Question

A mixture of argon and xenon gases at a total pressure of 704 mm Hg contains argon at a partial pressure of 466 mm Hg. If the gas mixture contains 4.53 grams ofargon, how many grams of xenon are present?

__(?)__g Xe

Answer 1

mole fraction x = Partial pressure of any individual gas / pressure of the total gas mixture

x = 466 mmHg / 704 mmHg = 0.662

mole fraction of Ar = 0.662

now one more formula

mole fraction of Ar = moles of Ar / total moles

moles of Ar = weihjt of Ar / molar mass of Ar = 4.53 g / 39.95 g /mol = 0.1134 moles

we know th emole fraction of Ar = 0.662

0.662 = 0.1134 moles / total moles

total moles = 0.1134 / 0.662 = 0.1713 moles

now we know the total moles which means moles of Ar+ moles of Xe

and we know the moles of Ar = 0.1134

0.1713 = molesof Xe + 0.1134

moles of Xe = 0.1713 - 0.1134 = 0.0579

now use the formula

moles = mass / molar mass

0.0579 = mass / 131.3 g /mol

mass= 0.579 mole x 131.3 g /mol

mass of Xe = 7.60 grams