Calculate percentage mass change and estimate the no-change concentration
Interpret osmosis data without treating an estimate as a universal cell constant.
Section B3.2 · Osmosis
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Recall signed changes, fractions and osmosis.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which mass is the denominator for percentage mass change?
What you will learn
- Calculate signed percentage mass changes for individual samples and compare concentration means.
- Estimate a no-net-change concentration and explain practical limitations using the supplied evidence.
Independent investigation data
Invented measurements for the independent task below. These are a separate investigation from the graph above. Samples came from the same plant type, spent the same time at the same temperature, and were blotted consistently. Sample dimensions were not recorded.
| Sample | Initial mass (g) | Final mass (g) |
|---|---|---|
| A | 2.00 | 2.12 |
| B | 3.00 | 3.12 |
| C | 4.00 | 4.20 |
| Sample | Initial mass (g) | Final mass (g) |
|---|---|---|
| D | 2.00 | 1.96 |
| E | 3.00 | 2.88 |
| F | 4.00 | 3.88 |
Calculate each sample’s signed percentage change before taking the mean for its concentration. Differences between samples help you judge the limits of the estimate.
Calculate signed percentage change
Percentage mass change = (final mass − initial mass) ÷ initial mass × 100. A positive value is a gain and a negative value is a loss. Use the initial mass in the denominator and retain the sign. Percentage change is dimensionless and reported as a percentage.
A concentration-versus-percentage-change plot can show where the response crosses zero. Use the solution concentration on the horizontal axis and signed percentage change on the vertical axis. Repeats help judge uncertainty rather than making one point look exact.
Interpret zero cautiously
Near zero mass change is consistent with no net water gain or loss over the measured interval under the model. Water molecules can still cross in both directions. It is not proof that water motion stopped or that no other process affected mass.
If neighbouring concentrations give positive and negative changes, the crossing lies between them. A straight-line interpolation is an estimate only when that assumption is stated; tissue variation, experimental error and nonlinear behaviour limit precision.
Use repeats without hiding the method
For each specimen, divide its own mass change by its own initial mass. Then calculate the mean of those percentages for each concentration. Dividing the mean mass change by the mean initial mass and multiplying by 100 gives a size-weighted percentage, which can differ from the mean of the individual percentages.
For the independent investigation in this lesson, state the range of individual percentage changes as a simple description of spread. Three specimens at each concentration do not establish an exact zero point. Percentage change helps compare different starting masses but does not control surface-area-to-volume ratio. Record and standardise tissue dimensions, use more concentrations near the estimated crossing and repeat the measurements before making a stronger claim.
Sources for this lesson
Pause and explain
A separate investigation gives +3% at 0.10 mol/dm³ and −9% at 0.50 mol/dm³. Assuming a straight line between these means, estimate the concentration giving zero mean change.
Worked example
A cylinder changes from 5.0 g to 5.5 g. Calculate percentage mass change.
Show the worked solution
- Change = 0.5 g.
- 0.5 ÷ 5.0 × 100 = 10.
- The sign is positive.
Answer +10%.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Initial mass 4.0 g; final mass 3.6 g. Calculate signed percentage change.
Give me a hint
Use (final − initial)/initial × 100.
Compare my reasoning
- Change = −0.4 g.
- −0.4 ÷ 4 × 100 = −10.
−10%.
Look for these in your work
- Retain the loss sign.
Initial mass 2.5 g; final mass 2.8 g. Calculate signed percentage change. Then use samples A–F in the independent investigation tables: calculate each signed percentage change, the mean and range at each concentration, and the zero crossing assuming a straight line. Explain one limit and a practical improvement.
Give me a hint
Start with the 2.5 g specimen. For the tables, calculate six percentages before taking two means. Locate zero using the distance from the positive mean divided by the total fall.
Compare my reasoning
- Gain = 0.3 g.
- 0.3 ÷ 2.5 × 100 = 12.
- Samples A, B and C change by +6%, +4% and +5%. Their mean is +5%, with values from +4% to +6%.
- Samples D, E and F change by −2%, −4% and −3%. Their mean is −3%, with values from −4% to −2%.
- Zero is 5/8 of the way from 0.20 to 0.40 mol/dm³: 0.20 + (5/8 × 0.20) = 0.325 mol/dm³, about 0.33 mol/dm³. This calculation assumes a straight line between the means.
- Dimensions were not recorded, so surface-area-to-volume differences may affect the changes in the fixed time. Standardise and record dimensions, then repeat at additional concentrations near 0.33 mol/dm³. Do not discard a repeat just because it differs.
+12%. Table means: +5% and −3%; ranges: +4% to +6% and −4% to −2%. Linear zero estimate: 0.325 mol/dm³ (about 0.33 mol/dm³).
Look for these in your work
- Calculate each percentage using its own initial mass and retain the sign.
- Show the two means, ranges and the interpolation fraction 5/8.
- State that the crossing is an estimate; link a practical improvement to a specific limitation.
Mean changes are +6% at 0.2 mol/dm³ and −3% at 0.5 mol/dm³. Estimate the zero point assuming a straight line.
Give me a hint
Zero is two-thirds of the way across the nine-percentage-point fall.
Compare my reasoning
- 6 ÷ 9 = 2/3 of the concentration interval.
- Interval = 0.3 mol/dm³.
- 0.2 + (2/3 × 0.3) = 0.4.
Approximately 0.4 mol/dm³, with the stated linear assumption.
Look for these in your work
- State estimate and assumption.
Common mistakes
- Divide by initial, not final, mass.
- Zero net change does not mean stationary water.
What can you explain now?
Changes are +4% at 0.2 mol/dm³ and −4% at 0.4 mol/dm³. With linear interpolation, where is zero?
Compare with the explanation
0.3 mol/dm³.
Equal opposite changes place the estimated crossing halfway.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Reconstruct the percentage-change formula and explain a zero crossing.
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.