Colligative Properties Worksheet Answer Key

Colligative Properties Worksheet: A Comprehensive Guide with Answers

Understanding colligative properties is crucial in chemistry, as they describe how the physical properties of a solution differ from those of its pure solvent. This comprehensive guide provides a detailed explanation of colligative properties, along with worked-out examples and answers to common worksheet questions. We'll explore vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure, ensuring you grasp the concepts and can confidently tackle any problem.

Introduction to Colligative Properties

Colligative properties depend solely on the concentration of solute particles in a solution, not on the identity of those particles. This means that whether you dissolve sugar or salt in water, the effect on the colligative properties will be the same if the concentration of solute particles is the same. The four main colligative properties are:

• Vapor Pressure Lowering: The presence of a non-volatile solute lowers the vapor pressure of the solvent.
• Boiling Point Elevation: The boiling point of a solution is higher than that of the pure solvent.
• Freezing Point Depression: The freezing point of a solution is lower than that of the pure solvent.
• Osmotic Pressure: The pressure required to prevent the flow of solvent across a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.

Understanding these properties requires a grasp of several key concepts, including mole fraction, molality, and the van't Hoff factor. We'll delve into these concepts in detail below.

1. Vapor Pressure Lowering

Raoult's Law governs vapor pressure lowering. It states that the partial pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution. Mathematically:

P_solvent = X_solvent * P°_solvent

where:

• P_solvent is the partial pressure of the solvent above the solution
• X_solvent is the mole fraction of the solvent
• P°_solvent is the vapor pressure of the pure solvent

Example: Calculate the vapor pressure of a solution containing 2 moles of glucose (a non-volatile solute) and 8 moles of water at 25°C. The vapor pressure of pure water at 25°C is 23.8 torr.

Solution:

1. Calculate the mole fraction of water: X_water = (moles of water) / (moles of water + moles of glucose) = 8 / (8 + 2) = 0.8

2. Apply Raoult's Law: P_water = X_water * P°_water = 0.8 * 23.8 torr = 19.04 torr

Therefore, the vapor pressure of the solution is 19.04 torr.

2. Boiling Point Elevation

Adding a non-volatile solute to a solvent elevates its boiling point. This elevation is directly proportional to the molality (moles of solute per kilogram of solvent) of the solution. The equation is:

ΔT_b = K_b * m * i

where:

• ΔT_b is the boiling point elevation
• K_b is the ebullioscopic constant (a solvent-specific constant)
• m is the molality of the solution
• i is the van't Hoff factor (accounts for the number of particles a solute dissociates into; i=1 for non-electrolytes)

Example: Calculate the boiling point of a solution containing 10 grams of sucrose (C₁₂H₂₂O₁₁) in 100 grams of water. K_b for water is 0.512 °C/m. The molar mass of sucrose is 342.3 g/mol.

Solution:

1. Calculate the molality:

   • Moles of sucrose = 10 g / 342.3 g/mol = 0.0292 moles
   • Molality (m) = 0.0292 moles / 0.1 kg = 0.292 m

2. Apply the boiling point elevation equation: Since sucrose is a non-electrolyte, i = 1. ΔT_b = 0.512 °C/m * 0.292 m * 1 = 0.149 °C

3. Calculate the new boiling point: The boiling point of water is 100 °C. Therefore, the boiling point of the solution is 100 °C + 0.149 °C = 100.149 °C

3. Freezing Point Depression

Similar to boiling point elevation, adding a solute lowers the freezing point of the solvent. The equation is:

ΔT_f = K_f * m * i

where:

• ΔT_f is the freezing point depression
• K_f is the cryoscopic constant (a solvent-specific constant)
• m is the molality of the solution
• i is the van't Hoff factor

Example: Determine the freezing point of a 0.5 m aqueous solution of NaCl. K_f for water is 1.86 °C/m.

Solution:

1. Consider the van't Hoff factor: NaCl dissociates into Na⁺ and Cl⁻ ions, so i = 2.

2. Apply the freezing point depression equation: ΔT_f = 1.86 °C/m * 0.5 m * 2 = 1.86 °C

3. Calculate the new freezing point: The freezing point of water is 0 °C. Therefore, the freezing point of the solution is 0 °C - 1.86 °C = -1.86 °C

4. Osmotic Pressure

Osmotic pressure (π) is the pressure required to stop osmosis. The equation is:

π = iMRT

where:

• π is the osmotic pressure
• i is the van't Hoff factor
• M is the molarity of the solution
• R is the ideal gas constant (0.0821 L·atm/mol·K)
• T is the temperature in Kelvin

Example: Calculate the osmotic pressure of a 0.1 M solution of glucose (a non-electrolyte) at 25°C.

