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
