Unit 2 · Topic 2.7 end-of-topic test
A bag made of a thin membrane that lets water through but not sucrose is filled with 0.40 mol/L sucrose (table sugar) solution, tied, weighed and lowered into a beaker of pure water. After an hour the bag is heavier. A student explains: 'Sucrose spread out of the bag into the water until it was even on both sides, and that is why the bag is heavier.'
What is wrong with the student's explanation?
Red blood cells, each 90 fL (femtoliters) in volume, were placed in three NaCl solutions. After ten minutes their volumes were: 0.10 mol/L NaCl, 120 fL; 0.15 mol/L NaCl, 90 fL; 0.30 mol/L NaCl, 60 fL. The membranes stayed intact and held the NaCl out.
How should the 0.30 mol/L solution be classified relative to the cells, and what did water do?
A hospital fluid is isotonic to red blood cells: after an hour in it, the volume of the cells is unchanged.
What is water doing at the cell membranes during that hour?
Red blood cells were placed in 0.50 mol/L urea and, separately, in 0.50 mol/L sucrose. Urea is a small molecule that crosses the cell membrane; sucrose cannot. After 2 minutes the cells in both solutions had shrunk from 100 units of volume to 72. After 12 minutes the cells in sucrose were still at 72, but the cells in urea had swollen to 118.
Why did the cells in urea shrink and then swell?
A red blood cell contains 0.30 mol/L of solute that cannot cross its membrane. It is placed in a solution of 0.10 mol/L of the same kind of solute. After a few minutes that solution is replaced by one of 0.40 mol/L. The cell stays intact throughout.
How does the cell's volume change, in order?
Paramecium, single-celled pond organisms with no cell wall, were kept in solutions of a solute that cannot cross their membranes. Their contractile vacuoles emptied 18 times a minute in pond water with almost no solute, 11 times a minute in 0.05 mol/L, and 4 times a minute in 0.10 mol/L.
What explains the pattern?
A marine fish lives in seawater that has far more solute than its body fluids. Its gill cells use active transport to push salt out into the sea, and it drinks seawater steadily. A freshwater fish, by contrast, takes salt in through its gills and makes large amounts of dilute urine.
What problem is the marine fish solving by drinking and pumping out salt?
A student reads that water potential (Ψ) is measured in bars and that pure water in an open beaker has Ψ = 0 bar. She asks what a water potential actually measures.
Which answer is right?
The figure shows three open beakers at the same temperature: pure water, 0.1 M sucrose and 0.3 M sucrose.
Rank their water potentials from highest to lowest.
A plant cell sits in an open beaker of solution. The cell's contents push outward on its wall at 3 bar and the wall pushes back, so the cell is turgid. Nothing presses on the solution in the beaker.
What are the pressure potentials of the cell contents and of the beaker solution?
A root cell has a pressure potential of +4 bar and a solute potential of −10 bar.
What is its water potential?
Two beakers at the same temperature hold 0.20 M solutions: one of NaCl, one of sucrose. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves; sucrose stays as whole molecules.
Which solution has more dissolved particles, and what value of i does each solute take?
A 0.20 M NaCl solution is at 27 °C. R = 0.0831 L·bar/(mol·K).
What is its solute potential?
A cell has a water potential of −6 bar. It is placed in a solution whose water potential is −9 bar.
What happens to the cell's water?
A root cell of a salt-marsh plant has a rigid wall and a solute potential of −8 bar; assume this does not change as water moves. It is placed in an open beaker of solution at Ψ = −3 bar and left until net water movement stops.
What pressure potential has the cell reached?
A leaf cell has a water potential of −4.9 bar. A student wants a sucrose solution at 22 °C, in an open beaker, with the same water potential. R = 0.0831 L·bar/(mol·K).
What sucrose concentration is needed?
A potato core has a mass of 12.5 g before it is placed in a salt solution and 11.0 g afterward.
What is the percent change in mass?
A student has the percent change in mass of potato cores soaked in 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 M sucrose, one value per concentration. The values do not fall exactly on a line: each carries some measurement scatter. The student wants a line graph that smooths out that scatter, shows the overall trend and lets the isotonic concentration be read off from the trend.
How should the graph be drawn?
(a) Explain why the core in 0.0 M sucrose gained mass. (1 point)
A full-credit answer: Water moved into the potato cells by osmosis, a net movement of water, because pure water has less solute than the cells, a higher water potential, and sucrose cannot cross the membranes. The extra water adds mass.
Check the box for each point your answer earns
Accept: "water moves toward the side with more solute, which is inside the cells". Do not award the point for "the potato absorbed water" with no reference to solute concentration or water potential, or for an answer in which sucrose moves.
Common slip: Saying the potato absorbed water, with no reference to solute or water potential, or having sucrose move. Sucrose stays put; water moves toward the side with more solute.
(b) Identify the independent variable, the dependent variable, and one variable the student kept the same. (1 point)
A full-credit answer: Independent variable: sucrose concentration (M). Dependent variable: percent change in mass. Kept the same: temperature (22 °C), time (24 hours), the potato used, the size and shape of the cores (same cutter), and blotting before weighing.
Check the box for each point your answer earns
Accept "mass of the core" as the dependent variable. Do not award the point if the independent and dependent variables are reversed.
Common slip: Reversing the independent and dependent variables. The concentration was set; the change in mass was measured.
(c) Estimate the sucrose concentration that is isotonic to the potato tissue, and calculate the water potential of the tissue at 22 °C. (1 point)
A full-credit answer: The change is +1.0% at 0.4 M and −8.1% at 0.6 M, so zero change falls just above 0.4 M: about 0.42 M is isotonic. The tissue is at about −10.3 bar (0.40 M read from the table gives −9.8 bar).
