# Exercise 10 | Colligative Properties of Solutions

Knowing that the molarity of the dissolved dioxygen at 25°C and 750 mmHg is 1.29 x 10-3 mol.L-1, calculate the solubility of dioxygen gas (in g.mL-1) in water at 25°C and a pressure of 1150 mmHg.

Henry’s Law:

Solubility of a gas is proportional to the pressure of the gas

Pgas = k x Mgas

Pgas = partial pressure of the gas (in atm)

Mgas = molarity of the dissolved gas (in mol.L-1)

k = proportionality constant (Henry’s law constant)

Let’s determine the Henry’s law constant for dioxygen:

k = $\frac{{\mathrm{P}}_{\mathrm{O}2}}{{M}_{O2}}$

PO2 = 750 mmHg = 0.987 atm

⇒ k = 7.65 x 102

Solubility of dioxygen gas at 25°C and a pressure of 1150 mmHg:

MO2 = $\frac{{\mathrm{P}}_{\mathrm{O}2}}{\mathrm{k}}$

PO2 = 1150 mmHg = 1.513 atm

⇒ MO2 =  = 1.97 x 10-3 mol.L-1

Solubility (in g.mL-1) = MO2 (in mol.mL-1) x MO2 (in g.mol-1) = 1.97 x 10-6 x 32.0 = 6.33 x 10-5 g.mL-1