The trouble states the we have actually aqueous remedies which way the solvent iswater. This way the value of Kf is thesame for all the given solutions.

You are watching: Rank these aqueous solutions from lowest freezing point to highest freezing point.

Therefore, ΔTf will simply depend top top theosmolality of every solution, offered by:


Ahigher osmolality or concentrationwill an outcome in alower freeze point.

I. 0.40 mC2H6O2

C2H6O2 is anon-electrolyte, which meansi = 1.

Solving for osmolality:

Osmolality = (1)(0.40) =0.40

II. 0.20 m Na3PO4

Na3PO4 is anelectrolyte, dissociating into 3 Na+ and PO43–, i m sorry meansi = 4.

Solving because that osmolality:

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Problem Details

Rank these aqueous services from lowest freezing allude to greatest freezing point.

I. 0.40 m C2H6O2

II. 0.20 m Na3PO4

III. 0.30 m KNO3

IV. 0.20 m C6H12O6

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