K f is the freezing point depression constant. It depends only on the SOLVENT. P

Question

Kf is the freezing point depression constant. It depends only on the SOLVENT.

Problem:

Match the following aqueous solutions with the appropriate letter from the column on the right.

Problem:

Match the following aqueous solutions with the appropriate letter from the column on the right.

Tf is the decrease in the freezing point = Tf(solution) - Tf(pure solvent) m is the molality of the solution = (# moles solute / kg solvent) i is the "van't Hoff" factor = (# moles of solute particles / mole of solute) For nonelectrolytes: i = 1 For strong electrolytes: i = number of (cations + anions) For weak electrolytes: 1 < i < number of (cations + anions)

Explanation / Answer

1)

Since only m*i are important then

Cr(CH3COO)2 --> 1+2 = 3 ions

m = 0.17 so

mi = 3*0.17 = 0.51

2)

CuI = 3 ions

0.15*3 = 0.45

NiI2 = 3 ions

0.16*3 = 0.48

E.G =1 ion

0.53

then

lowest to highes F. point

EG < Cr(CHCOO)2 < NiI2 < CuI2

2

same logic

1)

i = 1+1 = 2 ions

2*0.27 = 0.54

2)

mi = 0.15*3 = 0.45

3)

mi = 2*0.25 = 0.5

4)

0.47*1 = 0.47

then

lowest to highes F. point

AgNO3 <NaCl < EG < BaBr2

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