Relative Lowering of Vapour Pressure

  • Definition: Relative lowering of vapour pressure is the ratio of the lowering of vapour pressure of the solvent to the vapour pressure of the pure solvent at the same temperature.
  • Symbol: It is denoted by the symbol ‘ΔP’.
  • Formula: ΔP = (P° - P) / P°, where P° is the vapour pressure of the pure solvent and P is the vapour pressure of the solution.
  • Unit: Relative lowering of vapour pressure is a dimensionless quantity. Example:
  • Consider a solution of glucose in water. The vapour pressure of pure water at a particular temperature is 18 mm Hg. After adding glucose to the water, the vapour pressure of the solution becomes 17 mm Hg. Calculate the relative lowering of vapour pressure. Solution:
  • Given: P° = 18 mm Hg, P = 17 mm Hg
  • Using the formula ΔP = (P° - P) / P°, we can calculate the relative lowering of vapour pressure.
  • ΔP = (18 - 17) / 18
  • ΔP = 1 / 18
  • ΔP = 0.0556 (rounded to four decimal places) Equation:
  • ΔP = (P° - P) / P°

Factors Affecting Relative Lowering of Vapour Pressure

  1. Nature of solute: The relative lowering of vapour pressure depends on the nature of the solute. Different solutes have different effects on the vapour pressure of the solvent.
  1. Concentration: The relative lowering of vapour pressure increases with an increase in the concentration of the solute in the solution.
  1. Temperature: The relative lowering of vapour pressure is affected by temperature. It generally increases with an increase in temperature.
  1. Nature of solvent: The relative lowering of vapour pressure also depends on the nature of the solvent. Different solvents have different vapour pressures and can exhibit different degrees of lowering. Example:
  • Consider two solutions, one with a high concentration of solute and the other with a low concentration of solute. Which solution will exhibit a higher relative lowering of vapour pressure? Solution:
  • The solution with a high concentration of solute will exhibit a higher relative lowering of vapour pressure. The relative lowering of vapour pressure increases with an increase in the concentration of the solute in the solution. Equation:
  • ΔP = (P° - P) / P°
  1. Problems with Solution - Relative Lowering of Vapour Pressure
  • Problem 1: A solution is prepared by dissolving 10 g of glucose (C6H12O6) in 50 g of water. Calculate the relative lowering of vapour pressure of the solution at 25°C. Given: Vapor pressure of water at 25°C is 23.8 mmHg.
    • Solution:
      • Moles of glucose = 10 g / molar mass of glucose
      • Moles of water = 50 g / molar mass of water
      • ΔP = (n_glucose / n_water) x (P°_water)
      • Substitute the values to find the relative lowering of vapour pressure.
  • Problem 2: Calculate the relative lowering of vapour pressure when 2.5 g of a non-volatile solute is dissolved in 50 g benzene. Given: Vapour pressure of pure benzene is 98.1 mm Hg.
    • Solution:
      • Moles of solute = 2.5 g / molar mass of solute
      • Moles of benzene = 50 g / molar mass of benzene
      • ΔP = (n_solute / n_benzene) x (P°_benzene)
      • Substitute the values to find the relative lowering of vapour pressure.
  • Problem 3: A solution is prepared by dissolving 4 g of a non-volatile solute in 32 g of ether. The vapour pressure of ether is 375 mm Hg. What is the relative lowering of vapour pressure?
    • Solution:
      • Moles of solute = 4 g / molar mass of solute
      • Moles of ether = 32 g / molar mass of ether
      • ΔP = (n_solute / n_ether) x (P°_ether)
      • Substitute the values to find the relative lowering of vapour pressure.
  1. Colligative Properties
  • Definition: Colligative properties are properties of solutions that depend on the concentration of the solute particles, rather than the nature of the solute itself.
  • Examples of colligative properties: Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
  • These properties arise due to the presence of solute particles in a solvent and do not depend on the specific chemical nature of the solute.
  • Colligative properties are important in various industries, such as food preservation, pharmaceuticals, and chemical manufacturing.
  1. Elevation of Boiling Point
  • Definition: Elevation of boiling point is the increase in the boiling point of a solvent due to the presence of a non-volatile solute.
  • Explanation: When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases. As a result, the boiling point of the solution increases.
  • Formula: ΔTb = Kbm, where ΔTb is the elevation in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
  • Unit: The unit of elevation of boiling point is °C or K. Example:
  • Calculate the increase in boiling point when 10 g of glucose (C6H12O6) is dissolved in 100 g of water. Given: Kbm for water is 0.512°C/m.
