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freezing point of a solution?
What is the freezing point (in degrees C) of each of the solutions below? For water, K = 1.86. The vapor pressure of water at 45.0 degrees C is 71.93 mm Hg. A solution of 10.0 g of in 166 g of water at 45.0 degrees C, assuming complete dissociation. And can you show me how you got to the answer? Thank you.
The solution is 10 g LiCl in 166 g water.
1 Answer
- BobbyLv 71 decade agoFavorite Answer
ΔTf = Kf · mB x i
ΔTf, the freezing point depression, is defined as Tf (pure solvent) − Tf (solution) = ?
Kf, the cryoscopic constant = 1.86 for water
mB is the molality of the solution : 10 g of LiCl = 10 / 42.3944 = 0.236 moles in 166g of H2O
moles in 1000g = .236 x 1000 / 166 = 1.422m
i = vant hoff factor = 2 for LiCl ( 2 particles in solution)
ΔTf =1.86 x 1.422 x 2
= 5.29
freezing point is - 5.29 degrees C
