A gas mixture was prepared to contain 50% by mass of O₂ and Ne. When the total pressure of the mixture is 1.50 atm, what is the partial pressure of O₂?

Answer 1

I answered this already. Here it is again.

Total pressure = sum of partial pressures of O2 and Ne

P_t = P_O2 + P_Ne

Partial pressure of a gas = mol fraction x Pt

P_O2 = X_O2 x 1.50 atm and the mol fraction (X) of O₂ = 0.5 (50%)

Find mole fraction of O₂:

assume 100 g of gas mixture.

50% O₂ = 50 g O₂ x 1 mol O₂ / 32 g = 1.56 mols O₂

50% Ne = 50 g Ne x 1 mol Ne / 20.18 g = 2.48 mols Ne

Total mols gas = 4.04 moles

mole fraction O₂ = 1.56 mol / 4.04 mol = 0.386

Partial pressure O₂ = 0.386 x 1.75 atm = 0.676 atm (3 sig. figs.)