Colligative Properties

Overview

Colligative properties depend only on the number of solute particles, not their identity. These properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

The Four Colligative Properties

Property	Effect	Applications
Vapor Pressure Lowering	Decreases	Humidity control
Boiling Point Elevation	Increases	Antifreeze, cooking
Freezing Point Depression	Decreases	De-icing, antifreeze
Osmotic Pressure	Increases	Dialysis, IV solutions

Van't Hoff Factor (i)

Accounts for dissociation of electrolytes:

$$i = \frac{\text{moles of particles in solution}}{\text{moles of solute dissolved}}$$

Solute	Type	i (ideal)
Glucose	Nonelectrolyte	1
NaCl	Strong electrolyte	2
CaCl₂	Strong electrolyte	3
K₃PO₄	Strong electrolyte	4
Weak acid	Weak electrolyte	1 < i < 2

Vapor Pressure Lowering (Raoult's Law)

$$\Delta P = X_{\text{solute}} \times P°_{\text{solvent}}$$

Or:

$$P_{\text{solution}} = X_{\text{solvent}} \times P°_{\text{solvent}}$$

Where:

• $P°$ = vapor pressure of pure solvent
• $X$ = mole fraction

Example

What is the vapor pressure of a solution with $X_{\text{water}} = 0.90$ if $P°_{\text{water}} = 23.8$ mmHg?

$$P = 0.90 \times 23.8 = 21.4 \text{ mmHg}$$ $$\Delta P = 23.8 - 21.4 = 2.4 \text{ mmHg (lowering)}$$

Boiling Point Elevation

$$\Delta T_b = i \times K_b \times m$$

Where:

• $\Delta T_b$ = boiling point increase
• $K_b$ = ebullioscopic constant (solvent specific)
• $m$ = molality
• $i$ = van't Hoff factor

Common $K_b$ Values

Solvent	$K_b$ (°C/m)	Normal BP (°C)
Water	0.512	100.0
Benzene	2.53	80.1
Ethanol	1.22	78.4

Example

Find the boiling point of 0.50 m NaCl solution.

$$\Delta T_b = i \times K_b \times m = 2 \times 0.512 \times 0.50 = 0.51°C$$ $$\text{BP} = 100.0 + 0.51 = 100.51°C$$

Freezing Point Depression

$$\Delta T_f = i \times K_f \times m$$

Where:

• $\Delta T_f$ = freezing point decrease
• $K_f$ = cryoscopic constant (solvent specific)
• $m$ = molality
• $i$ = van't Hoff factor

Common $K_f$ Values

Solvent	$K_f$ (°C/m)	Normal FP (°C)
Water	1.86	0.0
Benzene	5.12	5.5
Camphor	40.0	176

Example

Find the freezing point of 0.30 m CaCl₂ solution.

$$\Delta T_f = i \times K_f \times m = 3 \times 1.86 \times 0.30 = 1.67°C$$ $$\text{FP} = 0.0 - 1.67 = -1.67°C$$

Osmotic Pressure (π)

$$\pi = iMRT$$

Where:

• $\pi$ = osmotic pressure
• $M$ = molarity
• $R$ = 0.0821 L·atm/(mol·K)
• $T$ = temperature (K)
• $i$ = van't Hoff factor

Example

Find the osmotic pressure of 0.10 M glucose at 25°C.

$$\pi = iMRT = 1 \times 0.10 \times 0.0821 \times 298 = 2.45 \text{ atm}$$

Determining Molar Mass

Colligative properties can be used to find molar mass of unknown solutes.

Using Freezing Point Depression

$$M = \frac{i \times K_f \times \text{mass of solute}}{\Delta T_f \times \text{kg of solvent}}$$

Example

2.00 g of unknown dissolved in 50.0 g water lowers FP by 0.62°C. Find M.

$$m = \frac{\Delta T_f}{i \times K_f} = \frac{0.62}{1 \times 1.86} = 0.333 \text{ mol/kg}$$ $$\text{mol} = 0.333 \times 0.050 = 0.0167 \text{ mol}$$ $$M = \frac{2.00 \text{ g}}{0.0167 \text{ mol}} = 120 \text{ g/mol}$$

Osmosis and Solutions

Types of Solutions

Type	Comparison	Cell Effect
Isotonic	Same concentration	No change
Hypotonic	Lower concentration	Cell swells
Hypertonic	Higher concentration	Cell shrinks

Applications

• IV solutions: Must be isotonic (0.9% NaCl)
• Food preservation: Hypertonic solutions dehydrate bacteria
• Dialysis: Selectively removes waste through semipermeable membrane

Ideal vs Real Solutions

Deviations from Raoult's Law

Positive Deviation:

• Weaker solute-solvent interactions
• Higher vapor pressure than predicted
• Example: Ethanol + Hexane

Negative Deviation:

• Stronger solute-solvent interactions
• Lower vapor pressure than predicted
• Example: Acetone + Chloroform

Summary Formulas

Property	Formula
Vapor Pressure	$P = X \times P°$
BP Elevation	$\Delta T_b = iK_bm$
FP Depression	$\Delta T_f = iK_fm$
Osmotic Pressure	$\pi = iMRT$

All increase with more solute particles (higher concentration or higher i).