Question

What is the normal boiling point of an aqueous solution that has a freezing point of 1.04 oC. Kf for water 1.86 oC/m (oC-kg/mol). Hint: Calculate the molality from the freezing point depression and use it to calculate the normal boiling point.

Answer 1

Given:

Freezing point of the solution = 1.04 ⁰C

Kf (water) = 1.86 ⁰C kg /mol

Step 1 : Calculation of molality

Delta Tf = m kf

m = delta Tf / kf

we know

Delta Tf = Freezing point of the solution – freezing point of the solvent.

Delta Tf = 1.04 deg C – 0 deg C

Delta Tf = 1.04 deg C

m = 1.04 deg C / 1.86 deg C kg /mol

= 0.55914 mol/Kg

Now we have to find the normal boiling point f the solution.

We know

Delta Tb = m kb

Here delta Tb is the elevation in boiling point , m is molality , kb is boiling point constant of solvent.

Kb for water = 0.512 deg C kg /mol

Lets plug the value of m and kb to calculate delta Tb

Delta Tb = 0.55914 x 0.512 = 0.28628 deg C

Delta Tb = boiling point of the solution – boiling point of pure solvent ‘

0.28628 deg C = boiling point of the solution – 100 deg C

Boiling point of the solution = 100 + 0.28628 = 100.3 deg C