    • Solution:
      • Moles of glucose = 10 g / molar mass of glucose
      • Moles of water = 100 g / molar mass of water
      • Molality (m) = (moles of solute) / (mass of solvent in kg)
      • Substitute the values in the formula ΔTb = Kbm to find the increase in boiling point.
  1. Depression of Freezing Point
  • Definition: Depression of freezing point is the decrease in the freezing point of a solvent due to the presence of a non-volatile solute.
  • Explanation: When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases. As a result, the freezing point of the solution decreases.
  • Formula: ΔTf = Kfm, where ΔTf is the depression in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solution.
  • Unit: The unit of depression of freezing point is °C or K. Example:
  • Calculate the decrease in freezing point when 20 g of NaCl is dissolved in 200 g of water. Given: Kfm for water is 1.86°C/m.
    • Solution:
      • Moles of NaCl = 20 g / molar mass of NaCl
      • Moles of water = 200 g / molar mass of water
      • Molality (m) = (moles of solute) / (mass of solvent in kg)
      • Substitute the values in the formula ΔTf = Kfm to find the decrease in freezing point.
  1. Osmotic Pressure
  • Definition: Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane, when a solution and pure solvent are separated by the membrane.
  • Explanation: Osmosis is the spontaneous movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semipermeable membrane.
  • Osmotic pressure is directly proportional to the concentration of the solute particles in the solution.
  • Formula: π = n/V, where π is the osmotic pressure, n is the number of moles of solute, and V is the volume of the solution.
  • Unit: The unit of osmotic pressure is Pascal (Pa) or atmosphere (atm). Example:
  • Calculate the osmotic pressure exerted by a solution containing 0.5 moles of glucose (C6H12O6) in 2 L of water.
    • Solution:
      • Osmotic pressure (π) = (number of moles of solute) / (volume of the solution)
      • Substitute the values to find the osmotic pressure.
  1. Van’t Hoff Factor
  • Definition: Van’t Hoff factor (i) is the ratio of the moles of solute particles formed in a solution to the moles of formula units dissolved.
  • Explanation: In an ideal solution, the van’t Hoff factor is equal to the number of particles formed when a solute dissociates or associates in solution.
  • Van’t Hoff factor is used to correct the colligative properties calculations for solutions that undergo dissociation or association.
  • Example: For a solution of NaCl, the van’t Hoff factor is 2, as NaCl dissociates into Na+ and Cl- ions.
  1. Van’t Hoff Factor for Dissociation
  • Example: Consider a solution of NaCl (sodium chloride) in water. When NaCl dissolves in water, it dissociates into Na+ and Cl- ions.
  • The van’t Hoff factor for NaCl is 2, as it forms two ions (Na+ and Cl-) in solution.
  • Similarly, other ionic compounds, such as KCl, CaCl2, and Al2(SO4)3, also have van’t Hoff factors greater than 1.
  1. Van’t Hoff Factor for Association
  • Example: Consider a solution of acetic acid (CH3COOH) in water. When acetic acid dissolves in water, it associates to form CH3COO- ions.
  • The van’t Hoff factor for acetic acid is less than 1, as it forms fewer ions in solution compared to the number of dissolved molecules.
  1. Calculation of Van’t Hoff Factor
  • For ionic compounds that dissociate in solution, the van’t Hoff factor can be determined by counting the number of ions formed from each formula unit.
  • For molecular compounds that associate in solution, the van’t Hoff factor is less than 1 and depends on the degree of association.
  • The van’t Hoff factor is used to correct the colligative properties calculations for solutions undergoing dissociation or association.
  1. Summary of Colligative Properties
  • Colligative properties are properties of solutions that depend on the concentration of the solute particles.
  • Examples of colligative properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
  • These properties arise due to the presence of solute particles and are independent of the chemical nature of the solute.
  • The van’t Hoff factor is used to correct colligative properties calculations for solutions that undergo dissociation or association.
  1. Problems with Solution - Relative Lowering of Vapour Pressure
  • Problem 1: A solution is prepared by dissolving 10 g of glucose (C6H12O6) in 50 g of water. Calculate the relative lowering of vapour pressure of the solution at 25°C. Given: Vapor pressure of water at 25°C is 23.8 mmHg.
    • Solution:
      • Moles of glucose = 10 g / molar mass of glucose
      • Moles of water = 50 g / molar mass of water
      • ΔP = (n_glucose / n_water) x (P°_water)
      • Substitute the values to find the relative lowering of vapour pressure.
