Introduction to Abnormal Molar Masses

  • In chemistry, molar mass is the mass of one mole of a substance.
  • Abnormal molar masses occur when the actual molar mass of a substance deviates from the expected molar mass based on its chemical formula.
  • This can happen due to various reasons such as association, dissociation, or hydration.

Abnormal Molar Masses - Association

  • Association refers to the combination of two or more molecules to form a larger molecule.
  • This leads to higher molar mass than expected.
  • Example: Dissociation of N2O4 into NO2.

Abnormal Molar Masses - Dissociation

  • Dissociation is the opposite of association and refers to the splitting of a compound into its constituent particles.
  • This leads to lower molar mass than expected.
  • Example: Dissociation of glucose (C6H12O6) into carbon dioxide (CO2) and water (H2O).

Abnormal Molar Masses - Hydration

  • Hydration is the process of a substance combining with water molecules.
  • This leads to higher molar mass than expected.
  • Example: Hydration of CuSO4 to form CuSO4·5H2O.

Calculating Abnormal Molar Masses - Association

  • To calculate the abnormal molar mass due to association, we consider the degree of association (n) and the formula weight of the associated molecule (M) as:
    • Abnormal Molar Mass = (n × M) / 1 mole

Calculating Abnormal Molar Masses - Dissociation

  • To calculate the abnormal molar mass due to dissociation, we consider the degree of dissociation (α) and the formula weight of the original compound (M) as:
    • Abnormal Molar Mass = (α × M) / 1 mole

Calculating Abnormal Molar Masses - Hydration

  • To calculate the abnormal molar mass due to hydration, we consider the number of water molecules per formula unit (n) and the formula weight of the hydrated compound (M) as:
    • Abnormal Molar Mass = (n × M) / 1 mole

Example - Abnormal Molar Masses - Association

  • Consider the association of acetic acid (CH3COOH) into dimers (CH3COOH)2 due to hydrogen bonding.
  • Molar Mass of acetic acid (CH3COOH): 60 g/mol
  • Degree of association (n): 2
  • Abnormal Molar Mass = (2 × 60 g/mol) / 1 mole
  • Abnormal Molar Mass = 120 g/mol

Example - Abnormal Molar Masses - Dissociation

  • Consider the dissociation of CaCO3 into CaO and CO2.
  • Molar Mass of CaCO3: 100 g/mol
  • Degree of dissociation (α): 0.7
  • Abnormal Molar Mass = (0.7 × 100 g/mol) / 1 mole
  • Abnormal Molar Mass = 70 g/mol

