# Problem: Rank these aqueous solutions from lowest freezing point to highest freezing point.I. 0.40 m C2H6O2II. 0.20 m Na3PO4III. 0.30 m KNO3IV. 0.20 m C6H12O6

###### FREE Expert Solution

change in freezing point (ΔT­f) is given by:

The change in freezing point is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{imK}}}_{{\mathbf{f}}}}$

where:

i = van’t Hoff factor

m = molality of the solution (in m or mol/kg)

Kf = freezing point depression constant (in ˚C/m) of the solvent

The problem states that we have aqueous solutions which means the solvent is water. This means the value of Kf is the same for all the given solutions.

Therefore, ΔTf will just depend on the osmolality of each solution, given by:

$\overline{){\mathbf{Osmolality}}{\mathbf{=}}{\mathbf{i}}{\mathbf{×}}{\mathbf{m}}}$

higher osmolality or concentration will result in a lower freezing point.

I. 0.40 m C2H6O2

C2H6O2 is a non-electrolyte, which means i = 1.

Solving for osmolality:

Osmolality = (1)(0.40) = 0.40

II. 0.20 m Na3PO4

Na3PO4 is an electrolyte, dissociating into 3 Na+ and PO43, which means i = 4.

Solving for osmolality:

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###### Problem Details

Rank these aqueous solutions from lowest freezing point to highest freezing point.

I. 0.40 m C2H6O2

II. 0.20 m Na3PO4

III. 0.30 m KNO3

IV. 0.20 m C6H12O6