Colligative Properties for AP Chemistry

Practice problems for these concepts can be found at:

• Solutions and Colligative Properties Multiple Choice Review Questions for AP Chemistry
• Solutions and Colligative Properties Free-Response Questions for AP Chemistry

Some of the properties of solutions depend on the chemical and physical nature of the individual solute. The blue color of a copper(II) sulfate solution and the sweetness of a sucrose solution are related to the properties of those solutes. However, some solution properties simply depend on the number of solute particles, not the type of solute. These properties are called colligative properties and include:

• vapor pressure lowering
• freezing-point depression
• boiling-point elevation
• osmotic pressure

Vapor Pressure Lowering

If a liquid is placed in a sealed container, molecules will evaporate from the surface of the liquid and eventually establish a gas phase over the liquid that is in equilibrium with the liquid phase. The pressure generated by this gas is the vapor pressure of the liquid. Vapor pressure is temperature-dependent; the higher the temperature, the higher the vapor pressure. If the liquid is made a solvent by adding a nonvolatile solute, the vapor pressure of the resulting solution is always less than that of the pure liquid. The vapor pressure has been lowered by the addition of the solute; the amount of lowering is proportional to the number of solute particles added and is thus a colligative property.

Solute particles are evenly distributed throughout a solution, even at the surface. Thus, there are fewer solvent particles at the gas–liquid interface where evaporation takes place. Fewer solvent particles escape into the gas phase, and so the vapor pressure is lower. The higher the concentration of solute particles, the less solvent is at the interface and the lower the vapor pressure. This relationship is referred to as Raoult's law.

Freezing-Point Depression

The freezing point of a solution of a nonvolatile solute is always lower than the freezing point of the pure solvent. It is the number of solute particles that determines the amount of the lowering of the freezing point. The amount of lowering of the freezing point is proportional to the molality of the solute and is given by the equation

ΔT_f = iK_f molality

where ΔT_f is the number of degrees that the freezing point has been lowered (the difference in the freezing point of the pure solvent and the solution); K_f is the freezing-point depression constant (a constant of the individual solvent); the molality is the molality of the solute; and i is the van't Hoff factor—the ratio of the number of moles of particles released into solution per mole of solute dissolved. For a nonelectrolyte, such as sucrose, the van't Hoff factor would be 1. For an electrolyte, such as sodium chloride, you must take into consideration that if 1 mol of NaCl dissolves, 2 mol of particles would result (1 mol Na⁺, 1 mol Cl^–). Therefore, the van't Hoff factor should be 2. However, because sometimes there is a pairing of ions in solution, the observed van't Hoff factor is slightly less (for example, it is 1.9 for a 0.05 m NaCl solution). The more dilute the solution, the closer the observed van't Hoff factor should be to the expected factor. If you can calculate the molality of the solution, you can also calculate the freezing point of the solution.

Let's learn to apply the preceding equation. Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.

The most common mistake is to forget to subtract the ΔT value from the normal freezing point.

The freezing-point depression technique is also commonly used to calculate the molar mass of a solute.

mass of a solute. For example, a solution is prepared by dissolving 0.490 g of an unknown compound in 50.00 mL of water. The freezing point of the solution is –0.201°C. Assuming the compound is a nonelectrolyte, what is the molecular mass of the compound? Use 1.00 g/mL as the density of water.

m = ΔT/K_f = 0.201 K/(1.86 K kg mol^–1) = 0.108 mol/kg

50.00 mL (1.00 g/mL) (1 kg/1000 g) = 0.0500 kg

(0.108 mol/kg) (0.0500 kg) = 0.00540 mol

0.490 g/0.00540 mol = 90.7 g/mol

Many students make the mistake of stopping before they complete this problem.