Water Potential

Unit 2 · Topic 2.7b end-of-topic test

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback tells you what a wrong choice assumed. For the two free-response questions, write your answer in full sentences (and show any calculation), then open the scoring guide and mark your own work against it. R = 0.0831 L·bar/(mol·K); temperatures in kelvin are °C + 273.

Question 1

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?

A. How strongly water tends to move: water moves from higher Ψ to lower Ψ B. How much water a solution holds: pure water scores the most, at 0 bar C. How much solute is dissolved: a higher Ψ means more solute D. How fast water is moving: 0 bar means the water is standing still

Question 2

Three open beakers at the same temperature, filled to the same level.

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. Beaker 3, then 2, then 1 B. Beaker 2, then 1, then 3 C. All equal: none is under pressure D. Beaker 1, then 2, then 3

Question 3

A beaker of sucrose solution stands open on the bench.

Which change lowers the water potential of the solution?

A. Adding pure water to the solution B. Dissolving more sucrose in the solution C. Pressing on the surface of the solution with a piston D. Pouring half of the solution away

Question 4

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. Cell −3 bar; beaker 0 bar B. Cell +3 bar; beaker 0 bar C. Cell 0 bar; beaker +3 bar D. Cell +3 bar; beaker +3 bar

Question 5

Two beakers of the same 0.2 M sucrose solution at the same temperature. The left is open; in the right a piston presses on the solution at 3 bar. Each dot is a sucrose molecule.

Two beakers hold the same 0.2 M sucrose solution at the same temperature. One is open; in the other a piston presses on the solution at 3 bar, as in the drawing.

Compare the water potentials of the two solutions.

A. The pressed solution has the higher water potential, by 3 bar B. The pressed solution has the lower water potential, because pressure squeezes water out of it C. The two water potentials are equal, because the two concentrations are equal D. The open solution has the higher water potential, 0 bar, because nothing presses on it

Question 6

A root cell has a pressure potential of +4 bar and a solute potential of −10 bar.

What is its water potential?

A. −14 bar B. −6 bar C. +6 bar D. −10 bar

Question 7

Two plant cells that touch, with the pressure potential and solute potential of each.

Two plant cells touch. Cell X has a pressure potential of +4 bar and a solute potential of −9 bar. Cell Y has a pressure potential of +1 bar and a solute potential of −7 bar, as in the drawing.

Which way does water move between the two cells?

A. From cell Y into cell X, because +4 bar is higher than +1 bar B. From cell Y into cell X, because −9 bar is lower than −7 bar C. From cell X into cell Y, because −5 bar is higher than −6 bar D. There is no net movement, because −5 bar and −6 bar are both negative

Question 8

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. Sucrose, twice as many; i = 2 for sucrose, 1 for NaCl B. The same number; i = 1 for both C. NaCl, twice as many; i = 2 for NaCl, 1 for sucrose D. NaCl, twice as many; i = 1 for both

Question 9

A 0.20 M NaCl solution is at 27 °C. R = 0.0831 L·bar/(mol·K).

What is its solute potential?

A. −9.97 bar B. −4.99 bar C. +9.97 bar D. −0.90 bar

Question 10

A 0.25 M glucose solution is at 30 °C. Glucose stays as whole molecules when it dissolves. R = 0.0831 L·bar/(mol·K).

What is the solute potential of the solution?

A. −12.6 bar B. −0.623 bar C. +6.29 bar D. −6.29 bar

Question 11

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. Net movement into the cell: 9 is bigger than 6 B. Net movement into the cell: the solution has more solute C. No net movement: both values are negative D. Net movement out of the cell: −9 is lower than −6

Question 12

A cell with a water potential of −5.0 bar is placed in an open beaker of 0.12 M NaCl at 20 °C. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves. R = 0.0831 L·bar/(mol·K).

What happens to the cell's water, and why?

A. The cell gains water: the solution, at −0.399 bar, is higher than the cell's −5.0 bar B. The cell gains water: the solution, at −2.92 bar, is higher than the cell's −5.0 bar C. The cell loses water: the solution, at −5.84 bar, is lower than the cell's −5.0 bar D. There is no net movement: the two values are both negative and within 1 bar of each other

Question 13

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. +5 bar B. +8 bar C. −5 bar D. +11 bar

Question 14

A walled cell has a solute potential of −9.5 bar; assume this does not change as water moves. It is placed in a solution with a water potential of −3.5 bar and left until water stops entering.

What pressure potential has the cell reached?

A. +9.5 bar B. −6.0 bar C. +13.0 bar D. +6.0 bar

Question 15

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. 0.10 M B. 0.20 M C. 0.40 M D. 2.7 M

Question 16

A piece of plant tissue has a water potential of −7.50 bar at 27 °C. A student wants a sucrose solution, in an open beaker at the same temperature, in which the tissue neither gains nor loses water. R = 0.0831 L·bar/(mol·K).

What sucrose concentration is needed?

A. 0.150 mol/L B. 3.34 mol/L C. 0.301 mol/L D. −0.301 mol/L

Question 17

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. −1.5% B. +12% C. −12% D. −14%

Question 18

Percent change in mass of a potato core after soaking in each sucrose solution.

The table shows a student's results: 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. Percent change on the x-axis, concentration (M) on the y-axis, one straight best-fit line B. Concentration (M) on the x-axis, percent change on the y-axis, points joined dot to dot C. Concentration (M) on the x-axis, final mass (g) on the y-axis, one straight best-fit line D. Concentration (M) on the x-axis, percent change on the y-axis, one straight best-fit line

Question 19

Percent change in mass of sweet potato cores after an hour in six sucrose solutions at 25 °C, one core per solution, with the best-fit straight line. The dashed line drops from the zero crossing to the concentration axis.

