Practice questions · Topic 2.7b

Unit 2 · Practice for the Topic 2.7b end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Two parts and one equation

Water potential, its two parts, Ψs = −iCRT, which way water goes and how hard the cell wall pushes.

Watch next: from six masses to a water potential

Sum up the potato lab: percent change, the graph, the isotonic point, −9.3 bar.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. 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. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, 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 dry bean seed has a water potential of about −100 bar. It sits in a dish of pure water.

Which way does water move, and why?

A. Into the seed, because 0 bar is a higher water potential than −100 bar B. Out of the seed, because 100 is a bigger number than 0 C. There is no net movement, because a dry seed holds no water to exchange D. Into the seed, because the water at 0 bar has the lower water potential

Question 2

A syringe holds sucrose solution at a concentration of 0.3 mol/L, written 0.3 M. A membrane across its tip lets water through but not sucrose, and the tip stands in an open beaker of 0.3 M sucrose at the same temperature. A student pushes the plunger in and holds it, pressing the solution inside at 2 bar.

Which way does water move across the membrane while the plunger is held?

A. There is no net movement of water B. Out of the syringe into the beaker C. Into the syringe from the beaker D. Sucrose moves out of the syringe through the membrane

Question 3

A leaf cell has a pressure potential of +7 bar and a solute potential of −15 bar.

What is its water potential?

A. −22 bar B. +8 bar C. −8 bar D. −15 bar

Question 4

A student has a beaker of 0.10 M NaCl at 25 °C. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves; glucose stays as whole molecules.

Which glucose solution at 25 °C has the same solute potential as the 0.10 M NaCl?

A. 0.20 M glucose B. 0.10 M glucose C. 0.05 M glucose D. 0.40 M glucose

Question 5

A 0.12 M NaCl solution sits in an open beaker at 24 °C. R = 0.0831 L·bar/(mol·K).

What is its solute potential?

A. −2.96 bar B. +5.92 bar C. −0.48 bar D. −5.92 bar

Question 6

Three cells in a row inside a root, from the soil side to the center, with the water potential of each.

The figure shows three cells in a row inside a root. Cell A, nearest the soil, has Ψ = −3 bar; cell B has Ψ = −6 bar; cell C, nearest the center of the root, has Ψ = −10 bar.

In which direction does water move along the row?

A. From C to B to A B. From B outward, to A and to C C. No net movement of water along the row D. From A to B to C

Question 7

A walled cell has a solute potential of −12.0 bar; assume this does not change as water moves. It sits in an open beaker of 0.10 M sucrose at 27 °C until water stops entering. R = 0.0831 L·bar/(mol·K).

What pressure potential has the cell reached?

A. +12.0 bar B. −9.51 bar C. +9.51 bar D. +14.5 bar

Question 8

A piece of plant tissue has a water potential of −9.00 bar at 25 °C. R = 0.0831 L·bar/(mol·K).

Which sucrose concentration, in an open beaker at 25 °C, has the same water potential as the tissue?

A. 0.363 mol/L B. 0.182 mol/L C. 4.33 mol/L D. −0.363 mol/L

Question 9

A cucumber core has a mass of 7.0 g before it is placed in a solution and 8.4 g an hour later.

What is the percent change in mass?

A. +1.4% B. +20% C. +17% D. −20%

Question 10

Percent change in mass of parsnip cores after an hour in six sucrose solutions at 23 °C, one core per solution, with the best-fit straight line.

A class cuts six cores from one parsnip, measures the mass of each, leaves each in a different sucrose solution at 23 °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 parsnip tissue?

A. −14.8 bar B. −7.38 bar C. −0.573 bar D. −4.92 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 · 6 points

A class investigates the water potential of apple tissue. The class cuts six cores from one apple with the same cutter, blots them dry and measures the mass of each. The class puts each core into a beaker of sucrose solution at 25 °C, at one of six concentrations from 0.0 M to 1.0 M. After two hours the class blots each core and measures its mass again. The membranes of the apple cells let water through but not sucrose. R = 0.0831 L·bar/(mol·K). The table shows the results; one value, for the 0.2 M core, is left for you to calculate.

Mass of each apple core before and after two hours in sucrose solution at 25 °C, and the percent change in mass.

(a) Identify the independent variable and the dependent variable in this investigation. (1 point)

Hint: Which quantity did the class set before the experiment began, and which did they measure at the end?

A full-credit answer: The independent variable is the sucrose concentration, in M.
The dependent variable is the percent change in mass of the core.

Check the box for each point your answer earns

independent variable, sucrose concentration (M); dependent variable, percent change in mass (accept: change in mass, or mass of the core).

Do not award the point if the two are reversed. Naming a controlled variable (temperature, time, the apple) in place of either earns nothing.

Common slip: Reversing the two, or naming a controlled variable such as temperature. The concentration was set; the change in mass was measured.

(b) Calculate the missing value for the 0.2 M core. (1 point)

Hint: A percent change compares the change to a starting value. Which of the two masses is the starting value here, and what sign should a gain carry?

