Boiling Point from Freezing point?

Question

An aqueous solution freezes at -2.47 degrees C. What is its boiling point?

I need help with this question, i feel like it's missing information..?

Answer 1

No you are not missing anything.

The solvent is water thus the freezing point of the solvent is 0 degC and the boiling point 100 deg C (for P=1 atm)
Thus ΔΤf= Tf0-Tf= 0-Tf =-Tf (1)
and ΔΤb=Tb-Tb0 =Tb-100=> Tb=100+ΔΤb (2)

ΔΤf= Kf*i*m (3)
ΔΤb= Kb*i*m (4)

where i is the van't Hoff coefficient and m the molality.
The solution is the same so i and m are the same. Thus when we divide equations (4) and (3) i and m are simplified and we get
ΔΤb / ΔΤf =Kb/Kf =>
ΔΤb= (Kb/Kf)*ΔΤf substitute ΔΤf using (1)
ΔΤb= (Kb/Kf)*(-Tf) (5)
Use (5) to substitute ΔΤb in (2) and you get

Tb= 100+ (Kb/Kf)*(-Tf) =100+ (0.512/1.858)*(-(-2.47)) =100.68 deg C

Answer 2

IMHO, you are missing some information here. There is no way to figure out the boiling point of a substance from it's melting point. There's gotta be more to this question.

Answer 3

im pretty sure the answer is 97.53 degrees C. if the 100 degree difference rule applies here which i think it does because im sure aqueous means water based. dont hunt me down if im wrong though lol

Answer 4

boiling point-the tempature when the liquid boils
freezing point-the tempature when the liquid turns to solid (like freezing water turning into ice)