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

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)

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.

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