What is the vapour density of 560cm$^3$ of a gas that weighs 0.4g at s.t.p? [Molar Volume of gas at s.t.p = 22.4 dm$^3$ ]

Analyze the question: We're given the volume (560 cm³) and mass (0.4 g) of a gas at s.t.p. We need its vapour density.

Concept:

• Molar volume at s.t.p = 22.4 dm³ = 22,400 cm³
• Molar mass = mass of 22.4 dm³ of the gas
• Vapour density (V.D.) = Molar mass ÷ 2

Solution:

Step 1: Find the mass of 22,400 cm³ of the gas.

560 cm³ weighs 0.4 g

22,400 cm³ weighs = (22,400 × 0.4) ÷ 560 = 8,960 ÷ 560 = 16 g

So the molar mass = 16 g/mol.

Step 2: Calculate the vapour density.

V.D. = Molar mass ÷ 2 = 16 ÷ 2 = 8.0

Check: Moles = 0.56 dm³ ÷ 22.4 dm³/mol = 0.025 mol. Molar mass = 0.4 ÷ 0.025 = 16 g/mol. V.D. = 8.

Tip: The most common mistake here is stopping at 16, which is the molar mass (option C is the trap). Always remember: V.D. = M ÷ 2, because vapour density compares the gas to hydrogen (H₂ = 2). Also convert cm³ to dm³ carefully (1 dm³ = 1000 cm³).