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16.3: Colligative Properties

Difficulty Level: At Grade Created by: CK-12

Lesson Objectives

• Define a colligative property and identify three colligative properties of solutions.
• Differentiate between the effects that an electrolyte has on the colligative properties of a solution compared to a nonelectrolyte.
• Calculate the freezing and boiling points of a solution of known molality.
• Use freezing or boiling point information to calculate the molar mass of an unknown solute.

Lesson Vocabulary

• boiling point elevation
• colligative property
• freezing point depression
• molal boiling-point elevation constant
• molal freezing-point depression constant

Recalling Prior Knowledge

• What is vapor pressure?
• How are melting and boiling points determined from a phase diagram?

Solvents have physical properties that can be affected by the process of dissolving a solute into the solvent. In this lesson, you will learn how the vapor pressure, freezing point, and boiling point of a solvent change when a solution is formed.

Vapor Pressure Lowering

A colligative property is a property of a solution that depends only on the number of solute particles dissolved in the solution and not on their identity. Recall that the vapor pressure of a liquid is determined by how easily its molecules are able to escape the surface of the liquid and enter the gaseous phase. When a liquid evaporates easily, it will have a relatively large number of its molecules in the gas phase and thus will have a high vapor pressure. Liquids that do not evaporate easily have a lower vapor pressure. Figure below shows the surface of a pure solvent compared to a solution. In the picture on the left, the surface is entirely occupied by liquid molecules, some of which will evaporate and form a vapor. On the right, a nonvolatile solute has been dissolved into the solvent. Nonvolatile means that the solute itself has little tendency to evaporate. Because some of the surface is now occupied by solute particles, there is less room for solvent molecules. This results in less solvent being able to evaporate. The addition of a nonvolatile solute results in a lowering of the vapor pressure of the solvent.

The solution on the right has had some of its solvent particles replaced by solute particles. Since the solute particles do not evaporate, the vapor pressure of the solution is lower than that of the pure solvent.

The lowering of the vapor pressure depends on the number of solute particles that have been dissolved. The chemical nature of the solute is not important because the vapor pressure is merely a physical property of the solvent. The only requirement is that the solute does not undergo a chemical reaction with the solvent and does not itself escape into the gaseous phase.

While the chemical nature of the solute is not a factor, it is necessary to take into account whether the solute is an electrolyte or a nonelectrolyte. Recall that ionic compounds are strong electrolytes that dissociate into ions when they dissolve. This results in a larger number of dissolved particles. For example, consider two different solutions of equal concentration. One is made from the ionic compound sodium chloride, while the other is made from the molecular compound glucose. The following equations show what happens when these solutes dissolve.

NaCl(s)Na+(aq)+Cl(aq)\begin{align*}\text{NaCl}{(s)} \rightarrow \text{Na}^+{(aq)}+\text{Cl}^-{(aq)}\end{align*} 2 dissolved particles
C6H12O6(s)C6H12O6(aq)\begin{align*}\text{C}_6\text{H}_{12}\text{O}_{6}{(s)} \rightarrow \text{C}_6\text{H}_{12}\text{O}_{6}{(aq)}\end{align*} 1 dissolved particle

The sodium chloride dissociates into two ions, while the glucose does not dissociate. Therefore, equal concentrations of each solution will result in twice as many dissolved particles in the case of the sodium chloride. The vapor pressure of the solvent in the sodium chloride solution will be lowered twice as much as that of the solvent in the glucose solution.

Freezing Point Depression

Figure below shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the vapor pressure of the solvent, resulting in a lower freezing point for the solution compared to the pure solvent. The freezing point depression is the difference in temperature between the freezing point of a pure solvent and that of a solution. On the graph, the freezing point depression is represented by ΔTf.

The vapor pressure of a solution (blue) is lower than the vapor pressure of a pure solvent (purple). As a result, the freezing point of a solvent decreases when any solute is dissolved into it.

