Worksheet

Colligative Properties: Fun Practice Sheet & Answers

Colligative Properties: Fun Practice Sheet & Answers
Colligative Properties Worksheet With Answers

Welcome to our exciting journey through the wonders of colligative properties! These properties, which depend on the number of solute particles in a solution rather than their nature, can make science class not just educational but also incredibly engaging. In this post, we’ll dive deep into what colligative properties are, explore different types, practice with real-world examples, and even tackle some fun practice problems to ensure you're mastering the concept. Let's jump right in!

What Are Colligative Properties?

Thermochemistry

Colligative properties refer to those characteristics of a solution that depend on the concentration of solute particles, not on their identity. Here's a rundown of the key colligative properties:

  • Vapor Pressure Lowering: The addition of a non-volatile solute decreases the vapor pressure of the solvent.
  • Boiling Point Elevation: The boiling point of a solvent increases when a solute is added.
  • Freezing Point Depression: The freezing point of a solvent decreases with the addition of a solute.
  • Osmotic Pressure: This is the pressure needed to stop osmosis through a semipermeable membrane.

Practice Sheet: Let's Dive Into Colligative Properties!

Chemistry

To make learning colligative properties a breeze, let's go through some practical problems together:

Problem 1: Vapor Pressure Lowering

Osmotic Pressure Worksheet With Answers

A solution is prepared by dissolving 25 grams of glucose (C6H12O6) in 100 grams of water. Calculate the vapor pressure of this solution at 25°C, given that the vapor pressure of pure water at this temperature is 23.8 mmHg.

🔬 Note: The molecular weight of glucose is approximately 180 g/mol.

Problem 2: Boiling Point Elevation

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What is the increase in boiling point of water when you dissolve 60 grams of NaCl in 500 grams of water? Use the molal boiling point elevation constant, Kb, of water which is 0.512°C/m.

Problem 3: Freezing Point Depression

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Determine the freezing point of a solution made by mixing 15 grams of ethylene glycol (C2H6O2) with 200 grams of water. The freezing point depression constant, Kf, for water is 1.86°C/m.

Exploring Each Property in Detail

Chemical Reaction

Vapor Pressure Lowering

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The principle behind vapor pressure lowering is Raoult’s Law, which states:

Psolution = Xsolvent * Psolvent

Where:

  • Psolution is the vapor pressure of the solution.
  • Xsolvent is the mole fraction of the solvent.
  • Psolvent is the vapor pressure of the pure solvent.

Boiling Point Elevation

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This property can be quantified using the equation:

ΔTb = i * Kb * m

Where:

  • ΔTb is the boiling point elevation.
  • i is the van’t Hoff factor, reflecting the number of particles the solute dissociates into in solution.
  • Kb is the boiling point elevation constant of the solvent.
  • m is the molality of the solution (moles of solute per kilogram of solvent).

Freezing Point Depression

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The formula for this colligative property is similar but focuses on freezing:

ΔTf = i * Kf * m

With the same variables as for boiling point elevation, only Kf is the freezing point depression constant.

Osmotic Pressure

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Osmotic pressure (π) can be calculated using:

π = i * MRT

Where:

  • M is the molarity of the solute.
  • R is the gas constant (0.0821 L⋅atm/mol⋅K).
  • T is the absolute temperature.

Answers and Explanations

Periodic Trends

Let’s solve the practice problems and understand the answers:

Answer for Problem 1: Vapor Pressure Lowering

Colligative Properties Youtube

Using Raoult’s Law:

Moles of glucose = (25g / 180g/mol) = 0.1389 mol

Moles of water = (100g / 18g/mol) ≈ 5.56 mol

Xsolvent = 5.56 / (5.56 + 0.1389) ≈ 0.976

Vapor Pressure of Solution = 0.976 * 23.8mmHg ≈ 23.24mmHg

Answer for Problem 2: Boiling Point Elevation

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Moles of NaCl = (60g / 58.5g/mol) ≈ 1.026 mol

Molality (m) = 1.026mol / 0.5kg ≈ 2.052m

For NaCl, i ≈ 2

ΔTb = 2 * 0.512°C/m * 2.052m ≈ 2.1°C

Answer for Problem 3: Freezing Point Depression

Solution Colligative Properties Quiz Studypool

Moles of ethylene glycol = (15g / 62g/mol) ≈ 0.2419 mol

Molality (m) = 0.2419mol / 0.2kg = 1.2095m

ΔTf = 1 * 1.86°C/m * 1.2095m ≈ 2.25°C

New Freezing Point = 0°C - 2.25°C = -2.25°C

Recap and Final Thoughts

Colligative Properties

Exploring colligative properties not only deepens our understanding of solution chemistry but also equips us with practical knowledge that spans across various scientific and industrial applications. Whether it’s understanding why adding salt to water speeds up boiling or designing desalination processes, the principles of colligative properties are fundamental. By working through these problems, you’ve taken important steps in mastering these concepts, which are crucial for anyone in the fields of chemistry, biology, and beyond.

What are some real-world applications of colligative properties?

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Colligative properties are used in antifreeze (freezing point depression), food preservation (boiling point elevation), and even in medical treatments like dialysis (osmotic pressure).

How do colligative properties change with different solvents?

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While the concepts remain the same, the constants (like Kb and Kf) vary with the solvent. For instance, the freezing point depression constant for water is different from that of benzene.

Why do colligative properties not depend on the nature of the solute?

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These properties are based on the number of solute particles, not their identity or chemical nature. Hence, what matters is how the solute particles change the solvent’s behavior, not what the solute itself is composed of.

Related Terms:

  • Thermochemistry
  • Chemistry
  • Chemical reaction
  • Periodic trends
  • Colligative properties
  • Osmotic pressure Worksheet with answers

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