Determine the boiling point of 8.6 grams of magnesium chloride in a water solution using 541 g of water .

Answer 1

Colligative properties:

Boiling point elevation... ∆T = imK

∆T = change in boiling point temperature = ?

i = van't Hoff factor = 3 for MgCl2 as it dissociates into 3 ions (1 Mg2+ and 2Cl-)

m = molality = moles solute/kg solvent = 8.6 g MgCl2 x 1 mol MgCl2/95.2 g = 0.090 moles/0.541 kg = 0.167 m

K = boiling point constant for water = 0.512º/m

Solving for ∆T we have...

∆T = (3)(0.167)(0.512) = 0.256ºC

The new boiling point would be 100º + 0.256 = 100.3º