Answer:
0.641 g of Nitrogen are present in the mixture.
Explanation:
We use the Ideal Gases Law, to solve this question.
For the mixture:
P mixture . V mixture = mol mixture . R . T
We convert the T° to K → 23°C + 273 = 296 K
R = Ideal gases constant → 0.082 L.atm/mol.K
1 atm . 2L = mol mixture . 0.082 L.atm/mol.K . 296K
2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles
We know that sum of partial pressure = 1
Partial pressure N₂ + Partial pressure O₂ = 1
1 - 0.722 atm = Partial pressure N₂ → 0.278 atm
We apply the mole fraction concept:
Partial pressure N₂ / Total pressure = Moles N₂ / Total moles
Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles
Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles
We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g
641 mg