When a pure solvent freezes, its particles become more ordered as the intermolecular forces that operate between the molecules fix each molecule in place. In the case of water, hydrogen bonds become more rigid and generate the hexagonally shaped network of molecules that characterizes the structure of ice. By dissolving a solute into the liquid solvent, this ordering process is disrupted. As a result, more energy must be removed from the solution in order to freeze it, and the freezing point of the solution is lower than that of the pure solvent.

In calculations involving colligative properties, the concentration of the solute will be measured by its molality. Molality is used rather than molarity because the molality of a solution is independent of its temperature. Recall that molality is defined as the moles of solute divided by the kilograms of solvent. Molarity is the moles of solute divided by the volume of the solution. While mass does not depend on temperature, the volume of the solution increases with temperature. As a liquid is heated, its molecules move slightly further apart from one another and the volume slightly increases.

The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation is:

ΔTf = Kf × m

The proportionality constant, Kf, is called the molal freezing-point depression constant. It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of Kf is −1.86 °C/m. Consequently, the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is −1.86°C. Every solvent has a unique molal freezing-point depression constant. A few of these are shown in Table below, along with a related value for the boiling point called Kb, which will be discussed in the next section.

Molal Freezing-Point and Boiling-Point Constants
Solvent Normal freezing point (°C) Molal freezing-point depression constant, Kf (°C/m) Normal boiling point (°C) Molal boiling-point elevation constant, Kb (°C/m)
Acetic acid 16.6 −3.90 117.9 3.07
Camphor 178.8 −39.7 207.4 5.61
Naphthalene 80.2 −6.94 217.7 5.80
Phenol 40.9 −7.40 181.8 3.60
Water 0.00 −1.86 100.00 0.512

Sample Problem 16.6 shows how the freezing point of a solution can be calculated.

Sample Problem 16.6: Freezing Point Depression in a Solution of a Nonelectrolyte

Ethylene glycol (C2H6O2) is a molecular compound that is used in many commercial anti-freezes. An aqueous solution of ethylene glycol is used in vehicle radiators to prevent the water in the radiator from freezing by lowering its freezing point. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water.

Step 1: List the known quantities and plan the problem.

Known

• mass of C2H6O2 = 400. g
• molar mass of C2H6O2 = 62.08 g/mol
• mass of H2O = 500. g = 0.500 kg
• Kf (H2O) = −1.86 °C/m

Unknown

• Tf of solution = ? °C

This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression.

Step 2: Solve.

400. g C2H6O2×1 mol C2H6O262.08 g C2H6O2=6.44 mol C2H6O2\begin{align*}\text{400. g C}_2\text{H}_6\text{O}_2 \times \dfrac{\text{1 mol C}_2\text{H}_6\text{O}_2}{62.08 \ \text{g C}_2\text{H}_6\text{O}_2}=6.44 \ \text{mol C}_2\text{H}_6\text{O}_2\end{align*}
6.44 mol C2H6O20.500 kg H2O=12.9 m C2H6O2\begin{align*}\mathrm{\dfrac{6.44 \ mol \ C_2H_6O_2}{0.500 \ kg \ H_2O}=12.9} \ m \ \mathrm{C_2H_6O_2}\end{align*}
ΔTf = Kf ×m = -1.86°C/m × 12.9 m = -24.0°C
Tf = -24.0°C

The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is −24.0°C.

The freezing point of the water decreases by a large amount, protecting the radiator form damage due to the expansion of water when it freezes. There are three significant figures in the result.

Practice Problem
1. Calculate the freezing point of a solution prepared by dissolving 27.56 g of glucose (C6H12O6) into 125 g of water.

Colligative properties have practical applications, such as the salting of roads in cold-weather climates (Figure below). By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly, making driving safer. Sodium chloride (NaCl) and either calcium chloride (CaCl2) or magnesium chloride (MgCl2) are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but it is less effective because it only dissociates into two ions instead of three.

Workers are pouring salt onto an icy road surface in order to lower the melting point of the ice.

