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What is the boiling point elevation for a solution with 0.5 moles of a non-volat

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Question: What is the boiling point elevation for a solution with 0.5 moles of a non-volatile solute in 1 kg of water? (Kb for water = 0.512 °C kg/mol) (2022)

Options:

  1. 0.256 °C
  2. 0.512 °C
  3. 1.024 °C
  4. 0.128 °C

Correct Answer: 0.256 °C

Exam Year: 2022

Solution: 

Boiling point elevation (ΔTb) = Kb * m = 0.512 °C kg/mol * 0.5 mol/kg = 0.256 °C.

What is the boiling point elevation for a solution with 0.5 moles of a non-volat

Practice Questions

Q1
What is the boiling point elevation for a solution with 0.5 moles of a non-volatile solute in 1 kg of water? (Kb for water = 0.512 °C kg/mol) (2022)
  1. 0.256 °C
  2. 0.512 °C
  3. 1.024 °C
  4. 0.128 °C

Questions & Step-by-Step Solutions

What is the boiling point elevation for a solution with 0.5 moles of a non-volatile solute in 1 kg of water? (Kb for water = 0.512 °C kg/mol) (2022)
  • Step 1: Identify the formula for boiling point elevation, which is ΔTb = Kb * m.
  • Step 2: Determine the value of Kb for water, which is given as 0.512 °C kg/mol.
  • Step 3: Find the number of moles of the solute, which is given as 0.5 moles.
  • Step 4: Calculate the molality (m) of the solution. Since we have 0.5 moles in 1 kg of water, m = 0.5 mol/kg.
  • Step 5: Substitute the values into the formula: ΔTb = 0.512 °C kg/mol * 0.5 mol/kg.
  • Step 6: Perform the multiplication: 0.512 * 0.5 = 0.256 °C.
  • Step 7: Conclude that the boiling point elevation is 0.256 °C.
  • Boiling Point Elevation – The increase in the boiling point of a solvent when a non-volatile solute is added, calculated using the formula ΔTb = Kb * m.
  • Molality – A measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent.
  • Colligative Properties – Properties that depend on the number of solute particles in a solution, not the identity of the solute.
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