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S4 Chemistry: Mole and Stoichiometry Paper 2 (Molar Volume of Gas, 24 dm3)
S4 Chemistry: Mole and Stoichiometry Paper 2 (Molar Volume of Gas, 24 dm3)
This paper drills the molar volume of 24 dm3/mol at RTP: moles to volume, volume to moles, mixed with mass and particle counts, plus gas mixtures and the dm3 versus cm3 trap.
Quick answer 1. 12 dm3 | 2. 0.2 mol | 3. 6 dm3 | 4. 48 dm3 and 56 g | 5. 0.0625 mol and 3.7625e22 | 6. 6 dm3 | 7. 7.2 dm3 | 8. 42 g | 9. 0.1 mol and 0.2 g | 10. 24 dm3, not cm3
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Chemistry – Mole and Stoichiometry |
| Level | Application |
02 Questions (10 items)
- FoundationAt room temperature and pressure (RTP), what volume does 0.5 mol of hydrogen occupy? (molar volume 24 dm3/mol)
- FoundationHow many moles are in 4.8 dm3 of gas at RTP?
- FoundationWhat volume does 11 g of carbon dioxide (44 g/mol) occupy at RTP?
- FoundationWhat are the volume and mass of 2 mol of nitrogen (28 g/mol) at RTP?
- IntermediateHow many moles and molecules are in 1.5 dm3 of oxygen at RTP?
- IntermediateWhat volume does 8 g of oxygen (32 g/mol) occupy at RTP?
- Intermediate0.1 mol of helium is mixed with 0.2 mol of argon. What total volume do they occupy at RTP?
- ChallengeWhat is the mass of 36 dm3 of nitrogen at RTP?
- ChallengeA 2.4 dm3 balloon at RTP holds how many moles, and what mass if it is hydrogen (2 g/mol)?
- Error redoA student answers that 1 mol of any gas at RTP is 24 cm3. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | V = 0.5 x 24 = 12 dm3. | At RTP always use 24 dm3/mol for any gas. |
| 2 | n = 4.8 / 24 = 0.2 mol. | Turn volume into moles by dividing by 24. |
| 3 | n = 11 / 44 = 0.25 mol; V = 0.25 x 24 = 6 dm3. | Mass to moles first, then multiply by molar volume. |
| 4 | V = 2 x 24 = 48 dm3; m = 2 x 28 = 56 g. | Volume uses molar volume; mass uses molar mass, from the same moles. |
| 5 | n = 1.5 / 24 = 0.0625 mol; molecules = 0.0625 x 6.02e23 = 3.7625e22. | Convert volume to moles before reaching for particle count. |
| 6 | n = 8 / 32 = 0.25 mol; V = 0.25 x 24 = 6 dm3. | Use 32 for O2, not 16. |
| 7 | Total moles 0.1 + 0.2 = 0.3 mol; V = 0.3 x 24 = 7.2 dm3. | Gas identity does not matter; only total moles count. |
| 8 | n = 36 / 24 = 1.5 mol; m = 1.5 x 28 = 42 g. | Volume, moles, mass: three steps, no skipping. |
| 9 | n = 2.4 / 24 = 0.1 mol; m = 0.1 x 2 = 0.2 g. | Hydrogen is light, so the same moles weigh very little. |
| 10 | Wrong: it is 24 dm3, which is 24,000 cm3; treating dm3 as cm3 makes it 1,000 times too small. | dm3 and cm3 differ by 1,000 times; settle units first. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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