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6.4 Colligative properties

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By the end of this section, you will be able to:
  • Express concentrations of solution components using mole fraction and molality
  • Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
  • Perform calculations using the mathematical equations that describe these various colligative effects
  • Describe the process of distillation and its practical applications
  • Explain the process of osmosis and describe how it is applied industrially and in nature

The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.

Mole fraction and molality

Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity ( M ) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:

M = mol solute L solution

Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality .

The mole fraction, X , of a component is the ratio of its molar amount to the total number of moles of all solution components:

X A = mol A total mol of all components

Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:

m = mol solute kg solvent

Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.

Calculating mole fraction and molality

The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C 2 H 4 (OH) 2 , in a solution prepared from 2.22 × 10 3 g of ethylene glycol and 2.00 × 10 3 g of water (approximately 2 L of glycol and 2 L of water)?

Solution

(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.

mol C 2 H 4 ( OH ) 2 = 2220 g × 1 mol C 2 H 4 ( OH ) 2 62.07 g C 2 H 4 ( OH ) 2 = 35.8 mol C 2 H 4 ( OH ) 2 mol H 2 O = 2000 g × 1 mol H 2 O 18.02 g H 2 O = 11.1 mol H 2 O X ethylene glycol = 35.8 mol C 2 H 4 ( OH ) 2 ( 35.8 + 11.1 ) mol total = 0.763

Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).

(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).

First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:

2220 g C 2 H 4 ( OH ) 2 ( mol C 2 H 2 ( OH ) 2 62.07 g ) = 35.8 mol C 2 H 4 ( OH ) 2

Then, convert the mass of the water from grams to kilograms:

2000 g H 2 O ( 1 kg 1000 g ) = 2 kg H 2 O

Finally, calculate molarity per its definition:

molality = mol solute kg solvent molality = 35.8 mol C 2 H 4 ( OH ) 2 2 kg H 2 O molality = 17.9 m

Check your learning

What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH 3 , dissolved in 125 g of water?

Answer:

7.14 × 10 −3 ; 0.399 m

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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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