This set covers choosing a separation method, percentage by mass, Rf values, elimination from candidates, observation versus inference and the limits of one test. The questions run from easier to harder, and every answer is worked in full. All numbers and observations are invented for practice.
Attempt each question before opening the answer. Use the mole and equation-ratio tutor or the equation balance reasoning trainer if you want to check any equation, and record repeated slips in the mistake log and retest queue. This set belongs to separation and chemical analysis.
Questions
Q1. A mixture is fine chalk powder stirred in water. The chalk does not dissolve. Name the method that separates it and say where the chalk ends up.
Show answer
Chalk is insoluble, so use filtration. The chalk stays on the filter paper as the residue. The water passes through as the filtrate.
Q2. Choose the most suitable method for each: (a) obtain solid salt from salt solution, (b) remove sand from a mixture of sand and water.
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(a) Evaporation (or crystallisation), because the salt is dissolved and passes through a filter. (b) Filtration, because sand is insoluble and is caught by the filter paper.
Q3. A 12.0 g mixture of sand and salt is added to water and stirred. After filtering and drying, the sand weighs 7.8 g. Calculate the percentage of salt in the mixture.
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Mass of salt = 12.0 − 7.8 = 4.2 g. Percentage = 4.2 ÷ 12.0 × 100 = 35%.
Check: 35% of 12.0 is 4.2, and 7.8 + 4.2 = 12.0.
Q4. On a chromatogram the solvent front moved 7.5 cm from the baseline. A spot moved 3.0 cm. Calculate its Rf value.
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Rf = 3.0 ÷ 7.5 = 0.40. An Rf value has no units.
Q5. A spot has Rf 0.64. The solvent front moved 12.5 cm. How far did the spot move from the baseline?
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Distance = 0.64 × 12.5 = 8.0 cm.
Check: 8.0 ÷ 12.5 = 0.64.
Q6. The solvent front moved 8.0 cm. Reference dyes have Rf values P = 0.30, Q = 0.75 and R = 0.50. An ink gives spots at 2.4 cm and 6.0 cm. What does the ink contain?
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Ink spot Rf values: 2.4 ÷ 8.0 = 0.30 and 6.0 ÷ 8.0 = 0.75. These match P and Q. No spot matches R (0.50).
Conclusion: the ink is a mixture containing at least two dyes, consistent with P and Q and showing no evidence of R.
Q7. A white solid could be zinc carbonate, sodium chloride, sodium carbonate or potassium chloride. It dissolves in water, and the solution fizzes when dilute acid is added. The gas turns limewater milky. Which candidate fits?
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Dissolving rules out zinc carbonate, which is insoluble. Fizzing with a gas that turns limewater milky shows a carbonate, which rules out both chlorides. Sodium carbonate remains, and it explains all three results.
Q8. Label each statement as an observation or an inference: (a) a blue precipitate forms, (b) copper(II) ions are present, (c) bubbles of a colourless gas form, (d) the gas is carbon dioxide.
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(a) observation, (b) inference, (c) observation, (d) inference. The observations can be seen. The inferences are conclusions drawn from them.
Q9. A colourless solution may contain aluminium, calcium or zinc ions. It gives a white precipitate with aqueous sodium hydroxide, which dissolves in excess. Aqueous ammonia also gives a white precipitate, which dissolves in excess. Which ion is present?
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Dissolving in excess sodium hydroxide rules out calcium. Dissolving in excess ammonia rules out aluminium, whose hydroxide does not dissolve. The ion consistent with both results is zinc.
Q10. A student adds dilute acid to a grey solid and sees bubbles. The student writes: “It is a carbonate.” Evaluate this conclusion.
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The bubbles show a gas is made, but more than one substance could do that. A carbonate gives carbon dioxide, and a reactive metal gives hydrogen. The gas should be tested: limewater turning milky suggests carbon dioxide, and a squeaky pop with a lit splint suggests hydrogen. Without that test, the conclusion is not supported.
Q11. A solution gives a white precipitate with silver nitrate solution. A student concludes the solution contains chloride ions. What step was missing, and why does it matter?
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Dilute nitric acid should be added before the silver nitrate. Carbonate ions also give a white precipitate with silver nitrate, and the acid removes them as carbon dioxide. A precipitate that remains is more reliably a halide. Even then, colour matters: silver chloride is white, silver bromide is cream and silver iodide is yellow.
If you got these wrong
- Q1 to Q3, choosing a method or a mass calculation: return to choosing filtration versus evaporation. Check whether the substance you want is dissolved.
- Q4 to Q6, Rf and matching spots: revisit interpreting a chromatogram. Measure from the baseline and divide spot by solvent front.
- Q7 and Q9, eliminating candidates: repeat using a stated result to narrow an unknown. Cross out candidates one result at a time.
- Q8, observation or inference: read distinguishing observation from inferred identity. If you can see it, it is an observation.
- Q10 and Q11, limits of one test: go to evaluating the limits of a single positive test. Ask what else would give the same result.
If the same error keeps returning, our online one-to-one Chemistry tuition gives a teacher the chance to watch you work and find the step where it happens.