What is the freezing point of an aqueous 2.65 m calcium chloride solution?
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The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m.
dT = i x Kf x m, where dT = temperature lowering, i = van't Hoff factor, Kf = freezing point depression constant, and m = molality.
For CaCl2, i = 3 because one mol CaCl2 releases 1 mol Ca2+ and 2 mol Cl- for a total of 3 mol of particles.
dT = 3 x 1.86 C/molal x 2.65 molal = 14.8 C
The freezing point of the solution:
0 - 14.8 C = -14.8 C
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The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m.
dT = i x Kf x m, where dT = temperature lowering, i = van't Hoff factor, Kf = freezing point depression constant, and m = molality.
For CaCl2, i = 3 because one mol CaCl2 releases 1 mol Ca2+ and 2 mol Cl- for a total of 3 mol of particles.
dT = 3 x 1.86 C/molal x 2.65 molal = 14.8 C
The freezing point of the solution:
0 - 14.8 C = -14.8 C
hope you understand
mark me as brilliant
thanks
Answered by
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The freezing point of pure water is 0.0ºC and Kf of pure water is -1.86ºC/m. dT = i x Kf x m, where dT =temperature lowering, i = van't Hoff factor, Kf = freezing point depression constant, and m = molality. For CaCl2, i = 3 because one mol CaCl2 releases 1 mol Ca2+ and 2 mol Cl- for a total of 3 mol of particles.
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