Boiling Point Elevation

Figure below shows again the phase diagram of a solution and the effect that the lowered vapor pressure has on the boiling point of the solution compared to the solvent. In this case, the solution has a higher boiling point than the pure solvent. Since the vapor pressure of the solution is lower, more heat must be supplied to the solution to bring its vapor pressure up to the pressure of the external atmosphere. The boiling point elevation is the difference in temperature between the boiling point of the pure solvent and that of the solution. On the graph, the boiling point elevation is represented by ΔTb.

The lowering of the vapor pressure in a solution causes the boiling point of the solution to be higher than that of the pure solvent.

The magnitude of the boiling point elevation is also directly proportional to the molality of the solution. The equation is:

ΔTb = Kb × m

The proportionality constant, Kb, is called the molal boiling-point elevation constant. It is a constant that is equal to the change in the boiling point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of Kb is 0.512 °C/m. The boiling temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is 100.512°C.

Electrolytes and Colligative Properties

As discussed earlier, ionic compounds are electrolytes that dissociate into two or more ions as they dissolve. This must be taken into account when calculating the freezing and boiling points of electrolyte solutions. Sample Problem 16.7 demonstrates how to calculate the freezing point and boiling point of a solution of calcium chloride. Calcium chloride dissociates into three ions according to the following equation:

CaCl2(s) → Ca2+(aq) + 2Cl-(aq)

The values of the freezing point depression and the boiling point elevation for a solution of CaCl2 will be three times greater than they would be for an equal molality of a nonelectrolyte.

Sample Problem 16.7: Freezing and Boiling Point of an Electrolyte

Determine the freezing and boiling point of a solution prepared by dissolving 82.20 g of calcium chloride in 400. g of water.

Step 1: List the known quantities and plan the problem.

Known

• mass of CaCl2 = 82.20 g
• molar mass of CaCl2 = 110.98 g/mol
• mass of H2O = 400. g = 0.400 kg
• Kf (H2O) = −1.86 °C/m
• Kb (H2O) = 0.512 °C/m
• CaCl2 dissociates into 3 ions

Unknown

• Tf = ? °C
• Tb = ? °C

The moles of CaCl2 is first calculated, followed by the molality of the solution. The freezing and boiling points are then determined, including multiplying by 3 for the three ions.

Step 2: Solve.

82.20 g CaCl2×1 mol CaCl2110.98 g CaCl2=0.7407 mol CaCl2\begin{align*}\mathrm{82.20 \ g \ CaCl_2 \times \dfrac{1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}=0.7407 \ mol \ CaCl_2}\end{align*}
0.7407 mol CaCl20.400 kg H2O=1.85 m CaCl2\begin{align*}\mathrm{\dfrac{0.7407 \ mol \ CaCl_2}{0.400 \ kg \ H_2O}=1.85} \ m \ \mathrm{CaCl_2}\end{align*}
ΔTf = Kf × m × 3 = -1.86°C/m × 1.85 m × 3 = -10.3°C
Tf = -10.3°C
ΔTb = Kb × m × 3 = 0.512°C/m × 1.85m × 3 = 2.84°C
Tb=102.84°C

Since the normal boiling point of water is 100.00°C, the calculated result for ΔTb must be added to 100.00 to find the new boiling point.

Practice Problem
1. Calculate the freezing and boiling points of a solution prepared by dissolving 20.0 g of Al(NO3)3 in 100.0 g of water.

Calculating Molar Mass

In the laboratory, freezing point or boiling point data can be used to determine the molar mass of an unknown solute. The Kf or Kb of the solvent must be known. It also must be known whether the solute is an electrolyte or a nonelectrolyte. If the solvent is an electrolyte, you would need to know the number of ions that are produced when it dissociates.

Sample Problem 16.8: Molar Mass from Freezing Point Depression

38.7 g of a nonelectrolyte is dissolved into 218 g of water. The freezing point of the solution is measured to be −5.53°C. Calculate the molar mass of the solute.

Step 1: List the known quantities and plan the problem.

