Examples of Colligative Properties and Raoult's Law

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.