During a titration experiment, 0.05 moles of carbon (lV) oxide is liberated. What is the volume of gas liberated?

Analyze the question: We are given the amount of CO₂ gas (0.05 mol) and asked for its volume. This is a molar volume problem.

Concept: At STP (s.t.p.), one mole of any gas occupies 22.4 dm³.

Volume = number of moles × molar volume

Solution:

V = 0.05 mol × 22.4 dm³/mol V = 1.12 dm³

Check against the options:

• A (22.40 dm³) is the volume of 1 mole, so it would be correct only if n = 1.
• B (11.20 dm³) would need 0.5 mol.
• C (2.24 dm³) would need 0.1 mol.
• D (1.12 dm³) matches 0.05 mol.

Tip: Quick shortcut: 0.1 mol gives 2.24 dm³, so 0.05 mol (half of that) gives 1.12 dm³. Remember that 0.1 mol is 2.24 dm³, and you can scale from there quickly in the exam. Also, the molar volume is 22.4 dm³ at s.t.p. only, so check the conditions the question states.