What concentration of aqueous CaCl2 solution freezes at -10.2ºC? The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m.
delta T = i*Kf*m
You know delta T, i is the van't Hoff factor (3 for CaCl2; i.e., the number of particles), Kf you have, solve for m.
To determine the concentration of the aqueous CaCl2 solution that freezes at -10.2ºC, we can use the equation:
ΔT = Kf * m
where ΔT is the change in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solution.
Given that the freezing point of pure water is 0.0ºC and the Kf of pure water is -1.86ºC/m, we can calculate the change in freezing point (ΔT):
ΔT = freezing point of solution - freezing point of pure water
ΔT = -10.2ºC - 0.0ºC
ΔT = -10.2ºC
Now, we can rearrange the equation to solve for m:
m = ΔT / Kf
m = -10.2ºC / (-1.86ºC/m)
m ≈ 5.48 mol/kg
Therefore, the concentration of the aqueous CaCl2 solution that freezes at -10.2ºC is approximately 5.48 mol/kg.