This set practises five skills: explaining diffusion, separating boiling from evaporation, reading a heating curve, linking gas pressure to collisions and deciding whether a substance has changed. The questions get harder as you go. All numbers and situations are invented for practice.
Answer on paper first. Then open the answer and compare your reasoning, not only your final result. The last section sends each kind of error back to a lesson.
Part A: Diffusion
Q1. A scent is released in one corner of a still room. Explain, using particles, why a person in the opposite corner smells it after some time.
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The scent particles are in constant random motion. There are more of them near the source, so more move away from that region than towards it. This net movement from high to low concentration carries them across the room until the scent is spread out.
Q2. Hydrogen (molar mass 2 g/mol) and oxygen (molar mass 32 g/mol) are at the same temperature. Which gas diffuses faster, and why?
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Hydrogen. At the same temperature the particles have the same average kinetic energy, so the lighter hydrogen molecules move faster and spread sooner.
Q3. An invented experiment: a dye takes 80 s to spread through a beaker at 15 °C and 40 s at 45 °C. State the conclusion and explain it.
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Diffusion is faster at the higher temperature. At 45 °C the particles have more kinetic energy and move faster, so they spread through the water in less time (40 s compared with 80 s).
Part B: Boiling and evaporation
Q4. Sort each statement as evaporation, boiling or both. (a) Occurs only at the surface. (b) Turns a liquid into a gas. (c) Bubbles form throughout the liquid. (d) Occurs at one fixed temperature for a given pressure.
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(a) Evaporation. (b) Both. (c) Boiling. (d) Boiling.
Q5. Explain why sweat on the skin makes you feel cooler.
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The sweat particles with the most kinetic energy escape from the surface by evaporation. The average kinetic energy of the remaining liquid falls, so it cools, and heat from the skin moves into it.
Part C: Heating curve
Use this invented data for substance Y, heated steadily.
| Time (min) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Temperature (°C) | 10 | 30 | 50 | 70 | 70 | 70 | 70 | 80 | 90 | 100 | 110 | 110 | 110 | 110 | 110 |
Q6. State the melting point and boiling point of Y, and the states present at 8 minutes and at 12 minutes.
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The flat section at 70 °C (3 to 6 min) is melting, so the melting point is 70 °C. The flat section at 110 °C (10 to 14 min) is boiling, so the boiling point is 110 °C. At 8 minutes Y is a liquid (between 6 and 10 min). At 12 minutes it is a mixture of liquid and gas.
Q7. Explain why the temperature stays at 70 °C from 3 to 6 minutes, although heating continues.
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The energy supplied is used to overcome the attractions between the particles so that they can move past each other. It is not increasing their average kinetic energy, so the temperature stays constant until all of Y has melted. Melting lasts 6 − 3 = 3 minutes.
Part D: Gas pressure
Q8. A gas of volume 500 cm³ is at 120 kPa. It is compressed to 250 cm³ at constant temperature. Explain the change, and calculate the new pressure if your course uses pressure × volume = constant.
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The particles are squeezed into a smaller space, so they hit the walls more often on each unit of area, and the pressure rises. 120 × 500 = 60 000, and 60 000 ÷ 250 = 240 kPa. Check: the volume is halved, so the pressure doubles, 120 × 2 = 240.
Q9. A sealed metal container of gas is heated. Explain why the pressure inside rises.
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The particles gain kinetic energy and move faster. They hit the walls more frequently and with greater force. The volume is fixed, so the force on each unit of wall area increases, which is a higher pressure.
Part E: Substance or spacing
Q10. Classify each as a change of state or a chemical change. Give a reason. (a) steam condensing on a cold window, (b) a candle wick burning, (c) dry ice (solid carbon dioxide) turning to gas.
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(a) Change of state: the H₂O molecules stay the same and move closer. (b) Chemical change: new substances form (burning rearranges atoms into different particles). (c) Change of state: the carbon dioxide particles are the same, only the spacing and movement change.
Q11. 27 g of water is turned into steam. Calculate the mass of steam and the amount in moles. Then state the amounts of hydrogen and oxygen formed if all that water were split by a chemical change, using 2H₂O → 2H₂ + O₂.
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Mass of steam = 27 g, since mass does not change in a change of state. Molar mass of H₂O = 2 × 1 + 16 = 18 g/mol, so amount = 27 ÷ 18 = 1.5 mol. From the equation, 2 mol H₂O gives 2 mol H₂ and 1 mol O₂, so 1.5 mol H₂O gives 1.5 mol H₂ and 0.75 mol O₂. Check the mass: 1.5 × 2 = 3 g and 0.75 × 32 = 24 g, and 3 + 24 = 27 g.
If you got these wrong
| Where you went wrong | What it usually means | Go to |
|---|---|---|
| Q1 to Q3: no particles, or said diffusion stops moving particles | You gave the observation without the mechanism | Explain diffusion using particle motion |
| Q4, Q5: mixed up the two processes | You treated boiling as fast evaporation | Distinguish boiling from evaporation |
| Q6, Q7: misread a flat section | You thought the heating had stopped | Interpret a heating curve |
| Q8, Q9: pressure without collisions | You named pressure but skipped frequency and force | Relate pressure to particle collisions |
| Q10, Q11: mass or substance confusion | You did not ask whether the particles changed | Separate a substance change from a change in spacing |
Record each error in the mistake log and retest queue. For the amounts and atom counting in Q11, try the mole and equation-ratio tutor and the equation balance reasoning trainer. Return to the module overview for the study order.
If the same error type keeps returning, it may help to talk it through with someone. That is something our teachers can do in online one-to-one Chemistry tuition.