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What is the freezing point of an aqueous solution that boils at 106.4 C

What is the freezing point of an aqueous solution that boils at 106.4 C
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Answer #1

Kf of water = 1.86 oC / m

Kb of water = 0.52 oC / m

Tf / Tb = Kf / Kb = 1.86 / 0.52

Tf / Tb = 3.58

Tf = 3.58 Tb

          = 3.58 x (106.4 -100)

         = 22.89

Tf   = 22.89

0 -Tf = 22.89

Tf = -22.89

freezing point of solution = -22.89 oC

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