point elevatiOn, Relative lowering of VP, Depression of FP — all depend on mole fraction / molality
Molality is better for colligative properties as it doesn't change with temperature
Camphor (Kf = 40) is best for precise molecular mass by Rast's method
Osmometry is chosen for proteins and polymers (very large M)
For NaCl: i = 1 + α (for partial dissociation), i → 2 (for complete dissociation)