Infer water movement and an unknown solution with limits

Predict water movement from the whole potential

A membrane can allow water movement while retaining sucrose. More internal solute then favors water entry if other contributions are matched, but pressure can oppose that tendency. Water potential, Ψ, expresses water’s potential relative to a reference in pressure units. In this bounded model, Ψ = Ψs + Ψp: solute potential plus pressure potential. Gravity and matrix effects are omitted. Water is driven from higher total potential toward lower total potential through a water-permeable boundary.

Begin with inside sucrose 0.30 M and zero pressure. Outside is fixed at 0.20 M and zero pressure in this panel. Predict direction before inspecting the ledger, then change only inside pressure. Capture a comparison that reverses or balances the prediction. The controls prescribe conditions; they do not simulate changing concentration, swelling, membrane rupture or an eventual volume.

Calculate the two contributions with consistent units

The solute expression is Ψs = −iCRT. Here i=1 for ideal nondissociating sucrose; C is concentration in mol/L (M); R=0.00831446261815324 L MPa/(mol K); and T=300 K. The product CRT has units MPa. The minus sign makes this solute contribution negative. Do not insert Celsius or mix a gas constant expressed in bars with pressure in MPa; 1 bar is 0.1 MPa.

At 0.20 M, Ψs = −1×0.20×R×300 ≈ −0.498868 MPa. Inside 0.30 M gives approximately −0.748302 MPa. With both pressures zero, outside has the higher potential, so inward water movement is favored. This supplies neither a measured flux nor a time to equilibrium.

The middle pressure option adds 0.10RT≈0.249434 MPa inside. At 0.30 M, the total becomes −0.748302+0.249434≈−0.498868 MPa, balancing outside despite unequal concentrations. The model uses unrounded values for this cancellation. With inside pressure 0.50 MPa, inside total is approximately −0.248302 MPa, now higher than outside, so outward movement is favored. Annotate how the pressure contribution changed the conclusion without changing the specified solute concentration.

On paper, predict inside 0.10 M at zero pressure: its potential is approximately −0.249434 MPa, so water is driven outward toward the more negative external potential. Equal total potentials mean no net water-potential drive, not the absence of molecular motion. The HS osmosis refresher offers a simpler bounded comparison; it does not supply this pressure calculation.

Read the unknown-solution packet independently

The second panel is a fixed synthetic inquiry packet, not measured classroom data and not output from the pressure controls. Separate bags contain known inside sucrose concentrations of 0.10, 0.20, 0.30 or 0.40 M and encounter the same unknown external sucrose solution. The stipulated membrane retains sucrose while allowing water; pressure differences are negligible. Bath size is sufficient to treat the external solution as approximately unchanged. Area, the prescribed 60-minute exposure and blotting method are matched. The packet does not establish that equilibrium was reached.

Each concentration has three independent synthetic bags with initial masses 10, 12 and 8 g. At 0.20 M the final masses are 9.70, 11.76 and 7.92 g. Calculate each percentage separately using 100×(final−initial)/initial. The results are −3%, −2% and −1%; their arithmetic mean is −2%. A negative sign indicates mass loss. Do not divide every bag’s change by the same initial mass.

The four group means are −5%, −2%, +2% and +5%. Sketch inside concentration on the horizontal axis and percentage mass change on the vertical axis. Plot all three points per group and distinguish them from the group mean. Under the packet assumptions, negative change at 0.20 M and positive change at 0.30 M bracket the external concentration between those values. The concentration setting belongs to the bag interior, so check the sign logic before writing the bracket.

Assuming a locally linear response, the zero crossing is 0.20 + [2/(2+2)]×0.10 = 0.25 M. This is a conditional estimate, not a hidden exact truth recovered by the browser. No bag was tested at 0.25 M. Zero mass change would not independently prove that solute leakage, pressure differences or handling effects were absent.

Keep variation and readout uncertainty separate

The three bag responses vary within every group. That variation is not the same as the stated ±0.005 g bound on each mass readout. Neither is automatically a confidence interval. For the 10.00→9.70 g bag, a conservative lowest percentage uses 9.695/10.005 and the highest uses 9.705/9.995, giving approximately −3.098% to −2.901%. These bounds address the specified readout uncertainty only.

As a separate sensitivity exercise, interpolate using the 0.20 M responses from −3% to −1% and the 0.30 M responses from +1% to +3%. Pairing their extremes gives crossings from 0.225 to 0.275 M before adding readout bounds. This is a range of chosen linear interpolations, not a probability statement or a substitute for new measurements. Propose additional concentrations near the crossing and more independently prepared bags. Repeated weighing of one bag addresses a different source of variation.

Write a practical measurement plan: specify solution preparation, membrane retention checks, matched temperature and duration, a consistent gentle blotting procedure, balance checks and randomized handling order. Explain how retained external droplets could bias a final mass upward. A teacher-supervised physical investigation could test these controls; completing this packet does not establish that you performed one.

Connect tonicity to cells and organisms

Tonicity describes an external solution’s effect on cell volume relative to a specified membrane and timescale. For a cell without a substantial opposing pressure difference, a hypotonic external solution has a lower effective concentration of nonpenetrating solutes and promotes water entry and swelling; a hypertonic solution has a higher effective concentration and promotes water loss and shrinkage. An isotonic solution produces no sustained net volume change under those conditions, although water molecules continue to move in both directions. These labels compare the external solution with the cell; they are not absolute properties of a solution alone.

Equal osmolarity counts equal total concentrations of dissolved particles, but does not guarantee isotonicity: permeating solutes can redistribute during the observation interval. Specify membrane permeability and time before predicting a sustained volume effect. A wall-bearing plant cell also develops opposing pressure as it gains water. It can therefore reach no net water movement in a hypotonic external solution when wall-supported pressure balances the solute contribution; that balance does not make the external solution isotonic or its solute concentration equal to the cell’s.

Under appropriate water-entry conditions, an animal cell can swell and may lyse; water loss can cause shrinkage. A plant wall resists expansion, allowing turgor pressure to rise, while strong water loss can separate the protoplast from the wall. A central vacuole contributes to plant water storage and turgor. These outcomes require properties absent from the prescribed-potential model.

At organism scale, water balance involves continuing exchanges. Contractile vacuoles help many freshwater protists expel excess water; animal kidneys regulate water and solute excretion. Plants regulate water loss through stomata as part of a wider water-transport system. These mechanisms support homeostasis rather than stopping molecular movement. The two lesson panels do not simulate these organs or organisms.

Transfer: challenge a zero-change conclusion

A new fictional bag shows zero mass change after one interval. A learner concludes that its inside and outside sucrose concentrations must be equal. Supply two alternative conditions that could undermine that claim, then specify a follow-up comparison before opening the response.

Public worked transfer: balance, permeability and handling

A pressure difference could balance unequal solute potentials. Solute leakage or inconsistent retained surface liquid could also complicate mass change. Measure or constrain pressure, verify solute retention and standardize blotting; repeat matched bags at nearby known concentrations with independent preparations. A single zero endpoint cannot separate these explanations. This public practice response is not independently assessed transfer or proof of physical laboratory technique.

Sources: the current AP Biology framework, Topic 2.7; OpenStax explanations of water potential, osmosis and tonicity, and organismal osmotic balance. All packet values and paper activities here are original teaching examples.