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

Let's assume we have 100 g of gas.

17.3% H2 = 100 g x 17.3% = 17.3 g H2 x 1 mol H2 / 1 g = 17.3 moles H2

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

Total moles of gas = 17.3 + 2.07 = 19.4 moles of gas

Calculate mole fractions:

mol fraction H2 = 17.3 / 19.4 = 0.892

mol fraction Ar = 2.07 / 19.4 = 0.107

Partial pressure = total pressure x mole fraction (see Raoults law)

Partial pressure Ar = 987 mbar x 0.107 = 106 mbar (3 sig.figs.)