Question

An unknown solute is dissolved in 800 grams of water. The boiling point of the solution is measured to be 101.8°C, or 1.8°C greater than that of pure water. What can be determined from this data and from known constants for water?

Answer 1

The boiling point of water = 100°C

The boiling point of the solution = 101.8°C

Elevation in boiling point = 1.8°C

Elevation of boiling point, ΔTb = Kb. m

where  ΔTb = Tb (solution) - Tb (pure solvent), m is the molality of the solution and Kb is the ebullioscopic constant of water

Kb of water = 0.512°C kg/mol

As we know the Kb of water and the ΔTb we can find the molality of the solution,

ΔTb = Kb. m

1.8°C = 0.512°C kg/mol x m

m = 1.8°C / 0.512°C kg/mol = 3.52 mol/kg  

Molality of the solution = 3.52 mol/kg

Mass of water = 800 g = 0.8 kg

Molality of the solution =  moles of solute / Kilograms of solvent

Moles of solute = Molality of the solution  x Kilograms of solvent

                         = 3.52 mol/kg x 0.8 kg = 2.816 mol