A 29.300% by mass solution of Strontium Fluoride will freeze at what temperature? Kf water is 1.86 C/m.?
- ChemTeamLv 73 months ago
We need the molality of the SrF2 solution. To do that, we first assume 100 g of the solution is present.
29.300%(w/w) means 29.3 g of SrF2 in the 100 g of solution as well as 70.7 g of water.
Since molality involves moles of solute, we do this:
29.3 g / 125.62 g/mol = 0.233243 mol
Now, for the molality:
0.233243 mol / 0.0707 kg = 3.29905 m
Now, the freezing point calculation:
Δt = (3) (1.86) (3.29905) <--- the 3 is the van 't Hoff factor
Δt = 18.408699
The solution will freeze at -18.4 C