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

$\mathsf{\heartsuit\;\; Density\;of\;Air = \frac{Mass\;of\;Air}{Volume\;of\;Air}}$

Given : Mass of Sample of Air = 50 grams

$\mathsf{\implies 1.19 = \frac{50}{Volume\;of\;Air}}$

$\mathsf{\implies Volume\;of\;Air = \frac{50}{1.19}}$

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

Given : Dry Air is about 20.95% Oxygen by Volume

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

$\mathsf{\implies Volume\;of\;Oxygen\;in\;42\;Liter\;of\;Air = (\frac{20.95}{100})\times 42}$

$\mathsf{\implies Volume\;of\;Oxygen\;in\;42\;Liter\;of\;Air = 0.2095 \times 42}$

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

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

$\mathsf{\implies Number\;of\;Oxygen\;Molecules\;in\;8.8\;Liter\;of\;Oxygen = (\frac{8.8 \times 6.023 \times 10^2^3}{22.4})}$

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