Water Potential

Unit 2 · Topic 2.7b end-of-topic test

Suggested time: about 48 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. 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

Water potential (Ψ) is measured in bars. Pure water in an open beaker has Ψ = 0 bar.

What does a water potential measure?

A. How strongly water tends to move B. How much water a solution holds C. How much solute is dissolved in a solution D. How fast the water is moving

Question 2

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

The figure shows three open beakers at the same temperature, with concentrations given in M, short for mol/L: 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 guard cell in a leaf sits in pond water in an open dish. The cell’s contents push outward on its cell wall at 7 bar and the cell wall pushes back, so the cell is turgid. Nothing presses on the pond water in the dish.

What are the pressure potentials of the cell contents and of the pond water?

A. Cell −7 bar; pond water 0 bar B. Cell +7 bar; pond water 0 bar C. Cell 0 bar; pond water +7 bar D. Cell +7 bar; pond water +7 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, and explain the difference.

A. The pressed solution has the higher water potential, by 3 bar, because pressure raises water potential 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 leaf cell has a pressure potential of +2.5 bar and a solute potential of −12 bar.

What is its water potential?

A. −14.5 bar B. −9.5 bar C. +9.5 bar D. −12 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, and why?

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 0.10 M solutions: one of potassium chloride, KCl, and one of fructose, a sugar. KCl splits completely into K⁺ and Cl⁻ when it dissolves; fructose stays as whole molecules.

Which solution has more dissolved particles, and what value of i does each solute take?

A. Fructose, twice as many; i = 2 for fructose, 1 for KCl B. The same number; i = 1 for both C. KCl, twice as many; i = 2 for KCl, 1 for fructose D. KCl, twice as many; i = 1 for both

Question 9

A 0.18 M NaCl solution is at 30 °C. R = 0.0831 L·bar/(mol·K).

What is its solute potential?

A. −9.06 bar B. −4.53 bar C. +9.06 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 sits in a solution whose water potential is −9 bar.

What happens to the cell's water?

A. Net movement of water into the cell B. Net movement of solute into the cell C. No net movement of water D. Net movement of water out of the cell

Question 12

A cell with a water potential of −5.0 bar sits in an open beaker of 0.14 M NaCl at 22 °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 −0.512 bar, is higher than the cell's −5.0 bar B. The cell gains water: the solution, at −3.43 bar, is higher than the cell's −5.0 bar C. The cell loses water: the solution, at −6.86 bar, is lower than the cell's −5.0 bar D. There is no net movement: the two values are both negative and within 2 bar of each other

Question 13

A root cell of a salt-marsh plant has a rigid cell wall and a solute potential of −8 bar; assume this does not change as water moves. It sits in an open beaker of solution at Ψ = −3 bar 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 −5.5 bar; assume this does not change as water moves. It sits in an open beaker of solution with a water potential of −5.5 bar until its water potential matches the solution’s.

What pressure potential has the cell reached?

A. +5.5 bar B. −5.5 bar C. Cannot tell: the equation gives no answer when the two values are equal D. 0 bar

Question 15

A cell from a moss leaf has a water potential of −3.0 bar. R = 0.0831 L·bar/(mol·K).

Which sucrose concentration, in an open beaker at 20 °C, has the same water potential as the moss cell?

A. 0.06 M B. 0.12 M C. 0.25 M D. 1.8 M

Question 16

A piece of plant tissue has a water potential of −7.50 bar at 27 °C. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves. R = 0.0831 L·bar/(mol·K).

Which NaCl concentration, in an open beaker at 27 °C, has the same water potential as the tissue?

A. 0.301 mol/L B. 1.67 mol/L C. 0.150 mol/L D. 6.65 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 carry some measurement scatter.

Which way of drawing the graph lets the isotonic concentration be read from these results?

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.

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

What is the water potential of the sweet potato tissue?

A. −24.8 bar B. −12.4 bar C. −1.04 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. The student cuts six cores from one potato with the same cutter, blots them dry and measures the mass of each. The student puts each core into a beaker of sucrose solution at 22 °C, at one of six concentrations from 0.0 M to 1.0 M. After 24 hours the student blots each core and measures its mass 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.

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

(a) 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.

(b) Explain why the core in 0.0 M sucrose gained mass. (1 point)

A full-credit answer: Pure water has no solute.
The potato cells have solute dissolved in their cytosol.
So the pure water has the higher water potential.
Water moves by osmosis from higher water potential to lower.
Sucrose cannot cross the membranes, so only water moves.
So water moved into the potato cells.
The extra water adds mass, so the core gained 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.

