Question 7.C-L.8: Calculate the boiling point, freezing point, and osmotic pre...
Calculate the boiling point, freezing point, and osmotic pressure of the following solutions. Assume that the osmotic pressure is measured at 27°C.
a. A 0.100 M solution of CaCl_2 in water. The CaCl_2 is a strong electrolyte.
b. A 0.100 M solution of ethylene glycol in water. Ethylene glycol does not dissociate in solution.
a. Because CaCl_2 is a strong electrolyte, it dissociates completely:
CaCl_2 → Ca^{2+} + 2Cl^{-}. Thus, n = 3.
ΔT_b = nK_bM = (3) (0.52°C/\cancel{M}) (0.100 \cancel{M}) = 0.16°CBecause the boiling point of a solution is higher than the boiling point of the pure solvent, ΔT_b is added to the boiling point of the pure solvent.
Solution B.P. = 100.00°C + 0.16°C = 100.16°C
ΔT_f = nK_f M = (3) (1.86°C/\cancel{M}) (0.100 \cancel{M}) = 0.558°CBecause the freezing point of a solution is lower than the freezing point of the pure solvent, ΔT_f is subtracted from the freezing point of the solvent.
Solution F.P. = 0.00°C – 0.558°C = 20.558°C = -0.56°C
π = nMRT \\ \\ = (3) (0.100 \cancel{mol}/\cancel{L}) (62.4 \cancel{L} torr/\cancel{K} \cancel{mol}) (300 \cancel{K}) \\ \\ = 5.62 × 10^{3} torr = 7.39 atmb. Because ethylene glycol does not dissociate in solution, n = 1.
ΔT_b = nK_bM = (1) (0.52°C/\cancel{M}) (0.100 \cancel{M}) = 0.052°CSolution B.P. = 100.00°C + 0.052°C = 100.05°C
ΔT_f = nK_f M = (1) (1.86°C/\cancel{M}) (0.100 \cancel{M}) = 0.186°CSolution F.P. = 0.00°C – 0.186°C = -0.186°C
= -0.19°C