Practice IB Biology Topic D2.3 Water Potential with authentic exam-style questions for both SL and HL students. This question bank focuses on the exact syllabus content for D2.3 Water Potential and mirrors Paper 1A, 1B, 2 style where relevant.
Get instant solutions, detailed explanations, and build confidence with questions aligned to IB examiner expectations.
An animal cell contains 0.30 mmol impermeant solute. Its volume changes as shown. Which description is supported?
After equilibrating at 0.40 mol dm⁻³, a cylinder enters the isotonic solution for untreated tissue. Which concentration and initial change follow?
| Stage | Medium | Cell volume / cm³ | Result |
|---|
| Initial | — | 1.00 | intact |
| Equilibrium | X | 0.75 | intact |
| After transfer | Y | — | lysis |
A protist's equal-volume contractions are measured at steady state. Cytoplasmic concentration and other conditions remain unchanged.
| Solution | Tonicity | Solute concentration / arbitrary units | Interval between contractions / s |
|---|---|---|---|
| X | hypotonic | 1 | 10 |
| Y | hypotonic | 3 | 25 |
Which Y:X water-entry rate is supported?
Table 1: Equilibrated cell states
| Solution | Cell state |
|---|---|
| X | Turgid |
| Y | Membrane just separating from wall |
The table describes equilibrated plant cells. Their walls are removed without changing cell contents or external solutions. Which initial water movements follow?
An osmosis experiment gives the results shown; solute cannot cross the membrane. Which statement about Y is correct on entry?
| Stage | Medium | Cell mass / g |
|---|---|---|
| Initial | — | 10 |
| Equilibrium | X | 8 |
| Equilibrium | Y | 9 |
X: 0.40 mol dm⁻³; Y: hypotonic
X: 0.225 mol dm⁻³; Y: hypertonic
X: 0.40 mol dm⁻³; Y: isotonic
X: 0.225 mol dm⁻³; Y: hypotonic
0.30 mol dm⁻³; mass loss
0.40 mol dm⁻³; no mass change
About 0.33 mol dm⁻³; mass gain
About 0.33 mol dm⁻³; mass loss
