Question

The boiling point elevation constant for water is 0.51 degrees C/m. What is the boiling point of a solution made by dissolving 21.6g of a non-ionizing solute with a molar mass of 103.5 g/mol in 187g of water?

Answer 1

Answer –

We are given, boiling point elevation constant for water, Kb = 0.51^oC/m

Mass of solute = 21.6 g , molar mass = 103.5 g/mol

Mass of solvent, water = 187 g = 0.187 kg

Now we need to first calculate moles of solute from given mass

We know formula,

Moles = given mass / molar mass

          = 21.6 g / 103.5 g.mol^-1

         = 0.209 moles

Now we ned to calculate molality and we know formula,

Molality= moles of solute / kg of solvent

              = 0.209 mol / 0.187 kg

              = 1.12 m

Now we need to calculate boiling point elevation, ΔTb

ΔTb = Kb * m

       = 0.51^oC/m * 1.12 m

        = 0.569^oC

We know, ΔTb = boiling point of solution – boiling point of pure solvent

boiling point of pure solvent = 100.0 ^oC

so, boiling point of solution = ΔTb + boiling point of pure solvent

                                        = 0.569^oC + 100.0 ^oC

                                        = 100.57^oC