Known

• ΔTf = −5.53°C
• mass of H2O = 218 g = 0.218 kg
• mass of solute = 38.7 g
• Kf (H2O) = −1.86°C/m

Unknown

• molar mass of solute = ? g/mol

Use the freeing point depression (ΔTf) to calculate the molality of the solution. Then use the molality equation to calculate the moles of solute. Then divide the grams of solute by the moles to determine the molar mass.

Step 2: Solve.

m=ΔTfKf=5.53C1.86C/m=2.97m\begin{align*}m=\mathrm{\dfrac{\Delta T_f}{K_f}=\dfrac{-5.53^\circ C}{-1.86^\circ C/m}=2.97\:}m\end{align*}
mol solute = m × kg H2O = 2.97 m × 0.218 kg = 0.648 mol
38.7 g0.648 mol=59.7 g/mol\begin{align*}\mathrm{\dfrac{38.7 \ g}{0.648 \ mol}=59.7 \ g/mol}\end{align*}

The molar mass of the unknown solute is 59.7 g/mol. Knowing the molar mass is an important step in determining the identity of an unknown substance. A similar problem could be done with the change in boiling point.

Practice Problem
1. 26.80 g of an unknown alkali metal chloride (MCl) is dissolved in 100. g of water. The boiling point of the solution is measured to be 103.7°C. Calculate the molar mass of the compound and determine the identity of M. (Hint: Make sure to account for the fact that MCl is an electrolyte.)

Lesson Summary

• The colligative properties of a solution depend only on the number of dissolved particles and not on the chemical nature of the solute. When a solute dissolves in a solvent, the vapor pressure decreases, the freezing point decreases, and the boiling point increases.
• Electrolytes are compounds that dissociate when they dissolve. Since more particles are produced, the effect on the colligative properties is greater for electrolyte solutions than for nonelectrolyte solutions.
• Freezing points and boiling points can be calculated for solutions of known molality by using the molal freezing-point depression and boiling-point elevation constants.
• The molar mass of an unknown solute can be determined from either its freezing point depression or boiling point elevation.

Lesson Review Questions

Reviewing Concepts

1. State the effect that dissolving a solute has on each of the following physical properties of a solvent.
1. vapor pressure
2. freezing point
3. boiling point
2. Explain why the colligative properties of a solvent are affected more by the dissolving of an electrolyte compared to an equal amount of a nonelectrolyte.
3. Are the values of Kf and Kb dependent on the identity of the solvent, the solute, or both? Explain.

Problems

1. Find the freezing and boiling point of a 0.760 m aqueous solution of a nonelectrolyte.
2. Calculate the freezing and boiling point of a solution prepared by dissolving 54.2 g of the nonelectrolyte urea, CO(NH2)2, in 455 g of water.
3. An aqueous solution of a nonelectrolyte boils at 100.82°C. What is the molality of this solution?
4. A solution is prepared by dissolving 114 grams of NaCl in 475 grams of water. Find the freezing and boiling points of this solution.
5. A solution of KBr is observed to freeze at −3.35°C. What would be the freezing temperature of an equal molality solution of K3PO4? Explain.
6. How many grams of CaCl2 should be added to 1.60 kg of water in order to prepare a solution that boils at 102.58°C?
7. Arrange the following in order from the highest freezing point to the lowest. All solutions are aqueous.
1. 0.35 m Zn(NO3)2
2. pure water
3. 0.30 m Al2(SO4)3
4. 0.60 m CsCl
5. 0.85 m C6H12O6
8. A certain aqueous solution is observed to boil at 104.14°C. What is the freezing point of this solution?
9. Calculate the molar mass of a nonelectrolyte solute if dissolving 4.87 g of it in 80.0 g of water results in a solution that freezes at −2.96°C.

Points to Consider

Many chemical reactions take place in aqueous solutions, and many of those reactions involve electrolytes that dissociate in solution.

• Which types of reactions most often involve ionic substances?
• How can reactions between ionic substances be shown most succinctly?

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