中文（繁體）
中文（简体）
English
S4 Chemistry: Mole Concept and Concentration (Intermediate, 10 Questions)
S4 Chemistry: Mole Concept and Concentration (Intermediate, 10 Questions)
Ten intermediate items on water of crystallisation, gas volume and mass, dilution, ion amounts, percentage by mass, empirical formula from percentages and mass calculations from equation ratios, each with an answer and a note.
Quick answer Answers in order: 0.2 mol, 265 g, 4.8 dm3, 0.2 mol/dm3, 0.3 mol, 36 per cent, CH2O, 8 g, 36 g, 3.01 x 10 to the 23. Three steps: convert to moles, read the coefficients, convert back.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 35 minutes |
| Difficulty mix | Intermediate 10 |
| Chapter | Chapter: S4 Chemistry: Mole Concept and Concentration |
| Level | S2 |
02 Questions (10 items)
- IntermediateHydrated copper sulfate CuSO4.5H2O has molar mass 250 g/mol. Find the moles in 50 g of the crystals.
- IntermediateSodium carbonate has molar mass 106 g/mol. Find the mass of 2.5 mol.
- IntermediateFind the volume of 8.8 g of carbon dioxide at room conditions, where the molar volume is 24 dm3/mol.
- Intermediate25 cm3 of 2 mol/dm3 solution is diluted with water to 250 cm3. Find the new concentration.
- Intermediate0.1 mol of aluminium sulfate, Al2(SO4)3, dissolves in water. How many moles of sulfate ions are produced?
- IntermediateFind the percentage by mass of water of crystallisation in CuSO4.5H2O of molar mass 250 g/mol.
- IntermediateA compound contains 40.0 per cent carbon, 6.7 per cent hydrogen and 53.3 per cent oxygen by mass. Find its empirical formula.
- IntermediateHow many grams of sodium hydroxide, molar mass 40, are needed to make 500 cm3 of 0.4 mol/dm3 solution?
- IntermediateHydrogen burns in oxygen: 2H2 + O2 gives 2H2O. Find the mass of water formed from 4 g of hydrogen.
- IntermediateAt standard conditions the molar volume is 22.4 dm3/mol. Find the number of molecules in 11.2 dm3 of oxygen, in units of 10 to the 23.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 0.2 mol | 50 / 250 = 0.2 mol, and the five water molecules contribute 90 to the molar mass. |
| 2 | 265 g | 2.5 x 106 = 265 g, remembering that the two sodium atoms contribute 46. |
| 3 | 4.8 dm3 | First 8.8 / 44 = 0.2 mol, then 0.2 x 24 = 4.8 dm3. |
| 4 | 0.2 mol/dm3 | The amount of solute stays at 0.05 mol, so dividing by the new volume of 0.25 dm3 gives 0.2 mol per dm3. |
| 5 | 0.3 mol | Each formula unit gives three sulfate ions, so 0.1 x 3 = 0.3 mol. |
| 6 | 36% | The five water molecules weigh 90 out of 250, which is 36 per cent. |
| 7 | CH2O | Dividing each percentage by the atomic mass gives the ratio 1 : 2 : 1, so the empirical formula is CH2O with molar mass 30. |
| 8 | 8 g | The amount needed is 0.5 x 0.4 = 0.2 mol, which weighs 0.2 x 40 = 8 g. |
| 9 | 36 g | 4 g of hydrogen is 2 mol, and the ratio 2 : 2 gives 2 mol of water, so the mass is 2 x 18 = 36 g. |
| 10 | 3.01 x 10 to the 23 | 11.2 / 22.4 = 0.5 mol, and multiplying by the Avogadro constant gives 3.01 in units of 10 to the 23. |
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.
Find a tutor freeTutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.