Understanding Colligative Properties

Oct 13, 2025 by magicalceylontours.com 31 views

Solutions Worksheet #3 (Colligative Properties)

Understanding Colligative Properties

Colligative properties are a set of properties that depend on the concentration of solute particles in a solution, rather than the identity of the solute. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. In this worksheet, we will focus on the first two colligative properties: boiling point elevation and freezing point depression.

Boiling Point Elevation

The boiling point of a solvent is the temperature at which the vapor pressure of the solvent equals the surrounding atmospheric pressure. When a solute is added to a solvent, it increases the boiling point of the solvent. This is known as boiling point elevation. The magnitude of boiling point elevation depends on the concentration of the solute particles in the solution.

Freezing Point Depression

The freezing point of a solvent is the temperature at which the solvent changes from a liquid to a solid. When a solute is added to a solvent, it decreases the freezing point of the solvent. This is known as freezing point depression. The magnitude of freezing point depression also depends on the concentration of the solute particles in the solution.

Colligative Properties and the Number of Solute Particles

Colligative properties depend on the number of solute particles in a solution, rather than the identity of the solute. This means that the magnitude of colligative properties will be the same for different solutes, as long as the concentration of solute particles is the same.

Solutions Worksheet #3 (Colligative Properties) - Problems

Solvent	Boiling Point (°C)	Freezing Point (°C)
Water	100	0
Benzene	80.1	5.5

Problem 1

A solution of sugar in water has a boiling point of 101.5°C. What is the molality of the solution?

Problem 2

A solution of salt in water has a freezing point of -1.5°C. What is the molality of the solution?

Problem 3

A solution of ethanol in benzene has a boiling point of 84.5°C. What is the molality of the solution?

Problem 4

A solution of glucose in water has a freezing point of -0.5°C. What is the molality of the solution?

Problem 5

A solution of sucrose in water has a boiling point of 102.5°C. What is the molality of the solution?

Problem 6

A solution of sodium chloride in water has a freezing point of -2.5°C. What is the molality of the solution?

Problem 7

A solution of acetone in benzene has a boiling point of 83.5°C. What is the molality of the solution?

Problem 8

A solution of fructose in water has a freezing point of -0.2°C. What is the molality of the solution?

Problem 9

A solution of glycerol in water has a boiling point of 103.5°C. What is the molality of the solution?

Problem 10

A solution of urea in water has a freezing point of -1.2°C. What is the molality of the solution?

Solutions to Problems

Problem 1

A solution of sugar in water has a boiling point of 101.5°C. To find the molality of the solution, we can use the formula:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTb / Kb

The boiling point elevation constant for water is 0.51 K kg/mol. Plugging in the values, we get:

m = (101.5 - 100) / 0.51 m = 1.0 mol/kg

Problem 2

A solution of salt in water has a freezing point of -1.5°C. To find the molality of the solution, we can use the formula:

ΔTf = Kf × m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTf / Kf

The freezing point depression constant for water is 1.86 K kg/mol. Plugging in the values, we get:

m = (-1.5 - 0) / 1.86 m = 0.8 mol/kg

Problem 3

A solution of ethanol in benzene has a boiling point of 84.5°C. To find the molality of the solution, we can use the formula:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTb / Kb

The boiling point elevation constant for benzene is 2.53 K kg/mol. Plugging in the values, we get:

m = (84.5 - 80.1) / 2.53 m = 1.4 mol/kg

Problem 4

A solution of glucose in water has a freezing point of -0.5°C. To find the molality of the solution, we can use the formula:

ΔTf = Kf × m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTf / Kf

The freezing point depression constant for water is 1.86 K kg/mol. Plugging in the values, we get:

m = (-0.5 - 0) / 1.86 m = 0.3 mol/kg

Problem 5

A solution of sucrose in water has a boiling point of 102.5°C. To find the molality of the solution, we can use the formula:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTb / Kb

The boiling point elevation constant for water is 0.51 K kg/mol. Plugging in the values, we get:

m = (102.5 - 100) / 0.51 m = 2.0 mol/kg

Problem 6

A solution of sodium chloride in water has a freezing point of -2.5°C. To find the molality of the solution, we can use the formula:

ΔTf = Kf × m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTf / Kf

The freezing point depression constant for water is 1.86 K kg/mol. Plugging in the values, we get:

m = (-2.5 - 0) / 1.86 m = 1.4 mol/kg

Problem 7

A solution of acetone in benzene has a boiling point of 83.5°C. To find the molality of the solution, we can use the formula:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTb / Kb

The boiling point elevation constant for benzene is 2.53 K kg/mol. Plugging in the values, we get:

m = (83.5 - 80.1) / 2.53 m = 1.2 mol/kg

Problem 8

A solution of fructose in water has a freezing point of -0.2°C. To find the molality of the solution, we can use the formula:

ΔTf = Kf × m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTf / Kf

The freezing point depression constant for water is 1.86 K kg/mol. Plugging in the values, we get:

m = (-0.2 - 0) / 1.86 m = 0.1 mol/kg

Problem 9

A solution of glycerol in water has a boiling point of 103.5°C. To find the molality of the solution, we can use the formula:

ΔTb = Kb × m

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution.

Rearranging the formula to solve for m, we get:

m = ΔTb / Kb

The boiling point elevation constant for water is 0.51 K kg/mol. Plugging in the values, we get:

m = (103.5 - 100) / 0.51 m = 3.2 mol/kg

Problem 10

A solution of ure