Problem 127P

What is the boiling point of an aqueous solution that freezes at −6.7 °C?

Solution: Here, we are going to determine the boiling point.

Step1:

The depression of freezing point of a solution is calculated by the equation:

Tf = m x Kf --------(1)

Where m is the molality of the solution and Kf is the freezing point depression constant for the solvent.

Similarly, the elevation of boiling point of a solution is calculated by the equation:

Tb = m x Kb --------(2)

Where m is the molality of the solution and Kb is the boiling point elevation constant for the solvent.

Step2:

The freezing point of water = 0oC

Freezing point of the solution = -6.7oC

Therefore, depression in freezing point(Tf ) = 0oC - (-6.7oC)

= 6.7oC

Freezing point depression constant for water(Kf) = 1.86oC kg mol-1

Substituting the values in equation (1), we get,

m = 6.7oC / 1.86oC kg mol-1

= 3.6 mol/kg

Step3:

Again, boiling point elevation constant of water(Kf) = 0.512oC kg mol-1

Substituting the values in equation (2), we get,

Tb = 3.6 mol/kg x 0.512oC kg mol-1

= 1.8432oC

Step4:

Tb = Tb - Tbo

Where, Tbo is the boiling point of the pure solvent.

For water, Tbo = 100oC

Therefore, boiling point of the solution = 100oC + 1.8432oC

= 101.8432oC

Thus, boiling point of the solution is 101.8432oC.

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