Vapour Pressure of Solutions of Solids in Liquids
- Introduction to the topic
- Definition of vapour pressure
- Explanation of solutions of solids in liquids
Raoult’s Law
- Statement of Raoult’s Law
- Explanation of ideal solutions
- Derivation of Raoult’s Law equation
Raoult’s Law Equation
- Presentation of the Raoult’s Law equation:
- P₁ = X₁P₀₁
- P₁: Vapour pressure of component 1 in solution
- X₁: Mole fraction of component 1 in solution
- P₀₁: Vapour pressure of pure component 1
Application of Raoult’s Law - Ideal Solutions
- Explanation of ideal solutions
- Calculation example: Dilute solutions Example:
- Component A mole fraction: 0.2
- Vapour pressure of component A: 50 mmHg
- Calculate the vapour pressure of the solution.
Non-Ideal Solutions
- Explanation of non-ideal solutions
- Departure from Raoult’s Law
- Explanation of positive and negative deviations
Positive Deviation
- Definition of positive deviation
- Explanation of intermolecular forces
- Example of positive deviation and its causes Example:
- Benzene and ethanol mixture
- Explanation of benzene-ethanol intermolecular forces
Negative Deviation
- Definition of negative deviation
- Explanation of intermolecular forces
- Example of negative deviation and its causes Example:
- Acetone and chloroform mixture
- Explanation of acetone-chloroform intermolecular forces
Colligative Properties
- Definition of colligative properties
- Explanation of how they depend on the number of particles
- Brief mention of four colligative properties:
- Vapour pressure lowering
- Boiling point elevation
- Freezing point depression
- Osmotic pressure
Vapour Pressure Lowering
- Explanation of how solutes lower the vapour pressure of solvents
- Explanation of vapor pressure lowering equation
- Example of calculating the decrease in vapour pressure Example:
- Solvent: Water
- Solute: NaCl
- Initial vapour pressure of water: 25 mmHg
- Calculate the final vapour pressure with the addition of NaCl
Boiling Point Elevation
- Explanation of how solutes elevate the boiling point of solvents
- Explanation of boiling point elevation equation
- Example of calculating the increase in boiling point Example:
- Solvent: Water
- Solute: NaCl
- Initial boiling point of water: 100°C
- Calculate the final boiling point with the addition of NaCl
Problems with Solution - Vapour Pressure of Solutions of Solids in Liquids
- Example problem:
- A solution is prepared by dissolving 5 grams of glucose (C6H12O6) in 200 grams of water. The vapour pressure of pure water is 23.8 mmHg at a certain temperature. Calculate the vapour pressure of the solution assuming ideal behaviour.
- Solution:
- Calculate the mole fraction of glucose: Xglucose = moles of glucose / total moles
- Calculate the moles of glucose: moles = mass / molar mass
- Calculate the moles of water: moles = mass / molar mass
- Calculate the total moles: moles of glucose + moles of water
- Calculate the vapour pressure using Raoult’s Law equation: P1 = X1 * P0
- Substitute values to find the solution’s vapour pressure
- Answer: The vapour pressure of the solution is X mmHg.
Problems with Solution - Non-Ideal Solutions
- Example problem:
- A solution is prepared by mixing 20 mL of methanol (CH3OH) with 30 mL of ethanol (C2H5OH). The vapour pressure of pure methanol is 93.0 mmHg and the vapour pressure of pure ethanol is 44.0 mmHg at the same temperature. Calculate the vapour pressure of the solution assuming ideal behaviour.
- Solution:
- Calculate the mole fraction of methanol: Xmethanol = moles of methanol / total moles
- Calculate the moles of methanol: moles = volume * density / molar mass
- Calculate the moles of ethanol: moles = volume * density / molar mass
- Calculate the total moles: moles of methanol + moles of ethanol
- Calculate the vapour pressure using Raoult’s Law equation: P1 = X1 * P0
- Substitute values to find the solution’s vapour pressure
- Answer: The vapour pressure of the solution is X mmHg.
Colligative Properties - Freezing Point Depression
- Explanation of how solutes affect the freezing point of solvents
- Explanation of freezing point depression equation
- Example of calculating the decrease in freezing point Example:
- Solvent: Water
- Solute: Sodium chloride (NaCl)
- Initial freezing point of water: 0°C
- Calculate the final freezing point with the addition of NaCl
- Answer: The freezing point of the solution is X°C.
Colligative Properties - Osmotic Pressure
- Explanation of osmotic pressure and its importance in biological systems
- Explanation of how solutes affect osmotic pressure
- Explanation of the van’t Hoff equation for osmotic pressure: π = iMRT
- π: Osmotic pressure
- i: Van’t Hoff factor (number of particles per formula unit)
- M: Molarity of the solute
- R: Gas constant
- T: Temperature in Kelvin
Osmotic Pressure Calculation
- Example problem:
- A solution is prepared by dissolving 0.5 moles of glucose (C6H12O6) in 2 liters of water at 25°C. Calculate the osmotic pressure of the solution.
