Question

What is the boiling point and freezing point of a 2.21m AlCl3 aqueous solution? (the kf=1.86C/m and kb=0.512C/m)

Answer 1

Colligative properties like changing BP and FP are based on\[\Delta T{_b} = i*K{_b}*m\] which will tell you how much the BP changes, but not what it changes to. Since the problem says the solution is aqueous, that means the solvent is water, which should boil at 100C. Plug in:\[\Delta T{_b} = (4)*(0.512 C/m)*(2.21m)\]
In this case, the ion-factor, i, is 4, because AlCl3 will dissociate into (approximately) 4 ions per formula unit of solute (1Al+3 and 3Cl-1's). It's probably closer to 3.3-3.5, but without having that # handy, we have to assume complete dissociation. Solve for DTb and add it to the BP of pure water. Solve for DTf the same way, but plug in the Kf instead of Kb, and subtract the DTf from 0C (the freezing point of pure water).

I get a new BP of 104.5C, and a new FP of -16.4C