Calculate the boiling point elevation of a solution containing 43.8 g of glucose (C6H12O6) dissolved in 359.9 g of water. Calculate the freezing point depression for the same solution.

Question

boiling point elevation = _____ degrees Celsius

freezing point depression = ____ degrees Celsius

Answer 1

Boiling point elevation and freezing point depression are both colligative properties.

∆T = imK

∆T = change in temperature

i = van't Hoff factor = 1 for glucose since it does not ionize or dissociate. It is a single particle.

m = molality = moles solut/kg solvent = 43.8g/180 g/mol = 0.243 mol / 0.3599 kg = 0.676 m

K = boiling or freezing constant = 0.512 and 1.86 respectively

Boiling point elevation:

∆T = (1)(0.676)(0.512) = 0.35º

Freezing point depression:

∆T = (1)(0.676)(1.86) = 1.26º