Question

Calculate the total volume occupied by 48 g oxygen and 42 g nitrogen at STP.

Options

• 67.2 L

• 22.4 L

• 33.6 L

• 56.2 L

Answer 1

67.2 L

Explanation:

Mol of oxygen = `48/32` = 1.5 mol

Mol of nitrogen = `42/28` = 1.5 mol

Total mol = 1.5 + 1.5 = 3 mol

Since 1 mole of gaseous mixture occupies 22.4 L at STP, three moles of gaseous mixture will occupy 67.2. L of volume.