Solution:

1. Convert temperature to Kelvin: T = 25°C + 273.15 = 298.15 K

2. Apply the osmotic pressure equation: Since glucose is a non-electrolyte, i = 1. π = 1 * 0.1 M * 0.0821 L·atm/mol·K * 298.15 K = 2.45 atm

Worksheet Problems and Answers

Here are some practice problems, followed by detailed solutions:

Problem 1: A solution is prepared by dissolving 15.0 g of urea, CO(NH₂)₂, in 120.0 g of water. The vapor pressure of pure water at 25°C is 23.8 torr. What is the vapor pressure of the solution? (Molar mass of urea = 60.06 g/mol)

Answer 1:

1. Calculate moles of urea: 15.0 g / 60.06 g/mol = 0.2497 moles

2. Calculate moles of water: 120.0 g / 18.02 g/mol = 6.66 moles

3. Calculate mole fraction of water: 6.66 / (6.66 + 0.2497) = 0.9636

4. Apply Raoult's Law: Vapor pressure of solution = 0.9636 * 23.8 torr = 22.9 torr

Problem 2: What is the boiling point elevation of a solution prepared by dissolving 5.00 g of NaCl in 100.0 g of water? K_b for water is 0.512 °C/m.

Answer 2:

1. Calculate moles of NaCl: 5.00 g / 58.44 g/mol = 0.0855 moles

2. Calculate molality: 0.0855 moles / 0.1 kg = 0.855 m

3. Apply boiling point elevation equation: ΔT_b = 0.512 °C/m * 0.855 m * 2 (i=2 for NaCl) = 0.876 °C

4. Boiling point of solution: 100 °C + 0.876 °C = 100.876 °C

Problem 3: A solution is made by dissolving 10.0 g of glucose (C₆H₁₂O₆) in 100.0 g of water. What is the freezing point of this solution? K_f for water is 1.86 °C/m.

Answer 3:

1. Calculate moles of glucose: 10.0 g / 180.16 g/mol = 0.0555 moles

2. Calculate molality: 0.0555 moles / 0.1 kg = 0.555 m

3. Apply freezing point depression equation: ΔT_f = 1.86 °C/m * 0.555 m * 1 (i=1 for glucose) = 1.03 °C

4. Freezing point of solution: 0 °C - 1.03 °C = -1.03 °C

Problem 4: Calculate the osmotic pressure of a solution containing 0.200 moles of sucrose in 500 mL of water at 20°C.

Answer 4:

1. Calculate molarity: 0.200 moles / 0.500 L = 0.400 M

2. Convert temperature to Kelvin: T = 20°C + 273.15 = 293.15 K

3. Apply osmotic pressure equation: π = 1 * 0.400 M * 0.0821 L·atm/mol·K * 293.15 K = 9.63 atm

Frequently Asked Questions (FAQ)

• What are some real-world applications of colligative properties? Colligative properties are essential in various applications, including antifreeze in cars (freezing point depression), intravenous solutions in medicine (osmotic pressure), and desalination processes (osmotic pressure).

• Why is the van't Hoff factor important? The van't Hoff factor accounts for the dissociation of ionic compounds in solution. It corrects for the fact that one mole of an ionic compound can produce more than one mole of particles in solution.

• What are some limitations of colligative property calculations? These calculations assume ideal solutions, which means there are no significant interactions between solute and solvent molecules. In reality, deviations from ideality can occur, especially at high concentrations.

• How does the nature of the solute affect colligative properties? The identity of the solute does not affect the magnitude of the colligative properties. Only the number of solute particles matters. However, the type of solute (electrolyte vs. non-electrolyte) influences the van't Hoff factor, thus indirectly impacting the magnitude of the colligative effect.

Conclusion

Colligative properties are a fundamental aspect of solution chemistry. Understanding Raoult's Law and the equations for boiling point elevation, freezing point depression, and osmotic pressure is crucial for solving various problems. By mastering these concepts and practicing with example problems, you'll build a solid foundation in this important area of chemistry. Remember to always consider the van't Hoff factor when dealing with ionic compounds. This comprehensive guide, along with the solved problems, provides a strong starting point for your journey into the world of colligative properties. Continue practicing and you will confidently navigate the complexities of solution chemistry.