>> View the other Key Equations and Concepts in this chapter

Vapor Pressure of Solutions

As discussed in Chapter 5, the addition of a solute modifies the physical properties of the solvent. Chapter 5 discusses the colligative properties of boiling point elevation and freezing point depression. Addition of a solute also modifies the vapor pressure of the solvent. This is also a colligative property, because it depends on the number of particles, not the identity of the particles.

The vapor pressure of a solution is calculated with Raoult's law

P_solution = X_solventP_solvent       (Equation 9.3)

X_solvent represents the mole fraction of the solvent. See Chapter 8 to review calculating mole fractions. P_solvent is the vapor pressure of the solvent. If the solvent is water, the vapor pressure at various temperatures can be found in Table 8.2.

>> Example 1

What is the vapor pressure of a mixture of 0.127 mol sugar and 100.0 g water at 25 °C?

Solution:

Since this refers to solution, where water is the solvent and sugar is the solute, Raoult's law must be used (Psolution = X_solventP_solvent). P_solvent is the vapor pressure of pure water. This value at 25 °C is 28.8 torr, according to Table 8.2. The mole fraction will be moles water/(moles water + moles sugar). According to the problem, there are 0.127 moles of sugar. Moles of water can be calculated from molar mass:

100.0 g water

1 mol 18.01 g

=

5.552 mol

So the mole fraction

X

=

5.552 5.552 + 0.127

=

0.9776

The four significant figures are correct, since the addition in the denominator gives a sum with four significant figures. Remember that the mole fraction does not have units.

P_solution = X_solventP_solvent = (0.9776)(28.8 torr) = 28.2 torr

The answer is limited to three significant figures by the vapor pressure. The vapor pressure of water is in torr (mm Hg), so the final answer has the same units. There is very little change from the vapor pressure of pure water; this is typical. If you do get a very low value for vapor pressure of a solution, check that you have not calculated the mole fraction of solute instead of the mole fraction of solvent.

If the solute is an ionic compound in water, you must use the moles of ions rather than the moles of compound as the moles of solute in the calculation of mole fraction. Recall that colligative properties depend on the number of particles in solution. When an ionic compound dissolves in water, the particles are ions, not the formula unit.

>> Example 2

What is the vapor pressure of a mixture of 3.44 g FeF₂ in 75.00 g water at 20 °C?

Solution:

The vapor pressure of water at 20 °C is 17.5 torr. Since FeF₂ is a soluble, ionic compound, when 1 mole of the compound dissolves, it makes 1 mole of Fe²⁺ and 2 moles of F^–. So the mole fraction is

Xwater

=

moles water moles water + moles ferrous ion + moles fluoride ion

mol water

=

75.00 g water

1 mol 18.01 g

=

4.164 mol

mol FeF2

=

3.44 g

1 mol FeF2 93.85 g

=

0.0367 mol

0.0367 mol FeF2

1 mol Fe2+ 1 mol FeF2

=

0.0367 mol Fe2+

0.0367 mol FeF2

2 mol F– 1 mol FeF2

=

0.0734 mol F–

X

=

4.164 4.164 + 0.0367 + 0.0734

=

4.164 4.274

=

0.9742

P_solution = (0.9742)(17.5) = 17.0 torr

>> View the other Key Equations and Concepts in this chapter