4CH1
States of Matter
Principles of Chemistry
Exam Frequency Analysis
Past paper frequency (2018 to 2024)
This topic accounts for approximately 6% of your exam marks.
stable
Low
Stable6%
Appears regularly as short-answer questions on particle diagrams and state changes.
Aim
- Determine the solubility (g per 100 g water) of a soluble salt, typically copper(II) sulfate or potassium nitrate, at a stated temperature
Apparatus
- Evaporating basin
- Boiling tube
- Water bath (a heated beaker of water will substitute)
- Thermometer
- Balance reading to 0.01 g
- Bunsen burner, tripod, gauze
- Stirring rod
- The chosen soluble solid
Method
- Weigh the empty evaporating basin; record its mass
- Half-fill a boiling tube with water and warm it in the water bath to just above the target temperature (say, just above 30 °C)
- Add the chosen solid in small portions, stirring after each addition, until further additions stop dissolving and a small amount of solid residue settles on the base of the tube
- Cool the boiling tube to exactly the target temperature (30 °C), stirring gently; let the residue settle
- Decant the clear saturated solution into the pre-weighed basin (leaving the residue behind in the tube); reweigh the basin
- Heat the basin gently to evaporate the water; reduce the heat before the basin runs dry, to prevent spitting
- Heat to constant mass: cool, weigh, reheat, cool, reweigh, repeating until two consecutive masses agree
- Calculate the solubility from the recorded masses
Sample calculation
| Reading | Mass / g |
|---|---|
| Empty basin | 30.00 |
| Basin + saturated solution | 90.00 |
| Basin + dry solid (constant mass) | 50.00 |
- Mass of solution: 90.00 − 30.00 = 60.00 g
- Mass of solute: 50.00 − 30.00 = 20.00 g
- Mass of water: 60.00 − 20.00 = 40.00 g
- Solubility at 30 °C = (20.00 / 40.00) × 100 = 50.0 g per 100 g of water
- Mass % of solute in the solution = (20.00 / 60.00) × 100 = 33.3 %
Sources of error
- Transferring undissolved residue into the basin → overestimates solubility; decant carefully, or filter at the target temperature
- Loss of solid by spitting during evaporation → heat gently near dryness; finish on residual heat
- Solution cooling below the target temperature before decanting → solubility falls; less solid in the decanted liquid; decant promptly
- Incomplete drying → trapped water inflates the apparent solute mass; always heat to constant mass