An aqueous solution has a boiling point of 101.27o

Question

An aqueous solution has a boiling point of 101.27oC.  Determine the freezing point of this solution (the molal boiling point constant for water is 0.52oC kg/mole while the molal freezing point constant is 1.86oC kg/mole)

Jules Bruno · Accepted Answer

Boiling point elevation:∆Tb=iKbm Freezing point depression:∆Tf=iKfmStep 1. Determine the molality•i. We’re given:Kb = 0.52 ˚C/mtb, solution = 101.27 °Ctb, pure solvent = 100 °CCalculating for molality:∆Tb=iKbmTb, solution-Tb, pure solvent=iKbm101.27 °C-100 °C=i(0.52 °C·m-1)m1.27 °C=0.52 °C·m-1(i·m)1.27 °C0.52 °C·mol-1=0.52 °C·m-1(i·m)0.52 °C·m-1i·m=1.27 °C0.52 °C·m-1 i•m = 2.44 mStep 2. Determine the freezing point of the solution. We’re given:Kf = 1.86 ˚C/mi•m = 2.44 mtf, pure solvent = 0 °CCalculating for molality:∆Tf=iKfmTf, pure solvent-Tf, solution=iKfm-Tf, solution=iKfm-Tf, pure solvent-Tf, solution=iKfm-Tf, solution=(2.44 m)(1.86 °C·m-1)-Tf, solution=4.54 °C-Tf, solution-1=4.54 °C-1Tf, solution=4.54 °C-1 Tf, solution = – 4.54 °C

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An aqueous solution has a boiling point of 101.27oC. Determine the freezing point of this solution (the molal boiling point constant for water is 0.52oC kg/mole while the molal freezing point constant is 1.86oC kg/mole)

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An aqueous solution has a boiling point of 101.27oC. Determine the freezing point of this solution (the molal boiling point constant for water is 0.52oC kg/mole while the molal freezing point constant is 1.86oC kg/mole)

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