(c) Determine 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 sucrose is isotonic to the tissue.
The tissue has the same water potential as that solution.
In an open beaker Ψp = 0 bar, so Ψ = Ψs; sucrose has i = 1 and T = 22 + 273 = 295 K.
Ψs = −(1)(0.42)(0.0831)(295) = −10.3 bar.
Therefore the tissue is at about −10.3 bar.

i = 1

C = 0.42 mol/L

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

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

the decision that the isotonic concentration lies between 0.40 and 0.45 M AND the reasoning it rests on (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 M 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) The student puts a seventh core from the same potato into 0.3 M sucrose at 22 °C. Make a claim about whether it gains or loses mass, and support your claim using water potential. (1 point)

A full-credit answer: The core gains mass.
In an open beaker Ψ = Ψs.
For 0.3 M sucrose at 295 K, Ψs = −(1)(0.3)(0.0831)(295) = −7.4 bar.
The tissue is at about −10 bar.
So −7.4 bar is higher than −10 bar.
Water moves from higher water potential to lower.
Therefore water moves from the solution into the cells, and the core gains mass.

i = 1

C = 0.3 mol/L

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

Ψs = −iCRT

Ψs = −(1)(0.3)(0.0831)(295)

Ψs = −7.4 bar

Check the box for each point your answer earns

the claim that the core gains mass, supported by evidence AND reasoning: 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 support 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 claim with no comparison of water potentials or concentrations.

Common slip: Making the claim 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. A student puts the cell into an open beaker of dilute solution whose water potential is −2 bar. At the moment the cell goes in, its contents rest against the cell wall without pushing on it. Later, once net water movement has stopped, the student pours the solution away and replaces it 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 water potential to lower.
At the start the contents rest against the cell wall without pushing on it, so Ψp = 0 bar. Ψ = Ψp + Ψs = 0 + (−8) = −8 bar.
So the cell’s water potential at the start is −8 bar.

Ψp = 0 bar (the contents rest against the cell 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 higher water potential to lower.
The beaker is at −2 bar and the cell at −8 bar, so water moves from the beaker into the cell.
As water enters, the contents press against the cell wall, the cell wall presses back, and the pressure potential rises, raising the cell’s water potential.
When the cell’s water potential reaches −2 bar, Ψp = +6 bar and no difference is left, so net movement stops.

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 cell 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) Calculate the sucrose concentration at 22 °C that would leave the cell's contents just resting against the cell 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 = 295 K (22 + 273)

Write down the equation:

Ψ_{s} = - i C R T

Make C the subject:

C = \frac{- Ψ_{s}}{i R T}

Substitute the values into the equation:

C = \frac{(8)}{(1) (0.0831) (295)}

Calculate:

C = 0.33 mol/L

A full-credit answer: The contents just rest against the cell wall, so Ψp = 0 bar.
So the cell’s water potential equals its solute potential, −8 bar.
Net water movement stops when the solution has the same water potential, −8 bar.
In an open beaker Ψ = Ψs, so the solution needs Ψs = −8 bar.
C = −Ψs ÷ (iRT) = (8) ÷ ((1)(0.0831)(295)) = 0.33 mol/L.
So 0.33 mol/L sucrose leaves the contents just resting against the cell wall.

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

(d) 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 0.5 M sucrose at 22 °C in an open beaker, so Ψ = Ψs = −(1)(0.5)(0.0831)(295) = −12.3 bar.
−12.3 bar is lower than the cell’s water potential, so water leaves the cell.
As water leaves, the contents stop pressing on the cell wall, so the pressure potential falls to zero and the cell loses turgor.
The contents shrink away from the cell wall (plasmolysis); the cell wall keeps its shape.

i = 1

C = 0.5 mol/L

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

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 cell wall falls to zero (the cell loses turgor), and the contents shrink away from the cell wall (plasmolysis) while the cell wall keeps its shape.

Accept: 'water leaves the cell and the cell plasmolyzes' reasoned from the new solution's water potential alone: −12.3 bar is below −8 bar, the lowest water potential this cell can hold (Ψs = −8 bar with Ψp no lower than 0 bar), so water must leave whatever the cell's state after part (b). The −2 bar step from part (b) is not required.

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 cell 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.

Free-response score: 0 of 4

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