A class cuts six cores from one sweet potato, weighs them, leaves each in a different sucrose solution at 25 °C for an hour, and weighs them again. The graph gives the percent change in mass of each core and the best-fit straight line, which crosses zero at 0.32 M. R = 0.0831 L·bar/(mol·K).

What is the water potential of the sweet potato tissue?

A. −15.8 bar B. −7.92 bar C. −0.665 bar D. −9.91 bar

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 4 points

A student investigates the water potential of potato tissue. Six cores are cut from one potato with the same cutter, blotted dry and weighed. Each core is placed in a beaker of sucrose solution at 22 °C: 0.0, 0.2, 0.4, 0.6, 0.8 or 1.0 M. After 24 hours each core is blotted and weighed again. The membranes of the potato cells are selectively permeable: they let water through but not sucrose. R = 0.0831 L·bar/(mol·K). The results are in the table: 0.0 M, 10.2 g to 11.7 g, +14.7%; 0.2 M, 10.0 g to 10.8 g, +8.0%; 0.4 M, 10.1 g to 10.2 g, +1.0%; 0.6 M, 9.9 g to 9.1 g, −8.1%; 0.8 M, 10.0 g to 8.3 g, −17.0%; 1.0 M, 10.3 g to 7.7 g, −25.2%.

Mass of each potato core before and after 24 hours in sucrose solution at 22 °C.

(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

water moved into the potato cells by osmosis (net water movement) because pure water has less solute than the cells (or a higher water potential), and sucrose cannot cross the membranes; the extra water adds mass.

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

all three: independent variable, sucrose concentration (M); dependent variable, percent change in mass (or change in mass); one controlled variable such as temperature (22 °C), time (24 hours), the potato used, the size and shape of the cores (same cutter), or blotting before weighing.

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).

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

Ψs = −iCRT

Ψs = −1 × 0.42 × 0.0831 × 295

Ψs = −10.3 bar

Check the box for each point your answer earns

an isotonic concentration between 0.40 and 0.45 M (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) AND the tissue's water potential from Ψs = −iCRT with i = 1 and T = 295 K: for 0.42 M, Ψ = −1 × 0.42 × 0.0831 × 295 = −10.3 bar (accept −9.8 to −11.0 bar, negative, in bars).

Accept 0.40 M read straight from the table, giving −9.8 bar. Score the water potential against the student's own estimate: any estimate from 0.40 to 0.45 M carried correctly into Ψs = −iCRT earns the point. 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.

i = 1

C = 0.3 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Ψs = −iCRT

Ψs = −1 × 0.3 × 0.0831 × 295

Ψs = −7.4 bar

Check the box for each point your answer earns

the core gains mass, because 0.3 M sucrose has Ψ = −1 × 0.3 × 0.0831 × 295 = −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.

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.

Free-response score: 0 of 4

Free response 2 · Conceptual Analysis · 4 points

A cell from the stem of a plant has a rigid cell wall. Its solute potential is −8 bar; assume this does not change as the cell gains or loses water. The cell is placed in an open beaker of dilute solution whose water potential is −2 bar. At the moment it is put in, its contents rest against the wall without pushing on it. Later, once net water movement has stopped, the solution is poured away and replaced with 0.5 M sucrose at 22 °C. R = 0.0831 L·bar/(mol·K).

(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.

Ψp = 0 bar (the contents rest against the wall without pushing)

Ψs = −8 bar

Ψ = Ψp + Ψs

Ψ = 0 + (−8)

Ψ = −8 bar

Check the box for each point your answer earns

water potential measures how strongly water tends to move (water moves from higher Ψ to lower Ψ), AND at the start Ψ = Ψp + Ψs = 0 + (−8) = −8 bar.

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.

Ψ = −2 bar (the solution’s, once net movement has stopped)

Ψs = −8 bar

Ψ = Ψp + Ψs

Ψp = Ψ − Ψs

Ψp = (−2) − (−8)

Ψp = +6 bar

Check the box for each point your answer earns

water moves from the beaker (Ψ = −2 bar, higher) into the cell (Ψ = −8 bar, lower); 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 (Ψp = +6 bar), when there is no longer a difference to drive net movement.

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.

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

Ψs = −iCRT

Ψs = −1 × 0.5 × 0.0831 × 295

Ψs = −12.3 bar

Check the box for each point your answer earns

the new solution has Ψ = Ψs = −1 × 0.5 × 0.0831 × 295 = −12.3 bar (accept −12.2 to −12.4 bar); water leaves the cell, the push against the wall falls to zero (the cell loses turgor), and the contents shrink away from the wall (plasmolysis) while the wall keeps its shape.

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.

Accept either treatment of the cell's solute potential as water leaves (held at −8 bar as the stem assumes, or becoming more negative as the contents concentrate); the prediction, water leaving until the contents pull away from the wall, is the same.

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) Calculate the sucrose concentration at 22 °C that would leave the cell's contents just resting against the wall, with no pressure on it, and justify your answer. (1 point)

mol/L

Write down the values in the question:

Ψ of the solution needed = −8 bar

i = 1 (sucrose does not split)

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

−8 = −1 \times C \times 0.0831 \times 295

C = \frac{8}{0.0831 \times 295}

C = 0.326 mol/L \approx 0.33 mol/L

A full-credit answer: 0.33 mol/L sucrose. With no pressure on the wall, Ψp = 0 bar, so the cell's water potential is its solute potential, −8 bar. Net water movement stops when the solution has the same water potential, −8 bar.

Do not award the point for using −12.3 bar (the 0.5 M solution) or for a concentration found with T = 22.

Free-response score: 0 of 4

Multiple choice checked: 0 of 19 correct.

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