Question

A mixture of gases with a total pressure of 1.20 atm is composed of 5.0 g of oxygen, 4.0 g of hydrogen and 2.0g of fluorine. Calculate the partial pressure of each gas in the mixture.

Answer 1

Ans. Moles of oxygen, O₂ = Mass / Molar mass

                                    = 5.0 g / (32.0 g/mol)

                                    = 0.15625 mol

Moles of hydrogen, H₂ = 4.0 g / (2.0 g/mol) = 2.0 mol

Moles of fluorine, F₂ = 2.0 g / (38.0 g/mol) = 0.05263 mol

Now,

Total number of moles in mixture = 0.15625 mol (O₂) + 2.0 mol (H₂) + 0.05263 mol (F₂)

                                    = 2.20888 mol

#1. Mole fraction of O₂ = Moles of O₂ / Total moles of all gases in mixture

                                    = 0.15625 mol / 2.20888 mol

                                    = 0.0707

Partial pressure of O₂ = mole fraction of O₂ x Total pressure in the vessel

                                    = 0.0707 x 1.20 atm

                                    = 0.085 atm

#2. Mole fraction of H₂ = 2 mol / 2.20888 mol = 0.9054

Partial pressure of H₂ = 0.9054 x 1.20 atm = 1.086 atm

#3. Mole fraction of F₂ = 0.05263 mol / 2.20888 mol = 0.0238

Partial pressure of H₂ = 0.0238 x 1.20 atm = 0.028 atm