A gas mixture contains 17.3% hydrogen and 82.7% argon by mass. What is the partial pressure of argon in the mixture if the total pressure is 987 mbar ?

Question

Express the partial pressure in millibars to three significant figures.

Answer 1

Raoult's law and Dalton's law

Let us assume we have 100 g of gas.

moles of H₂ = 17.3% x 100 g = 17.3 g H₂ x 1 mol / 2 g = 8.65 moles H₂

moles Ar = 82.7% x 100 g = 82.7 g Ar x 1 mol / 39.95 g = 2.07 moles Ar

Total moles of gas = 8.65 + 2.07 = 10.7 moles

Mole fraction of H₂ = 8.65 mols / 10.7 mols = 0.808

Mole fraction Ar = 2.07 mols / 10.7 mols = 0.193

Partial pressure of H₂ = mol fraction H₂ x Total pressure = 0.808 x 987 mbar = 797 mbar

Partial pressure of Ar = mol fraction Ar x Total pressure = 0.193 x 987 mbar = 190 mbar