How to Determine Molar Mass of a Solute from Relative Lowering in Vapour Pressure?

You can determine the molar mass of a solute from the relative lowering of vapor pressure by utilizing Raoult's Law. Here's how:

Understanding Raoult's Law and Relative Lowering of Vapor Pressure

Raoult's Law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. For a solution containing a non-volatile solute, the presence of the solute lowers the vapor pressure of the solution compared to the pure solvent. The "relative lowering of vapor pressure" is defined as:

(P₀ - Pₛ) / P₀ = χ₂

Where:

• P₀ = Vapor pressure of the pure solvent
• Pₛ = Vapor pressure of the solution
• χ₂ = Mole fraction of the solute in the solution

Steps to Determine Molar Mass

1. Experimental Determination of Vapor Pressures: Experimentally measure the vapor pressure of the pure solvent (P₀) and the vapor pressure of the solution (Pₛ) at a specific temperature.

2. Calculate the Relative Lowering of Vapor Pressure: Calculate (P₀ - Pₛ) / P₀ using the experimentally obtained values.

3. Relate to Mole Fraction of Solute: The relative lowering of vapor pressure is equal to the mole fraction of the solute (χ₂). Therefore:

   (P₀ - Pₛ) / P₀ = χ₂

4. Express Mole Fraction in Terms of Moles and Molar Masses: The mole fraction of the solute (χ₂) can be expressed as:

   χ₂ = n₂ / (n₁ + n₂)

   Where:

   • n₂ = number of moles of solute
   • n₁ = number of moles of solvent

   Often, for dilute solutions, n₂ << n₁, so we can approximate χ₂ ≈ n₂ / n₁.

5. Express Moles in Terms of Mass and Molar Mass: The number of moles of solute (n₂) and solvent (n₁) can be calculated using their respective masses and molar masses:

   n₂ = W₂ / M₂ (where W₂ is the mass of solute and M₂ is the molar mass of solute – this is what we want to find)
   n₁ = W₁ / M₁ (where W₁ is the mass of solvent and M₁ is the molar mass of solvent)

6. Substitute and Solve for Molar Mass of Solute (M₂): Substitute the expressions for n₂ and n₁ into the equation for χ₂ (or the approximation for dilute solutions). You will then have an equation with known values (P₀, Pₛ, W₁, M₁, W₂) and one unknown, M₂. Solve for M₂.

   (P₀ - Pₛ) / P₀ = (W₂ / M₂) / (W₁ / M₁) = (W₂ M₁) / (M₂ W₁)

   Rearranging to solve for M₂:

   M₂ = (W₂ M₁) / (W₁ [(P₀ - Pₛ) / P₀])

Example:

Let's say:

• P₀ (pure water at 25°C) = 23.76 mmHg
• Pₛ (solution of 5g of unknown solute in 100g of water at 25°C) = 23.3 mmHg
• W₂ (mass of solute) = 5 g
• W₁ (mass of solvent, water) = 100 g
• M₁ (molar mass of water) = 18.015 g/mol

Then:

M₂ = (5 g 18.015 g/mol) / (100 g [(23.76 mmHg - 23.3 mmHg) / 23.76 mmHg])
M₂ = (90.075 g²/mol) / (100 g * [0.46/23.76])
M₂ = (90.075 g²/mol) / (1.936 g)
M₂ ≈ 46.53 g/mol

Therefore, the approximate molar mass of the solute is 46.53 g/mol.

Important Considerations:

• Non-Volatile Solute: This method is applicable only when the solute is non-volatile.
• Ideal Solutions: Raoult's Law works best for ideal solutions, where the interactions between solute and solvent molecules are similar to those within the pure substances. Deviations from ideality can lead to inaccuracies in the determined molar mass.
• Dilute Solutions: For concentrated solutions, the approximation n₂ << n₁ may not be valid, and the full equation for mole fraction should be used.
• Units: Ensure consistent units are used throughout the calculations.