A sample of 1.44 g of helium and an unweighed quantity of O₂ are mixed in a flask at room temperature. The partial pressure of helium in the flask is 42.0 torr, and the partial pressure of oxygen is 159 torr. What is the mass of the oxygen?

Mass of O₂ present in the mixture = 14.7 g

Explanation:

We apply the mole fraction to solve this excersise:

Mole fractio of a gas is:

Moles of X gas / Total moles = Partial pressure of X gas / Total pressure

We determine total pressure:

42 Torr + 159 Torr = 201 Torr

Sum of mole fraction = 1

Mole fraction of O₂ = 159 Torr / 201 Torr → 0.791

1 - Mole fraction of O₂ = Mole fraction of He → 1 - 0.791 = 0.209

We convert the mass of He, in order to determine the moles

1.44 g / 4g/mol = 0.36 moles

Mole fraction He = Moles of He / Total moles

0.36 moles / Total moles = 0.209

0.209 / 0.36 moles = Total moles → 0.580 moles

Mole fraction of O₂ = 0.791 = Moles of O₂ / Total moles

0.791 . 0.580 moles = Moles of O₂ → 0.459 moles

We convert the moles of O₂ to mass → 0.459 mol . 32g / 1mol = 14.7 g

wher is the rest of the question