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.