Question

What is the boiling point of an aqueous solution that has a vapor pressure of 23.0 torr at 25 ∘C? (P∘H2O=23.78 torr; Kb= 0.512 ∘C/m).

Answer 1

Answer:

Boiling point of the solution is 100.964°C

Explanation:

In this problem, first, you must use Raoult's law to calculate molality of the solution. When you find the molality you can obtain the boiling point elevation because of the effect of the solute in the solution (Colligative properties).

Using Raoult's law:

Psol = Xwater × P°water.

As vapour pressure of the solution is 23.0torr and for the pure water is 23.78torr:

23.0torr= Xwater × 23.78torr.

0.9672 = Xwater.

The mole fraction of water is:

$0.9672 = \frac{X_{H_2O}}{X_{H_2O}+X_{solute}}$

Also,

$1 = X_{H_2O}+X_{solute}$

You can assume moles of water are 0.9672 and moles of solute are 1- 0.9672 = 0.0328 moles

Molality is defined as the ratio between moles of solute (0.0328moles) and kg of solvent. kg of solvent are:

$09672mol *\frac{18.01g}{1mol}* \frac{1kg}{1000g} = 0.01742kg$

Molality of the solution is:

0.0328mol Solute / 0.01742kg = 1.883m

Boiling point elevation formula is:

ΔT = Kb×m×i

Where ΔT is how many °C increase the boiling point regard to pure solvent, Kb is a constant (0.512°C/m for water), m molality (1.883m) and i is Van't Hoff factor (Assuming a i=1).

Replacing:

ΔT = 0.512°C/m×1.882m×1

ΔT = 0.964°C

As the boiling point of water is 100°C,

Boiling point of the solution is 100.964°C