Question

Dry air is about 20.95 % oxygen by volume. Assuming STP, how many oxygen molecules are in a 50.0 g sample of air? The density of air is 1.19 g/L

Answer 1

Given : Density of Air is 1.19 gram per liter

`\mathsf{\heartsuit\;\; Density\;of\;Air = (Mass\;of\;Air)/(Volume\;of\;Air)}`

Given : Mass of Sample of Air = 50 grams

`\mathsf{\implies 1.19 = (50)/(Volume\;of\;Air)}`

`\mathsf{\implies Volume\;of\;Air = (50)/(1.19)}`

`\mathsf{\implies Volume\;of\;Air = 42\;Liter}`

Given : Dry Air is about 20.95% Oxygen by Volume

`\mathsf{\implies Volume\;of\;Oxygen\;in\;42\;Liter\;of\;Air = 20.95\%\;of\;42\;Liter}`

`\mathsf{\implies Volume\;of\;Oxygen\;in\;42\;Liter\;of\;Air = ((20.95)/(100))* 42}`

`\mathsf{\implies Volume\;of\;Oxygen\;in\;42\;Liter\;of\;Air = 0.2095 * 42}`

`\mathsf{\implies Volume\;of\;Oxygen\;in\;42\;Liter\;of\;Air = 8.8\;Liter}`

At STP : 22.4 Liter of Oxygen contains 6.023 × 10²³ Molecules of O₂

`\mathsf{\implies Number\;of\;Oxygen\;Molecules\;in\;8.8\;Liter\;of\;Oxygen = ((8.8 * 6.023 * 10^2^3)/(22.4))}`

`\mathsf{\implies Number\;of\;Oxygen\;Molecules\;in\;8.8\;Liter\;of\;Oxygen = 2.366 * 10^2^3}}`