Question

what is the expected boiling point of a solution made by dissolving 15.65g of glycerol in 49.00mL of water? the boiling point elevation constant for water if 0.512 degrees C/ molal. 

Answer 1

given problem is based on the colligative property, elevation in boiling point.

∆Tb = i × kb × m

∆Tb is elevation in boiling point, i is vanthoff factor for glycerol i = 1 ,

Kb = elevation constant = 0.512 deg C / molal

m is molality of solution,

molality = number of moles of gycerol ÷ mass of water (kg)

Number of moles of glycerol = mass ÷ molarmass = 15.65 g ÷ 90 g /mol = 0.170 moles

Mass of water = volume × density = 49 ml × 1 g/ml = 49 g = 0.049 kg

Molality = 0.170moles ÷ 0.049 kg = 3.469 molal

Now , ∆ Tb = 1 × 0.512 deg°C/ molal × 3.469 molal = 1.78 °C

∆ Tb = 1.78 °C = boiling point of water after addition of solute - boiling point of water water

1.78°C = T f - 100°C

Tf = 101.78 °c