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S4 Chemistry: The Mole (Paper 2)
S4 Chemistry: The Mole (Paper 2)
This paper focuses on concentration: making solutions from mass, dilution and titration volumes.
Direct answer Concentration times volume gives moles; on dilution or mixing the moles stay the same but the total volume is used.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Chemistry: The Mole, Concentration and Percentage Yield |
| Level | Four basic, three advanced, two challenge and one error-repair item |
02 Questions (10 items)
- Foundation0.50 mol of sodium hydroxide is dissolved to make 250 cm3 of solution. Find the concentration.
- FoundationHow many moles of solute are in 0.50 dm3 of 2.0 mol dm-3 hydrochloric acid?
- FoundationHow many moles of solute are in 500 cm3 of a 0.10 mol dm-3 solution?
- Foundation4.0 g of sodium hydroxide, with NaOH = 40, is made up to 500 cm3. Find the concentration.
- Intermediate25.0 cm3 of 2.0 mol dm-3 solution is diluted to 250 cm3. Find the new concentration.
- IntermediateTo make 250 cm3 of 0.50 mol dm-3 solution, what volume of 2.0 mol dm-3 stock is needed?
- IntermediateWhat volume of 0.0500 mol dm-3 hydrochloric acid neutralises 25.0 cm3 of 0.100 mol dm-3 sodium hydroxide in a one to one ratio?
- ChallengeWhat volume of 0.100 mol dm-3 sodium hydroxide neutralises 20.0 cm3 of 0.0500 mol dm-3 sulphuric acid, which has two hydrogen ions?
- Challenge100 cm3 of 0.20 mol dm-3 solution is mixed with 100 cm3 of 0.10 mol dm-3 of the same solute. Find the concentration.
- Error redoError repair: 250 cm3 of water is added to 250 cm3 of 2.0 mol dm-3 solution and a student answers 0.10 mol dm-3. What is correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2.0 mol dm-3 | 250 cm3 is 0.250 dm3, so 0.50 divided by 0.250 gives 2.0 mol dm-3. |
| 2 | 1.0 mol | Moles equal concentration times volume, so 2.0 times 0.50 is 1.0 mol. |
| 3 | 0.050 mol | 500 cm3 is 0.500 dm3, so 0.10 times 0.500 is 0.050 mol. |
| 4 | 0.20 mol dm-3 | 4.0 divided by 40 is 0.10 mol, and dividing by 0.500 dm3 gives 0.20 mol dm-3. |
| 5 | 0.20 mol dm-3 | The moles do not change on dilution, so 0.050 mol divided by 0.250 dm3 gives 0.20 mol dm-3. |
| 6 | 62.5 cm3 | The solution needs 0.125 mol, which is 0.0625 dm3 or 62.5 cm3 of the stock, then made up to 250 cm3. |
| 7 | 50.0 cm3 | The sodium hydroxide is 0.00250 mol and the acid needs the same, so 0.0500 dm3 of acid is used. |
| 8 | 20.0 cm3 | The acid gives 0.00200 mol of hydrogen ions, so 0.00200 mol of sodium hydroxide or 0.0200 dm3 is needed. |
| 9 | 0.15 mol dm-3 | The solute totals 0.030 mol in 0.200 dm3, so the concentration is 0.15 mol dm-3. |
| 10 | 1.0 mol dm-3 | The solute is still 0.50 mol but the total volume is now 0.500 dm3, giving 1.0 mol dm-3; the water added is not the new volume. |
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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