States of Matter
8 marks · HardA student is investigating the solubility of a salt, ammonium nitrate (NH₄NO₃), in water. They perform an experiment and obtain the following results:
| Item | Mass / g |
|---|---|
| Empty evaporating basin | 26.47 |
| Evaporating basin + solution | 62.12 |
| Evaporating basin + dry crystals | 40.11 |
Using the results above, calculate the solubility of ammonium nitrate at 35°C in g per 100g of water.
Calculate the mass of crystals and mass of water used.
Calculate the solubility (g per 100g of water).
The student repeats this experiment at different temperatures and obtains the following solubility data:
| Temperature/°C | Solubility / g per 100g water |
|---|---|
| 10 | 65 |
| 20 | 77 |
| 35 | 98 |
| 55 | 138 |
Explain, using your result from part (a), how the student could check that their measurement at 35°C fits the trend shown by the other data points. What pattern would you expect to see on a solubility curve for this salt?
What does the solubility curve show about the relationship between temperature and solubility for ammonium nitrate?
State how a result that does NOT fit the trend would appear on a solubility curve.
Give the correct term for a solution in which no more solute can dissolve at a particular temperature, and undissolved solid is present.
State the equation used to calculate solubility in g per 100g of water from experimental mass data.
Explain briefly why it is important to pour off only the solution and not any undissolved solid during the experiment.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a(i)) Mass of crystals = 40.11 - 26.47 = 13.64g; Mass of water = 62.12 - 40.11 = 22.01g · must show correct mass of crystals, correct mass of water
(a(ii)) Solubility = (13.64 / 22.01) × 100 = 61.97 g per 100g water (accept 62 g/100g to 2 sf) g per 100g water · must show use of the correct formula, correct calculation, final answer in correct units
(b(i)) Solubility increases as temperature increases · must show trend described
(b(ii)) A point that does not fit would appear as an outlier and not lie on the smooth curve joining the other points · must show description of an outlier
(c) Saturated solution · must show correct term
(d) Solubility (g per 100g water) = (mass of solute / mass of solvent) × 100 · must show correct formula
(e) Pouring off only the solution ensures you do not add any extra solid, so the mass of dissolved solute is measured accurately · must show accuracy of measuring dissolved solute
Common mistakes: Using wrong difference in masses; Using incorrect formula for solubility
Show mark scheme and worked solution
Marking points
M1 (a(i)) · 1 mark · Correct calculation of mass of crystals and mass of water (both needed)
M2 (a(ii)) · 1 mark · Correct use of solubility formula (follow-through)
A1 (a(ii)) · 1 mark · Correct final value for solubility (allow 62 g/100g water, 2 sf) (follow-through)
B1 (b(i)) · 1 mark · Correct trend stated for solubility and temperature
B2 (b(ii)) · 1 mark · Identifies outlier/not on curve for result that does not fit
B3 (c) · 1 mark · Correctly gives term saturated solution
B4 (d) · 1 mark · States correct solubility formula
B5 (e) · 1 mark · Explains the purpose of pouring off only solution clearly
Worked solution
(a)(i) Mass of crystals = mass of evaporating basin + dry crystals - mass of empty evaporating basin.
Mass of water = mass of evaporating basin + solution - mass of evaporating basin + dry crystals.
(a)(ii) Use the solubility equation:
(b)(i) The solubility curve shows that as temperature increases, solubility increases.
(b)(ii) If a result does not fit the trend, it appears as an outlier and does not lie on the curve formed by the other points.
(c) When no more solute can dissolve and undissolved solid remains, the solution is called a saturated solution.
(d) Solubility (g per 100g water) = (mass of solute / mass of solvent) × 100
(e) If any undissolved solid is poured into the evaporating basin, it will be included in the final mass of solute, giving an incorrect (too high) value of solubility.