Learning Outcomes

At the end of this section, you should be able to:

  • Define the colligative properties of solutions
  • Explain the Lowering of vapour pressure, depression of freezing, elevation of ‬boiling point and osmotic pressure as colligative properties.
  • Differentiate between ideal and non-ideal solutions and state Raoult’s Law.
  • Determine the molar mass using osmotic pressure. (The derivation is not required).

Definition
Colligative properties are properties of solutions that depend on the concentration of solute molecules or ions but not upon the identity of the solute.

Vapour Pressure Lowering

What is Vapour Pressure?

Vapour pressure is the pressure of a solvent’s vapour in equilibrium with its liquid phase.

It occurs when molecules escape from the liquid to form vapour.

Lowering the Vapour Pressure

The vapour pressure of a solution is lower than that of a pure solvent. This is because solute particles reduce the number of solvent molecules escaping into the vapour phase.

When non-volatile solute particles are introduced into a solvent, they occupy space on the surface where solvent molecules typically evaporate. This leads to a decrease in the quantity of solvent molecules on the surface, lowering the speed at which they can transition into the vapour phase. As a result, the solvent’s vapour pressure decreases.

Raoult’s Law

Raoult’s Law is fundamental in understanding how solute particles affect the vapour pressure of a solvent. This law is directly related to the colligative properties of solutions.


Raoult's Law
Raoult’s Law states that the vapour pressure of a solvent in a solution (P1) is proportional to the mole fraction of the solvent (X1) in the solution and the vapour pressure of the pure solvent (P1°).

P1 = X1 × P1°

  • P1: Vapour pressure of the solvent in the solution
  • X1: Mole fraction of the solvent in the solution
  • P1°: Vapour pressure of the pure solvent

Mole Fraction

Mole fraction is the ratio of the number of moles of a given component to the total number of moles of solution.

For a two-component solution, where na and nb represent the number of moles of the two components,

Mole Fraction

Mole fraction of component A = X_a = \frac{n_a}{n_a+n_b}

Example: 30 g of glucose is dissolved in 500 ml water at 25°C. What is the vapour pressure of the solution? (The vapour pressure of water is 23.8 torr at 25°C, and the density of water is 1 g/mL.)

Solution:

As you can see, the vapour of the solution is slightly lower than the vapour pressure of the pure solvent.

  1.  

Boiling Point Elevation

Boiling Point
The boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid is equals the external atmospheric pressure. At this temperature, the liquid solvent turns into vapour.

How Adding a Solute Elevates the Boiling Point

When a solute is added to a solvent, the solution exhibits a phenomenon called boiling point elevation. This happens because solute particles disrupt solvent molecules, requiring more heat to boil.

Mechanism of Boiling Point Elevation

  1. Lower Vapour Pressure:
    • Adding a non-volatile solute to a solvent decreases the vapour pressure of the solvent.
    • This is because solute particles occupy space at the surface, reducing the number of solvent molecules that can escape into the vapour phase.
  2. Increased Boiling Temperature:
    • To reach boiling, the vapour pressure of the solution must equal the external pressure.
    • Since the presence of solute particles lowers the vapour pressure, the solution must be heated to a higher temperature to achieve the necessary vapour pressure for boiling.

Freezing Point Depression

Freezing Point
The freezing point of a solvent is the temperature at which the liquid phase and the solid phase of the solvent are in equilibrium. At this temperature, the solvent transitions from liquid to solid.

How Adding a Solute Depresses the Freezing Point

When a solute is added to a solvent, the solution exhibits a phenomenon called freezing point depression. This effect occurs because the solute particles disrupt the orderly structure needed for the solvent to freeze.

Mechanism of Freezing Point Depression

  1. Disruption of Crystalline Structure:
    • Pure solvents freeze by forming a crystalline structure.
    • Solute particles interfere with the formation of this structure, making it more difficult for the solvent to solidify.
  2. Lowering the Freezing Temperature:
    • Because the solute particles disrupt the freezing process, the solution must be cooled to a lower temperature to achieve the same degree of crystallization.
    • This means the freezing point of the solution is lower than that of the pure solvent.

Osmotic Pressure

Osmotic Pressure
The Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane from a dilute solution into a concentrated solution. This flow, known as osmosis, occurs naturally to equalize solute concentrations on both sides of the membrane.

How Osmotic Pressure Arises

  1. Semipermeable Membrane:
    • A semipermeable membrane allows only solvent molecules (e.g., water) to pass through, but not solute particles (e.g., salt or sugar).
  2. Concentration Gradient:
    • When two solutions of different concentrations are separated by a semipermeable membrane, solvent molecules move from the region of lower solute concentration to the region of higher solute concentration.
  3. Equilibrium:
    • This movement continues until the solute concentrations on both sides of the membrane are equal, or until an external pressure (osmotic pressure) is applied to stop the flow of solvent molecules.

Determining Molar Mass from Osmotic Pressure

Osmotic pressure is a useful tool for characterizing solutions and determining molar masses, similar to other colligative properties. However, osmotic pressure is especially advantageous because even a small concentration of solute generates a relatively large osmotic pressure.

Experiments have shown that the relationship between osmotic pressure and solution concentration can be expressed by the following equation:

 \Pi = MRT
where  \Pi is the osmotic pressure in atmospheres, M is the molarity of the solution, R is
the gas law constant, and T is the Kelvin temperature.

Example:

The osmotic pressure of a solution containing 5.87 mg of an unknown protein per 10.0 mL of solution is 2.45 Torr at 25 °C. Find the molar mass of the unknown protein.


Ideal and Nonideal Solutions

Ideal Solutions:

  • Obey Raoult’s Law over the entire range of concentration.
  • The interactions between solute and solvent molecules are similar to those between solvent-solvent and solute-solute molecules.
  • No heat is absorbed or evolved when mixing the solute and solvent.
  • Volume of mixing is zero (no expansion or contraction upon mixing).

Nonideal Solutions:

  • Do not obey Raoult’s Law over the entire range of concentration.
  • The interactions between solute and solvent molecules are different from those between solvent-solvent and solute-solute molecules.
  • Heat is absorbed or evolved when mixing the solute and solvent.
  • Volume of mixing is not zero (expansion or contraction occurs upon mixing).

Differences Between Ideal and Nonideal Solutions

FeatureIdeal SolutionsNonideal Solutions
Obedience to Raoult’s LawObeys Raoult’s Law at all concentrationsDeviates from Raoult’s Law
Intermolecular ForcesSimilar between solute-solvent, solvent-solvent, and solute-soluteDifferent between solute-solvent compared to solvent-solvent and solute-solute
Heat of MixingZero (no heat absorbed or evolved)Non-zero (heat is absorbed or evolved)
Volume Change on MixingZero (no expansion or contraction)Non-zero (expansion or contraction occurs)
ExamplesBenzene and tolueneWater and ethanol, chloroform and acetone

Explanation with Examples

  • Ideal Solution Example: Benzene and toluene
  • Both are nonpolar hydrocarbons with similar molecular structures and sizes.
  • Their interactions (van der Waals forces) are similar, leading to ideal behaviour.
  • Nonideal Solution Example: Water and ethanol
  • Water molecules form strong hydrogen bonds with each other.
  • When ethanol is added, the interactions between water and ethanol molecules differ significantly from water-water and ethanol-ethanol interactions.
  • This leads to deviations from Raoult’s Law, heat absorption or evolution, and volume changes.