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

The molar masses of O₂ and Ne are approximately 31.998 g/mol and 20.180 g/mol, respectively. Let's redo the calculations with these values:

Assume we have 100 grams of the gas mixture, which means 50 grams of O₂ and 50 grams of Ne.

moles_O₂ = 50 g / 31.998 g/mol ≈ 1.5626 mol

moles_Ne = 50 g / 20.180 g/mol ≈ 2.4772 mol

The mole fraction of O₂ (χ_O₂) can be calculated as:

χ_O₂ = moles_O₂ / (moles_O₂ + moles_Ne) ≈ 1.5626 mol / (1.5626 mol + 2.4772 mol) ≈ 0.3866

Now, we can find the partial pressure of O₂ using Dalton's Law:

P_O₂ = χ_O₂ × P_total ≈ 0.3866 × 1.50 atm ≈ 0.5799 atm

So, the partial pressure of O₂ in the gas mixture using the exact molar masses is approximately 0.5799 atm. The competitor's result of 0.676 atm (to 3 significant figures) is incorrect.

Answer 2

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 1.50 atm = 0.579 atm (3 sig. figs.)