Interpret repeated limewater cloudiness timings
Calculate means and keep the carbon-dioxide conclusion within the method’s limits.
Section 11.1 · Gas exchange in humans
Exam scope and assessment notes
Core and Extended · Biology 0970, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0970 has the same subject content as IGCSE 0610, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Recall means and the fair-test variables.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How is a three-value mean calculated?
What you will learn
- Read the supplied teaching results
- Keep ratios distinct from calibration
Interpret repeated limewater cloudiness timings
| Air sample | Trial 1 / s | Trial 2 / s | Trial 3 / s | Mean / s |
|---|---|---|---|---|
| Inspired | 30 | 31 | 29 | 30 |
| Expired | 6 | 5 | 7 | 6 |
Read the supplied teaching results
In the invented fixed-flow investigation, inspired-air timings are 30, 31 and 29 seconds; expired-air timings are 6, 5 and 7 seconds. All use fresh equal limewater and the same cloudiness endpoint. Add each group of three values and divide by three: the means are 30 seconds and 6 seconds respectively.
The expired sample reaches the endpoint sooner in all three paired trials. Under the matched method this supports the conclusion that expired air has more carbon dioxide. Repeats show some variation and make a single unusual endpoint less decisive; they do not remove a systematic error such as an unnoticed difference in flow rate.
Keep ratios distinct from calibration
Thirty divided by six gives a time ratio of five. That describes the observed timings, not automatically a fivefold carbon-dioxide concentration. The reagent reaction, endpoint and apparatus response would need calibration before timing could be translated into an exact gas concentration.
These numbers are teaching data, not a measurement from a real learner. Describe their pattern and method limits without diagnosing lung function. If sample delivery is not matched or one tube begins cloudy, the same numerical means cannot establish a fair air-composition comparison. Check the design before treating arithmetic as sufficient evidence.
Pause and explain
What does the matched 30 s versus 6 s result support?
Worked example
Calculate the inspired-air mean for 30, 31 and 29 seconds.
Show the worked solution
- Add: 30 + 31 + 29 = 90.
- Divide by three trials.
Answer 30 seconds.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate both mean times from the supplied three trials per sample.
Give me a hint
Find 90/3 and 18/3.
Compare my reasoning
- Inspired total: 90 seconds.
- Expired total: 18 seconds.
Inspired: 30 s. Expired: 6 s.
Look for these in your work
- Use all repeats
- Include units
Write the observed pattern and the supported air-composition conclusion.
Give me a hint
Separate observation from inference.
Compare my reasoning
- Expired air clouds comparable limewater sooner.
- At matched flow this supports a higher carbon-dioxide fraction.
Observation: expired-air mean is shorter. Inference: expired air contains more carbon dioxide under the controlled comparison.
Look for these in your work
- State both levels
- Retain the method conditions
Evaluate: 30/6 = 5, therefore expired air has exactly five times the carbon-dioxide concentration.
Give me a hint
A timing ratio is not automatically a concentration calibration.
Compare my reasoning
- The quotient correctly describes the time ratio.
- No calibration linking time and concentration is supplied.
The time ratio is five, but the exact concentration ratio is not established without calibration.
Look for these in your work
- Accept the correct arithmetic
- Reject the unsupported conversion
Common mistakes
- Dividing a sum by the wrong number of trials.
- Treating an inverse time ratio as calibrated concentration.
What can you explain now?
What does the matched 30 s versus 6 s result support?
Compare with the explanation
Expired air reaches the endpoint faster and contains more CO₂
It is a comparative inference under the controlled method.
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Make it stick
Recalculate the means and distinguish time ratio from concentration ratio.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.