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.

A. 5.48 m
B. 11.0 m
C. 1.83 m
D. 6.12 m
E. 2.74 m

delta T = i*Kf*m

Solve for m (i = 3)

To find the concentration of aqueous CaCl2 solution that freezes at -10.2ºC, we can use the formula for freezing point depression:

ΔTf = -Kf * m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant for the solvent (water in this case), and m is the molality of the solute.

We are given the following information:
- ΔTf = -10.2ºC
- Kf (for water) = -1.86ºC/m

We can rearrange the formula to solve for m:

m = ΔTf / -Kf

Plugging in the values we know:

m = -10.2ºC / -1.86ºC/m

m = 5.48 m

Therefore, the concentration of the aqueous CaCl2 solution that freezes at -10.2ºC is 5.48 m.

Hence, the correct answer is A. 5.48 m.