Colligative Properties Worksheet Answers Explained
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Understanding colligative properties is essential for students studying chemistry, especially in areas related to solutions and their behaviors. In this article, we will explore the concept of colligative properties, examine different types, and provide clear explanations of common worksheet answers related to this topic. Let’s dive in! 🧪
What are Colligative Properties?
Colligative properties are physical properties of solutions that depend primarily on the number of solute particles present in a given amount of solvent, rather than the identity of those particles. These properties include:
- Vapor Pressure Lowering
- Boiling Point Elevation
- Freezing Point Depression
- Osmotic Pressure
These properties are critical for understanding how solutions behave when solutes are added, influencing various real-world applications from cooking to biological systems.
The Four Major Colligative Properties
1. Vapor Pressure Lowering 🌧️
When a non-volatile solute is added to a solvent, the vapor pressure of the resulting solution is lower than that of the pure solvent. This occurs because solute particles occupy space at the surface, preventing some solvent molecules from escaping into the vapor phase.
Formula: [ \Delta P = P^0 - P = X_{solute} \cdot P^0 ] Where:
- ( \Delta P ) is the decrease in vapor pressure.
- ( P^0 ) is the vapor pressure of the pure solvent.
- ( P ) is the vapor pressure of the solution.
- ( X_{solute} ) is the mole fraction of the solute.
2. Boiling Point Elevation 🔥
Adding a solute to a solvent raises the boiling point of the solution. The change in boiling point (( \Delta T_b )) is proportional to the molality of the solute.
Formula: [ \Delta T_b = i \cdot K_b \cdot m ] Where:
- ( \Delta T_b ) is the increase in boiling point.
- ( i ) is the van't Hoff factor (number of particles the solute dissociates into).
- ( K_b ) is the ebullioscopic constant.
- ( m ) is the molality of the solution.
3. Freezing Point Depression ❄️
Similarly, the addition of a solute lowers the freezing point of the solvent. This is particularly important in scenarios such as de-icing roads.
Formula: [ \Delta T_f = i \cdot K_f \cdot m ] Where:
- ( \Delta T_f ) is the decrease in freezing point.
- ( K_f ) is the cryoscopic constant of the solvent.
4. Osmotic Pressure 🌊
Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. It is particularly relevant in biological systems, where it influences cell function.
Formula: [ \Pi = i \cdot C \cdot R \cdot T ] Where:
- ( \Pi ) is the osmotic pressure.
- ( C ) is the molar concentration of the solute.
- ( R ) is the ideal gas constant.
- ( T ) is the temperature in Kelvin.
Example Problems and Answers Explained
Let’s consider some example problems that might appear on a worksheet focused on colligative properties, along with their answers and explanations.
Example 1: Vapor Pressure Lowering
Problem: A solution is made by dissolving 1 mole of NaCl in 1000 g of water. Calculate the vapor pressure lowering if the vapor pressure of pure water is 23.8 mmHg.
Answer: To calculate the vapor pressure lowering, we first find the mole fraction of the solute:
- Moles of water: ( \frac{1000 , g}{18 , g/mol} \approx 55.56 , moles )
- Total moles: ( 1 + 55.56 = 56.56 )
- Mole fraction of NaCl: ( X_{NaCl} = \frac{1}{56.56} \approx 0.0177 )
Now, apply the formula:
[ \Delta P = 0.0177 \cdot 23.8 \approx 0.42 , mmHg ]
Example 2: Boiling Point Elevation
Problem: What is the boiling point of a solution made by dissolving 0.5 moles of glucose (C₆H₁₂O₆) in 500 g of water? The ebullioscopic constant ( K_b ) for water is 0.512 °C kg/mol.
Answer: Calculate the molality:
- Moles of water: ( \frac{500 , g}{18 , g/mol} \approx 27.78 , moles )
- Molality (m): ( \frac{0.5}{0.5} = 1 , mol/kg )
Now calculate the boiling point elevation:
[ \Delta T_b = 1 \cdot 0.512 \cdot 1 = 0.512 , °C ]
So, the boiling point of the solution is ( 100 + 0.512 = 100.512 , °C ).
Important Notes
Remember: The van't Hoff factor ( i ) is crucial when determining boiling point elevation and freezing point depression. For ionic compounds like NaCl, ( i = 2 ) since it dissociates into two ions. For non-ionic compounds like glucose, ( i = 1 ).
Summary of Key Formulas
To facilitate your understanding of colligative properties, here’s a summary table of the key formulas involved:
<table> <tr> <th>Property</th> <th>Formula</th> </tr> <tr> <td>Vapor Pressure Lowering</td> <td>ΔP = X<sub>solute</sub> • P<sup>0</sup></td> </tr> <tr> <td>Boiling Point Elevation</td> <td>ΔT<sub>b</sub> = i • K<sub>b</sub> • m</td> </tr> <tr> <td>Freezing Point Depression</td> <td>ΔT<sub>f</sub> = i • K<sub>f</sub> • m</td> </tr> <tr> <td>Osmotic Pressure</td> <td>Π = i • C • R • T</td> </tr> </table>
With a solid understanding of these properties and how to apply the formulas correctly, you'll be well-prepared for any chemistry challenge related to solutions! Keep practicing, and don't hesitate to refer to your notes when needed. Happy studying! 📚