Example - Abnormal Molar Masses - Hydration

  • Consider the hydration of Na2CO3 to form Na2CO3·10H2O.
  • Molar Mass of Na2CO3·10H2O: 286 g/mol
  • Number of water molecules per formula unit (n): 10
  • Abnormal Molar Mass = (10 × 286 g/mol) / 1 mole
  • Abnormal Molar Mass = 2860 g/mol
  1. Problems with Solution - Abnormal Molar Masses
  • Problem 1:
    • A certain substance has an expected molar mass of 100 g/mol, but its actual molar mass is found to be 150 g/mol. What is the degree of association, assuming it undergoes association?
    • Solution:
      • Abnormal Molar Mass = (n × M) / 1 mole
      • 150 g/mol = (n × 100 g/mol) / 1 mole
      • n = 1.5
  • Problem 2:
    • A compound has a molar mass of 80 g/mol. When heated, it undergoes dissociation with a degree of dissociation of 0.6. Calculate the abnormal molar mass.
    • Solution:
      • Abnormal Molar Mass = (α × M) / 1 mole
      • Abnormal Molar Mass = (0.6 × 80 g/mol) / 1 mole
      • Abnormal Molar Mass = 48 g/mol
  • Problem 3:
    • A hydrated compound has a molar mass of 322 g/mol. It contains 6 water molecules per formula unit. Calculate the abnormal molar mass.
    • Solution:
      • Abnormal Molar Mass = (n × M) / 1 mole
      • Abnormal Molar Mass = (6 × 322 g/mol) / 1 mole
      • Abnormal Molar Mass = 1932 g/mol
  1. Factors Affecting Degree of Association/Dissociation
  • Temperature: Higher temperature promotes dissociation, whereas lower temperature promotes association.
  • Concentration: Higher concentration promotes association, whereas lower concentration promotes dissociation.
  • Presence of Catalyst: Some substances can act as catalysts to enhance or reduce the degree of association/dissociation.
  • Inert Atmosphere: In some cases, the presence of an inert gas can affect the degree of association/dissociation.
  1. Colligative Properties and Abnormal Molar Masses
  • Colligative properties are properties of a solution that depend only on the number of solute particles, not on their nature.
  • Abnormal molar masses can affect colligative properties by altering the effective number of solute particles.
  • Examples of colligative properties: boiling point elevation, freezing point depression, osmotic pressure.
  1. Boiling Point Elevation
  • Boiling point elevation occurs when a solute is added to a pure solvent, increasing the boiling point of the solution.
  • The degree of boiling point elevation depends on the concentration of the solute and the abnormal molar mass.
  • Equation: ΔTb = Kb × m × i
    • ΔTb = boiling point elevation
    • Kb = molal boiling point elevation constant
    • m = molality of the solute
    • i = van’t Hoff factor (number of particles after dissociation/association)
  1. Boiling Point Elevation - Example
  • Question:
    • A solution is prepared by dissolving 10 g of glucose (C6H12O6) in 500 g of water. Given that the molar mass of glucose is 180 g/mol and the molal boiling point elevation constant (Kb) for water is 0.52 °C/m, calculate the boiling point elevation of the solution.
  • Solution:
    • Number of moles of glucose = 10 g / 180 g/mol = 0.0556 mol
    • Molality of the solute (glucose) = 0.0556 mol / 0.5 kg = 0.1112 mol/kg
    • As glucose does not dissociate or associate, i = 1
    • ΔTb = 0.52 °C/m × 0.1112 mol/kg × 1
    • ΔTb = 0.0576 °C
  1. Freezing Point Depression
  • Freezing point depression occurs when a solute is added to a pure solvent, decreasing the freezing point of the solution.
  • The degree of freezing point depression depends on the concentration of the solute and the abnormal molar mass.
  • Equation: ΔTf = Kf × m × i
    • ΔTf = freezing point depression
    • Kf = molal freezing point depression constant
    • m = molality of the solute
    • i = van’t Hoff factor (number of particles after dissociation/association)
  1. Freezing Point Depression - Example
  • Question:
    • A solution is prepared by dissolving 20 g of sodium chloride (NaCl) in 200 g of water. Given that the molar mass of NaCl is 58.5 g/mol and the molal freezing point depression constant (Kf) for water is 1.86 °C/m, calculate the freezing point depression of the solution.
  • Solution:
    • Number of moles of NaCl = 20 g / 58.5 g/mol = 0.342 mol
    • Molality of the solute (NaCl) = 0.342 mol / 0.2 kg = 1.71 mol/kg
    • NaCl dissociates into 2 ions (Na+ and Cl-), so i = 2
    • ΔTf = 1.86 °C/m × 1.71 mol/kg × 2
    • ΔTf = 6.37 °C
  1. Osmotic Pressure
  • Osmotic pressure is the pressure required to prevent the flow of solvent across a semipermeable membrane, caused by differences in solute concentration.
  • Abnormal molar masses affect osmotic pressure by altering the effective number of solute particles.
  • Equation: π = i × n × R × T
    • π = osmotic pressure
    • i = van’t Hoff factor (number of particles after dissociation/association)
    • n = molarity of solute
    • R = ideal gas constant
    • T = temperature in Kelvin
  1. Osmotic Pressure - Example
  • Question:
    • A solution is prepared by dissolving 0.5 mol of glucose (C6H12O6) in 1 L of water at 25 °C. Given that the ideal gas constant (R) is 0.0821 L·atm/(mol·K), calculate the osmotic pressure of the solution.
  • Solution:
    • Molarity of glucose = 0.5 mol / 1 L = 0.5 M
    • Glucose does not dissociate or associate, so i = 1
    • T = 25 °C + 273.15 = 298.15 K
    • π = 1 × 0.5 M × 0.0821 L·atm/(mol·K) × 298.15 K
    • π = 12.2 atm
  1. Conclusion
  • Abnormal molar masses occur when the actual molar mass of a substance deviates from the expected molar mass based on its chemical formula.
  • Factors affecting degree of association/dissociation include temperature, concentration, presence of catalyst, and inert atmosphere.
  • Abnormal molar masses can affect colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure.
  • Understanding abnormal molar masses is important in various fields of chemistry, including solution chemistry and physical chemistry. Sure! Here are slides 21 to 30: ``markdown

Factors Affecting Degree of Association/Dissociation

  • Temperature: Higher temperature promotes dissociation, whereas lower temperature promotes association.
  • Concentration: Higher concentration promotes association, whereas lower concentration promotes dissociation.
  • Presence of Catalyst: Some substances can act as catalysts to enhance or reduce the degree of association/dissociation.
  • Inert Atmosphere: In some cases, the presence of an inert gas can affect the degree of association/dissociation.