- Solution:
- Calculate the van’t Hoff factor (i) for glucose
- Calculate the molarity of the solution: Molarity = moles of solute / volume of solution
- Convert temperature to Kelvin: T(Kelvin) = 25 + 273
- Substitute values into the van’t Hoff equation to calculate osmotic pressure
- Answer: The osmotic pressure of the solution is X atm.
Problems with Solution - Vapour Pressure Lowering
- Example problem:
- A solution is prepared using 10 grams of sucrose (C12H22O11) in 500 mL of water. The vapour pressure of pure water is 23.8 mmHg at a certain temperature. Calculate the vapour pressure of the solution assuming ideal behaviour.
- Solution:
- Calculate the mole fraction of sucrose: Xsucrose = moles of sucrose / total moles
- Calculate the moles of sucrose: moles = mass / molar mass
- Calculate the moles of water: moles = volume * density / molar mass
- Calculate the total moles: moles of sucrose + moles of water
- Calculate the vapour pressure using Raoult’s Law equation: P1 = X1 * P0
- Substitute values to find the solution’s vapour pressure
- Answer: The vapour pressure of the solution is X mmHg.
Colligative Properties - Boiling Point Elevation
- Explanation of how solutes affect the boiling point of solvents
- Explanation of boiling point elevation equation
- Example of calculating the increase in boiling point Example:
- Solvent: Water
- Solute: Sodium chloride (NaCl)
- Initial boiling point of water: 100°C
- Calculate the final boiling point with the addition of NaCl
- Answer: The boiling point of the solution is X°C.
Colligative Properties - Reverse Osmosis
- Explanation of reverse osmosis and its applications
- Difference between osmosis and reverse osmosis
- Explanation of pressure required for reverse osmosis
- Example of reverse osmosis in water purification
Applications of Colligative Properties
- Introduction to various applications of colligative properties in everyday life and industries
- Mention of applications such as antifreeze, food preservation, inkjet printing, and more
- Brief explanation of how each application utilizes the colligative properties
Summary and Conclusion
- Recap of key concepts covered in the lecture:
- Raoult’s Law and ideal solutions
- Positive and negative deviations from Raoult’s Law
- Vapour pressure lowering
- Boiling point elevation
- Freezing point depression
- Osmotic pressure
- Reverse osmosis
- Importance of understanding these concepts in chemistry and related fields
- Encouragement for further exploration and study in the topic
Applications of Colligative Properties - Antifreeze
- Explanation of how antifreeze works
- Antifreeze lowers the freezing point of the coolant in a car’s radiator
- This prevents the coolant from freezing in cold temperatures and damaging the engine
- Example: Ethylene glycol used as antifreeze in automobiles
Applications of Colligative Properties - Food Preservation
- Explanation of how colligative properties are used in food preservation
- Addition of solutes (such as salt or sugar) to foods lowers the water activity, inhibiting microbial growth
- Explanation of the process of osmosis in food preservation
- Example: Canning, salting, and sugaring methods
Applications of Colligative Properties - Inkjet Printing
- Explanation of how colligative properties are used in inkjet printing
- The ink used in inkjet printers has a specific viscosity and surface tension to ensure printing accuracy
- Addition of solutes or adjusting the solvent composition alters the colligative properties of the ink
- Example: Glycol, glycerol, and surfactants used in ink formulations
Applications of Colligative Properties - Osmotic Pressure in Biology
- Explanation of how osmotic pressure is crucial for biological systems
- Cells maintain osmotic balance through osmosis
- Osmotic pressure helps in water uptake, nutrient absorption, and waste removal in cells
- Example: Plant cells taking up water from the soil through osmosis
Applications of Colligative Properties - Reverse Osmosis in Water Purification
- Explanation of reverse osmosis as a water purification process
- High pressure is applied to force water molecules through a semipermeable membrane, leaving impurities behind
- Reverse osmosis removes contaminants such as salts, bacteria, and organic matter from water
- Example: Desalination of seawater to obtain freshwater
Summary and Conclusion
- Recap of key concepts covered in the lecture:
- Vapour pressure of solutions of solids in liquids
- Raoult’s Law and ideal solutions
- Positive and negative deviations from Raoult’s Law
- Colligative properties: vapour pressure lowering, boiling point elevation, freezing point depression, osmotic pressure, reverse osmosis
- Applications of colligative properties in various fields
- Emphasis on the importance of understanding these concepts for 12th Boards exam and beyond
- Encouragement for further practice and exploration in the topic
Vapour Pressure of Solutions of Solids in Liquids Introduction to the topic Definition of vapour pressure Explanation of solutions of solids in liquids