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S4 Chemistry Mole and Concentration Paper 4: Error Review
S4 Chemistry Mole and Concentration Paper 4: Error Review
Ten original error-review items, each with an answer and a one-line explanation.
Direct answers Items 1-3 revisit the direction of mass, mole and particle conversions, 4-6 apply molar volume and the concentration formula, 7-9 cover preparing solutions, dilution and neutralisation volumes, and item 10 clears up the cm3 to dm3 conversion. Cover the answers first, then check each note.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Chemistry: Mole and Concentration |
| Level | S4 |
02 Questions (10 items)
- FoundationError review: how many moles are there in 36 g of water (Mr = 18)?
- FoundationError review: what is the mass of 0.5 mol of carbon dioxide (Mr = 44)?
- FoundationError review: how many molecules are there in 3 mol of hydrogen (take 6.02 x 10^23)?
- IntermediateError review: at room temperature and pressure (24 dm3 per mole) what volume do 0.25 mol of gas occupy?
- IntermediateError review: 0.5 mol of solute is dissolved to make 250 cm3 of solution. Find the concentration in mol/dm3.
- IntermediateError review: how many moles of solute are there in 100 cm3 of a 2 mol/dm3 solution?
- ChallengeError review: how many grams of sodium hydroxide (Mr = 40) are needed for 100 cm3 of a 0.1 mol/dm3 solution?
- ChallengeError review: 50 cm3 of 2 mol/dm3 solution is diluted to 250 cm3. Find the new concentration.
- ChallengeError review: what volume of 0.100 mol/dm3 hydrochloric acid neutralises 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide?
- Error redoError review: a pupil treats 100 cm3 as 1 dm3 and reports 0.100 mol in a 0.100 mol/dm3 solution. What is correct and what went wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2 mol | Moles equal mass divided by molar mass, 36 / 18 = 2; taking 36 as moles skips the division. |
| 2 | 22 g | Mass equals moles times molar mass, 0.5 x 44 = 22 g; multiplying by the reciprocal gives 0.011 g. |
| 3 | 1.806 x 10^24 molecules | 3 x 6.02 x 10^23 = 1.806 x 10^24; quoting 6.02 x 10^23 counts only one mole. |
| 4 | 6 dm3 | 0.25 x 24 = 6 dm3; 22.4 belongs to the older convention and does not apply here. |
| 5 | 2 mol/dm3 | 250 cm3 is 0.250 dm3, so 0.5 / 0.250 = 2; dividing by 250 is a thousand times too small. |
| 6 | 1/5 mol | 100 cm3 is 0.100 dm3, so 2 x 0.100 = 1/5 mol; forgetting the conversion is the usual slip. |
| 7 | 4 g | 0.1 x 0.100 = 0.010 mol is needed, and 0.010 x 40 = 4 g; assuming 0.1 mol is ten times too much. |
| 8 | 2/5 mol/dm3 | The number of moles does not change: 2 x 0.050 = 0.100 mol, and dividing by 0.250 dm3 gives 2/5 mol/dm3. |
| 9 | 25.0 cm3 | The alkali contains 0.00250 mol and reacts 1:1 with the acid, so the same concentration needs the same volume. |
| 10 | 1/100 mol | 100 cm3 is 0.100 dm3, so 0.100 x 0.100 = 1/100 mol; a tenfold volume error gives a tenfold answer error. |
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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