Question

An aqueous solution boils at 106.3 ºC. What is its freezing point? (You will need to look up the two ks for water.) Bonus for the experts: do you have enough information to calculate the vapor pressure at 25 ºC? If not, what else would you need?

Answer 1

use equation for boiling point elevation
∆Tb = Kb x m x i

and this equation for freezing point depression
∆Tf = Kf x m x i

Kb = ebullioscopic constant (aka boiling point elevation constant) for the solvent (water in this case) = 0.512 C/m

Kf = cryoscopic constant (aka freezing point depression constant) for the solvent. = 1.86 C / m for water

m = molality of the solution = moles solute / kg solvent
i = van't hoff factor = # ions produced in solution / molecule solvent

got that?... the idea here is m and i are constant for the solution. the concentration doesn't change. neither does the amount of ions in solution.

so from the equations above...

∆Tb / Kb = m x i = ∆Tf / Kf

rearranging
∆Tf = ∆Tb x Kf / Kb

since the normal boiling point of water = 100 ºC

∆Tf = (106.3 C - 100 C) x (1.86 C/m) / (0.512 C/m)

= 22.88 ºC

since the normal freezing point of water = 0 ºC and ∆Tf = 22.8 ºC,

the depressed freezing point = -22.88 ºC