Write down the values in the question:
i = 1 C = 0.42 mol/L R = 0.0831 L·bar/(mol·K) T = 22 + 273 = 295 K Ψp = 0 bar (open beaker), so Ψ = Ψs
Write down the equation:
Ψs = −iCRT
Substitute in the values, and calculate:
Ψs = −1 × 0.42 × 0.0831 × 295 Ψs = −10.3 bar
Check the box for each point your answer earns
Accept 0.40 M read straight from the table, giving −9.8 bar. Do not award the point for a positive value, for a value not in bars, or for 22 used in place of 295 K.
Common slip: Using 22 in place of 295 K, or dropping the minus sign. T must be in kelvin, and a solute potential is negative.
(d) A seventh core from the same potato is placed in 0.3 M sucrose at 22 °C. Predict whether it gains or loses mass, and justify your prediction using water potential. (1 point)
A full-credit answer: The core gains mass. 0.3 M sucrose is at −7.4 bar, which is higher, less negative, than the tissue’s water potential of about −10 bar, so net water movement is from the solution into the cells.
Write down the values in the question:
i = 1 C = 0.3 mol/L R = 0.0831 L·bar/(mol·K) T = 22 + 273 = 295 K
Write down the equation:
Ψs = −iCRT
Substitute in the values, and calculate:
Ψs = −1 × 0.3 × 0.0831 × 295 Ψs = −7.4 bar
Check the box for each point your answer earns
Accept a justification from the table: 0.3 M lies between 0.2 M (+8.0%) and 0.4 M (+1.0%), both of which gained, so the core gains. Do not award the point for the right prediction with no comparison of water potentials or concentrations.
Common slip: Giving the right prediction with no comparison of water potentials. The point needs the two values, or the two table rows, side by side.
(a) Describe what water potential measures, and calculate the cell's water potential at the moment it is placed in the beaker. (1 point)
A full-credit answer: Water potential measures how strongly water tends to move; water moves from higher Ψ to lower Ψ. At the start the cell’s water potential is −8 bar.
Write down the values in the question:
Ψp = 0 bar (the contents rest against the wall without pushing) Ψs = −8 bar
Write down the equation:
Ψ = Ψp + Ψs
Substitute in the values, and calculate:
Ψ = 0 + (−8) Ψ = −8 bar
Check the box for each point your answer earns
Accept "the tendency of water to leave a region" as the description. Do not award the point for Ψ = +8 bar, or for −8 bar with no statement of what Ψ measures.
Common slip: Giving Ψ = +8 bar, or −8 bar with no statement of what Ψ measures. Both parts are needed.
(b) State which way water moves at first, and explain why the net movement later stops. (1 point)
A full-credit answer: Water moves from the beaker, at −2 bar, into the cell, at −8 bar. As water enters, the contents press against the wall and the pressure potential rises, raising the cell’s water potential until it equals −2 bar, with Ψp = +6 bar. Then there is no difference left to drive net movement.
Write down the values in the question:
Ψ = −2 bar (the solution’s, once net movement has stopped) Ψs = −8 bar
Write down the equation:
Ψ = Ψp + Ψs
Make Ψp the subject:
Ψp = Ψ − Ψs
Substitute in the values, and calculate:
Ψp = (−2) − (−8) Ψp = +6 bar
Check the box for each point your answer earns
Accept "the cell becomes turgid and its Ψ rises to match the solution's" without the +6 bar figure. Do not award the point for "water moves in until the concentrations are equal": the cell's solute potential stays at −8 bar, and it is pressure that closes the gap.
Common slip: Saying water moves in until the concentrations are equal. The cell’s solute potential stays at −8 bar; it is pressure that closes the gap.
(c) Predict what happens to the cell after the solution is replaced with 0.5 M sucrose at 22 °C. Include the water potential of the new solution. (1 point)
A full-credit answer: The new solution is at −12.3 bar. Water leaves the cell, the push against the wall falls to zero and the cell loses turgor, and the contents shrink away from the wall (plasmolysis) while the wall keeps its shape.
Write down the values in the question:
i = 1 C = 0.5 mol/L R = 0.0831 L·bar/(mol·K) T = 22 + 273 = 295 K Ψp = 0 bar (open beaker), so Ψ = Ψs
Write down the equation:
Ψs = −iCRT
Substitute in the values, and calculate:
Ψs = −1 × 0.5 × 0.0831 × 295 Ψs = −12.3 bar
Check the box for each point your answer earns
Accept "the cell goes limp" or "loses turgor" for the effect on the cell. Do not award the point for a positive Ψ or for water entering the cell. No final pressure potential is required: with Ψs fixed at −8 bar the cell cannot reach −12.3 bar, so net water movement out continues and no equilibrium is reached; an answer saying so is correct, and an answer that leaves the end point open is not penalized.
Common slip: Giving a positive Ψ for the solution, or having water enter the cell. −12.3 bar is lower than anything the cell can reach.
(d) Justify your prediction using the water potentials of the cell and of the new solution. (1 point)
A full-credit answer: Water moves toward the lower water potential. The cell’s Ψ is −2 bar when the change is made and can fall no lower than −8 bar once its pressure potential has dropped to 0; both are higher than −12.3 bar, so net water movement is out of the cell. The wall can push on the contents but cannot pull water in, so once Ψp is 0 the contents pull away from it.
Check the box for each point your answer earns
Accept a comparison in words: the sucrose solution's water potential is lower than any value the cell can have, so water leaves. Do not award the point for "the wall keeps the water in" or for a comparison in which water moves toward the higher water potential. Do not require a final Ψp or an equilibrium value: none exists here.
Common slip: Saying the wall keeps the water in, or sending water toward the higher water potential. A wall pushes; it cannot pull.