Colligative Properties and Abnormal Molar Masses

  • Colligative properties are properties of a solution that depend only on the number of solute particles, not on their nature.
  • Abnormal molar masses can affect colligative properties by altering the effective number of solute particles.
  • Examples of colligative properties: boiling point elevation, freezing point depression, osmotic pressure.

Boiling Point Elevation

  • Boiling point elevation occurs when a solute is added to a pure solvent, increasing the boiling point of the solution.
  • The degree of boiling point elevation depends on the concentration of the solute and the abnormal molar mass.
  • Equation: ΔTb = Kb × m × i
    • ΔTb = boiling point elevation
    • Kb = molal boiling point elevation constant
    • m = molality of the solute
    • i = van’t Hoff factor (number of particles after dissociation/association)

Boiling Point Elevation - Example

  • Question:
    • A solution is prepared by dissolving 10 g of glucose (C6H12O6) in 500 g of water. Given that the molar mass of glucose is 180 g/mol and the molal boiling point elevation constant (Kb) for water is 0.52 °C/m, calculate the boiling point elevation of the solution.
  • Solution:
    • Number of moles of glucose = 10 g / 180 g/mol = 0.0556 mol
    • Molality of the solute (glucose) = 0.0556 mol / 0.5 kg = 0.1112 mol/kg
    • As glucose does not dissociate or associate, i = 1
    • ΔTb = 0.52 °C/m × 0.1112 mol/kg × 1
    • ΔTb = 0.0576 °C

Freezing Point Depression

  • Freezing point depression occurs when a solute is added to a pure solvent, decreasing the freezing point of the solution.
  • The degree of freezing point depression depends on the concentration of the solute and the abnormal molar mass.
  • Equation: ΔTf = Kf × m × i
    • ΔTf = freezing point depression
    • Kf = molal freezing point depression constant
    • m = molality of the solute
    • i = van’t Hoff factor (number of particles after dissociation/association)

Freezing Point Depression - Example

  • Question:
    • A solution is prepared by dissolving 20 g of sodium chloride (NaCl) in 200 g of water. Given that the molar mass of NaCl is 58.5 g/mol and the molal freezing point depression constant (Kf) for water is 1.86 °C/m, calculate the freezing point depression of the solution.
  • Solution:
    • Number of moles of NaCl = 20 g / 58.5 g/mol = 0.342 mol
    • Molality of the solute (NaCl) = 0.342 mol / 0.2 kg = 1.71 mol/kg
    • NaCl dissociates into 2 ions (Na+ and Cl-), so i = 2
    • ΔTf = 1.86 °C/m × 1.71 mol/kg × 2
    • ΔTf = 6.37 °C

Osmotic Pressure

  • Osmotic pressure is the pressure required to prevent the flow of solvent across a semipermeable membrane, caused by differences in solute concentration.
  • Abnormal molar masses affect osmotic pressure by altering the effective number of solute particles.
  • Equation: π = i × n × R × T
    • π = osmotic pressure
    • i = van’t Hoff factor (number of particles after dissociation/association)
    • n = molarity of solute
    • R = ideal gas constant
    • T = temperature in Kelvin

Osmotic Pressure - Example

  • Question:
    • A solution is prepared by dissolving 0.5 mol of glucose (C6H12O6) in 1 L of water at 25 °C. Given that the ideal gas constant (R) is 0.0821 L·atm/(mol·K), calculate the osmotic pressure of the solution.
  • Solution:
    • Molarity of glucose = 0.5 mol / 1 L = 0.5 M
    • Glucose does not dissociate or associate, so i = 1
    • T = 25 °C + 273.15 = 298.15 K
    • π = 1 × 0.5 M × 0.0821 L·atm/(mol·K) × 298.15 K
    • π = 12.2 atm

Conclusion

  • Abnormal molar masses occur when the actual molar mass of a substance deviates from the expected molar mass based on its chemical formula.
  • Factors affecting degree of association/dissociation include temperature, concentration, presence of catalyst, and inert atmosphere.
  • Abnormal molar masses can affect colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure.
  • Understanding abnormal molar masses is important in various fields of chemistry, including solution chemistry and physical chemistry. `` I hope these slides help with your lecture on abnormal molar masses. Let me know if you need any further assistance!