  • Problem 2: Calculate the relative lowering of vapour pressure when 2.5 g of a non-volatile solute is dissolved in 50 g benzene. Given: Vapour pressure of pure benzene is 98.1 mm Hg.
    • Solution:
      • Moles of solute = 2.5 g / molar mass of solute
      • Moles of benzene = 50 g / molar mass of benzene
      • ΔP = (n_solute / n_benzene) x (P°_benzene)
      • Substitute the values to find the relative lowering of vapour pressure.
  • Problem 3: A solution is prepared by dissolving 4 g of a non-volatile solute in 32 g of ether. The vapour pressure of ether is 375 mm Hg. What is the relative lowering of vapour pressure?
    • Solution:
      • Moles of solute = 4 g / molar mass of solute
      • Moles of ether = 32 g / molar mass of ether
      • ΔP = (n_solute / n_ether) x (P°_ether)
      • Substitute the values to find the relative lowering of vapour pressure.
  1. Colligative Properties
  • Definition: Colligative properties are properties of solutions that depend on the concentration of the solute particles, rather than the nature of the solute itself.
  • Examples of colligative properties: Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
  • These properties arise due to the presence of solute particles in a solvent and do not depend on the specific chemical nature of the solute.
  • Colligative properties are important in various industries, such as food preservation, pharmaceuticals, and chemical manufacturing.
  1. Elevation of Boiling Point
  • Definition: Elevation of boiling point is the increase in the boiling point of a solvent due to the presence of a non-volatile solute.
  • Explanation: When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases. As a result, the boiling point of the solution increases.
  • Formula: ΔTb = Kbm, where ΔTb is the elevation in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
  • Unit: The unit of elevation of boiling point is °C or K.
  1. Depression of Freezing Point
  • Definition: Depression of freezing point is the decrease in the freezing point of a solvent due to the presence of a non-volatile solute.
  • Explanation: When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases. As a result, the freezing point of the solution decreases.
  • Formula: ΔTf = Kfm, where ΔTf is the depression in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solution.
  • Unit: The unit of depression of freezing point is °C or K.
  1. Osmotic Pressure
  • Definition: Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane, when a solution and pure solvent are separated by the membrane.
  • Explanation: Osmosis is the spontaneous movement of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semipermeable membrane.
  • Osmotic pressure is directly proportional to the concentration of the solute particles in the solution.
  • Formula: π = n/V, where π is the osmotic pressure, n is the number of moles of solute, and V is the volume of the solution.
  • Unit: The unit of osmotic pressure is Pascal (Pa) or atmosphere (atm).
  1. Van’t Hoff Factor
  • Definition: Van’t Hoff factor (i) is the ratio of the moles of solute particles formed in a solution to the moles of formula units dissolved.
  • Explanation: In an ideal solution, the van’t Hoff factor is equal to the number of particles formed when a solute dissociates or associates in solution.
  • Van’t Hoff factor is used to correct the colligative properties calculations for solutions that undergo dissociation or association.
  • Example: For a solution of NaCl, the van’t Hoff factor is 2, as NaCl dissociates into Na+ and Cl- ions.
  1. Van’t Hoff Factor for Dissociation
  • Example: Consider a solution of NaCl (sodium chloride) in water. When NaCl dissolves in water, it dissociates into Na+ and Cl- ions.
  • The van’t Hoff factor for NaCl is 2, as it forms two ions (Na+ and Cl-) in solution.
  • Similarly, other ionic compounds, such as KCl, CaCl2, and Al2(SO4)3, also have van’t Hoff factors greater than 1.
  1. Van’t Hoff Factor for Association
  • Example: Consider a solution of acetic acid (CH3COOH) in water. When acetic acid dissolves in water, it associates to form CH3COO- ions.
  • The van’t Hoff factor for acetic acid is less than 1, as it forms fewer ions in solution compared to the number of dissolved molecules.
  1. Calculation of Van’t Hoff Factor
  • For ionic compounds that dissociate in solution, the van’t Hoff factor can be determined by counting the number of ions formed from each formula unit.
  • For molecular compounds that associate in solution, the van’t Hoff factor is less than 1 and depends on the degree of association.
  • The van’t Hoff factor is used to correct the colligative properties calculations for solutions undergoing dissociation or association.
  1. Summary of Colligative Properties
  • Colligative properties are properties of solutions that depend on the concentration of the solute particles.
  • Examples of colligative properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
  • These properties arise due to the presence of solute particles and are independent of the chemical nature of the solute.
  • The van’t Hoff factor is used to correct colligative properties calculations for solutions that undergo dissociation or association.