Question

What is the boiling point of a solution made using 759 g of sucrose, c12h22o11, in 0.300 kg of water, h2o?

Answer 1

According to the boiling point elevation formula:

ΔTb = i * Kb * m

- when i is the van't Hoff factor for sucrose = 1 

as the sucrose is an organic molecule so, there is no dissociation in water so i = 1
- Kb is the ebullioscopic constant of the solvent (water) = 0.512°C Kg mol-1

 - and when m is the molality = moles sucrose / Kg water

so, we need first to get moles sucrose = mass/molar mass

                                                                 = 759 g / 342.3 g/mol

                                                                 = 2.22 moles 

∴ m = 2.22 moles / 0.3 Kg

       = 7.4 mol/kg

by substitution on ΔTb:

∴ΔTb = 1 * 0.512 * 7.4 

          = 3.8 

∴Tb = 100 °C + 3.8 = 103.8 °C