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S4 Chemistry: Mole Concept and Concentration (Foundation, 10 Questions)
S4 Chemistry: Mole Concept and Concentration (Foundation, 10 Questions)
Ten foundation items on the first conversions of the mole concept: mass to moles, moles to mass, particle number, molar volume and the link between concentration and volume, each with an answer and a note.
Quick answer Answers in order: 2 mol, 22 g, 1.505 x 10 to the 23, 0.1 mol, 48 dm3, 2 mol/dm3, 0.02 mol, 10 g, 7 mol, hydrogen. Write the molar mass first, then decide whether to multiply or divide.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 25 minutes |
| Difficulty mix | Foundation 10 |
| Chapter | Chapter: S4 Chemistry: Mole Concept and Concentration |
| Level | S2 |
02 Questions (10 items)
- FoundationThe molar mass of water is 18 g/mol. Find the number of moles in 36 g of water.
- FoundationThe molar mass of carbon dioxide is 44 g/mol. Find the mass of 0.5 mol of carbon dioxide.
- FoundationTake the Avogadro constant as 6.02 x 10 to the 23 per mol. Find the number of particles in 0.25 mol, in units of 10 to the 23.
- FoundationThe molar mass of sodium hydroxide is 40 g/mol. Find the number of moles in 4 g.
- FoundationTake the molar volume of a gas at room conditions as 24 dm3/mol. Find the volume of 2 mol of gas under those conditions.
- Foundation0.5 mol of sodium chloride is dissolved to make 250 cm3 of solution. Find the concentration.
- FoundationHow many moles of hydrogen chloride are in 100 cm3 of 0.2 mol/dm3 hydrochloric acid?
- FoundationThe molar mass of calcium carbonate is 100 g/mol. Find the mass of 0.1 mol.
- FoundationHow many moles of atoms are there in 1 mol of sulfuric acid, H2SO4?
- FoundationCompare the number of molecules in 1 g of hydrogen (molar mass 2) with 1 g of oxygen (molar mass 32).
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2 mol | Moles equal mass divided by molar mass, so 36 / 18 = 2 mol. |
| 2 | 22 g | Mass is moles times molar mass, so 0.5 x 44 = 22 g. |
| 3 | 1.505 x 10 to the 23 | Multiply the number of moles by the Avogadro constant: 0.25 x 6.02 gives 1.505 in units of 10 to the 23. |
| 4 | 0.1 mol | Dividing 4 by 40 gives 0.1 mol, and the molar mass is confirmed by 23 + 16 + 1 = 40. |
| 5 | 48 dm3 | Volume is moles times molar volume, so 2 x 24 = 48 dm3, valid only at the stated conditions. |
| 6 | 2 mol/dm3 | 250 cm3 is 0.25 dm3, so the concentration is 0.5 / 0.25 = 2 mol per dm3. |
| 7 | 0.02 mol | 100 cm3 is 0.1 dm3, so the amount is 0.1 x 0.2 = 0.02 mol. |
| 8 | 10 g | 0.1 x 100 = 10 g, with the molar mass built from 40 + 12 + 48. |
| 9 | 7 mol | Each molecule has 2 hydrogen, 1 sulfur and 4 oxygen atoms, so 1 mol of molecules contains 7 mol of atoms. |
| 10 | Hydrogen has more | 1 g of hydrogen is 0.5 mol while 1 g of oxygen is only 0.03125 mol, so the lighter molar mass gives more molecules. |
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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