Examples of Colligative Properties and Raoult's Law

In this article, we will explore the concept of colligative properties and how Raoult's Law applies to vapor pressure calculations. We'll provide clear examples to help you grasp these fundamental concepts in chemistry.
By Jamie

Understanding Colligative Properties and Raoult’s Law

Colligative properties are properties of solutions that depend on the number of solute particles in a given amount of solvent, rather than the identity of those solute particles. Raoult’s Law is a key principle that helps us understand how the addition of a solute affects the vapor pressure of a solvent.

What is Raoult’s Law?

Raoult’s Law states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent. Mathematically, it can be expressed as:

\[ P_{solution} = X_{solvent} \times P^0_{solvent} \]

Where:

  • \( P_{solution} \) is the vapor pressure of the solution.
  • \( X_{solvent} \) is the mole fraction of the solvent.
  • \( P^0_{solvent} \) is the vapor pressure of the pure solvent.

Example 1: Calculating Vapor Pressure Reduction

Scenario:

You have a solution made by dissolving 1 mole of sodium chloride (NaCl) in 3 moles of water (H₂O). The vapor pressure of pure water at the given temperature is 23.8 mmHg.

Step-by-Step Calculation:

  1. Determine the total number of moles in the solution:

    • Moles of water = 3 moles
    • Moles of sodium chloride = 1 mole (NaCl dissociates into two ions: Na⁺ and Cl⁻, so total ions = 2 moles)
    • Total moles = 3 moles (H₂O) + 2 moles (Na⁺ and Cl⁻) = 5 moles

  2. Calculate the mole fraction of the solvent (water):
    \[ X_{H2O} = \frac{3 \text{ moles of H₂O}}{5 \text{ total moles}} = 0.6 \]

  3. Apply Raoult’s Law:
    \[ P_{solution} = X_{H2O} \times P^0_{H2O} = 0.6 \times 23.8 \text{ mmHg} = 14.28 \text{ mmHg} \]

  4. Calculate the vapor pressure reduction:

    • Vapor pressure reduction = \( P^0_{H2O} - P_{solution} = 23.8 \text{ mmHg} - 14.28 \text{ mmHg} = 9.52 \text{ mmHg} \)

Conclusion:

The addition of sodium chloride decreased the vapor pressure of water by 9.52 mmHg.

Example 2: Boiling Point Elevation

Scenario:

You have a solution of 1 mole of glucose (C₆H₁₂O₆) dissolved in 4 moles of water. The boiling point elevation constant (K_b) for water is 0.512 °C/m.

Step-by-Step Calculation:

  1. Determine the total moles:

    • Moles of water = 4 moles
    • Moles of glucose = 1 mole
    • Total moles = 4 + 1 = 5 moles

  2. Calculate the mole fraction of the solute (glucose):
    \[ X_{glucose} = \frac{1 \text{ mole}}{5 \text{ total moles}} = 0.2 \]

  3. Calculate the boiling point elevation (ΔT_b):
    \[ \Delta T_b = i \times K_b \times m \]

    • For glucose, the van ‘t Hoff factor (i) = 1 (does not dissociate)
    • Molarity (m) = 1 mole of glucose / 1 kg of water (assuming density of water is 1 kg/L) = 1 m
    • \[ \Delta T_b = 1 \times 0.512 \times 1 = 0.512 \text{ °C} \]

  4. Calculate the new boiling point:

    • New boiling point = 100 °C + 0.512 °C = 100.512 °C

Conclusion:

The addition of glucose raised the boiling point of the water by 0.512 °C, resulting in a new boiling point of 100.512 °C.

Summary

Colligative properties, such as vapor pressure reduction and boiling point elevation, are crucial for understanding the behavior of solutions. Raoult’s Law provides a method to calculate these properties based on the concentration of solutes, highlighting the importance of the number of particles over their identity.