A full-credit answer: The starting mass is 6.2 g.
The final mass is 6.8 g.
percent change = ((6.8) − (6.2)) ÷ (6.2) × 100 = +9.7%.
So the 0.2 M core changed by +9.7%.

initial mass = 6.2 g

final mass = 6.8 g

percent change = \frac{final - initial}{initial} \times 100

percent change = \frac{(6.8) - (6.2)}{(6.2)} \times 100

percent change = +9.7 %

Check the box for each point your answer earns

+9.7% (accept +9.6 to +9.8%, with the positive sign and the % sign).

Do not award the point for +0.6 (the change in grams), for +8.8% (divided by the final mass) or for a negative value.

Common slip: Giving +0.6, the change in grams, or dividing by the final mass 6.8 g. Divide the change by the starting mass.

(c) Describe the pattern in the results. Determine the sucrose concentration that is isotonic to the apple tissue. (1 point)

Hint: Where in the table does the change in mass switch from a gain to a loss?

A full-credit answer: As the sucrose concentration rises, the gain in mass shrinks.
Above some concentration the gain turns into a loss.
The change is +3.9% at 0.4 M and −1.3% at 0.6 M.
So zero change falls between 0.4 M and 0.6 M.
−1.3% is nearer to zero than +3.9%, so the crossing lies nearer to 0.6 M.
Therefore about 0.55 M sucrose is isotonic to the tissue.

Check the box for each point your answer earns

the pattern (as the sucrose concentration rises, the gain in mass shrinks and becomes a loss) AND the decision that the isotonic concentration lies between 0.4 M (+3.9%) and 0.6 M (−1.3%), with the reasoning it rests on: zero change falls between a gain and a loss, nearer the smaller change. Any estimate from 0.50 M to 0.60 M earns the point; interpolation gives about 0.55 M.

Do not award the point for 0.4 M or 0.6 M read straight from the table with no interpolation, or for the pattern alone. Carry the student’s estimate forward into parts (e) and (f).

Common slip: Picking 0.4 M or 0.6 M straight from the table. Zero lies between the two rows; judge where.

(d) Explain why the cores in the more concentrated solutions lost mass. (1 point)

Hint: Which side has more solute per liter, and which way does water move across a membrane that lets water through but not sucrose?

A full-credit answer: In the concentrated solutions, the beaker held more solute per liter than the apple cells.
So the solution had the lower water potential.
Water moves by osmosis from higher water potential to lower.
So water left the apple cells for the solution.
Sucrose cannot cross the membranes, so only water moved.
The water lost is the mass lost.
So the cores lost mass.

Check the box for each point your answer earns

water left the apple cells by osmosis (net water movement), because those solutions held more solute than the cells (had a lower water potential) and sucrose cannot cross the membranes; the water lost is the mass lost.

Accept: 'water moves toward the side with more solute, which was the beaker'. Do not award the point for 'sucrose entered the cores' or for 'the cores dried out' with no reference to solute or water potential.

Common slip: Having sucrose move into the cores. Sucrose stays put; water moves toward the side with more solute.

(e) Calculate the water potential of the apple tissue at 25 °C, using your estimate from part (c). (1 point)

Hint: The formula is Ψs = −iCRT. What is Ψp for a solution in an open beaker, what is i for sucrose, and which unit does T need?

A full-credit answer: The apple tissue has the same water potential as the isotonic solution, 0.55 M sucrose.
In an open beaker Ψp = 0 bar, so Ψ = Ψs.
Sucrose has i = 1, and T = 298 K.
Ψs = −(1)(0.55)(0.0831)(298) = −13.6 bar.
So the apple tissue is at about −13.6 bar.

i = 1

C = 0.55 mol/L

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψp = 0 bar (open beaker), so Ψ = Ψs

Ψ_{s} = - i C R T

Ψ_{s} = - (1) (0.55) (0.0831) (298)

Ψ_{s} = −13.6 bar

Check the box for each point your answer earns

Ψ = Ψs of the isotonic solution = −1 × C × 0.0831 × 298, using the student’s own estimate from part (c) with i = 1 and T = 298 K. For C = 0.55 M this is −13.6 bar. Any value from −12.4 bar (C = 0.50 M) to −14.9 bar (C = 0.60 M) earns the point, provided it matches the student’s part (c) estimate; negative, in bars.

Do not award the point for a positive value, for 25 used in place of 298 K, or for i = 2.

Common slip: Using 25 in place of 298 K, or dropping the minus sign. T must be in kelvin, and a solute potential is negative.

(f) A seventh core from the same apple is placed in 0.70 M sucrose at 25 °C. Make a claim about whether it gains or loses mass, and support your claim using water potentials. (1 point)

Hint: Find the water potential of the 0.70 M solution, then ask which way water moves between two water potentials.

A full-credit answer: The core loses mass.
In an open beaker Ψ = Ψs.
For 0.70 M sucrose at 298 K, Ψs = −(1)(0.70)(0.0831)(298) = −17.3 bar.
The tissue is at about −13.6 bar.
So −17.3 bar is lower than −13.6 bar.
Water moves from higher water potential to lower.
Therefore water moves from the cells into the solution, and the core loses mass.

i = 1

C = 0.70 mol/L

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψ of the tissue = −13.6 bar (from part e)

Ψ_{s} = - i C R T

Ψ_{s} = - (1) (0.70) (0.0831) (298)

Ψ_{s} = −17.3 bar

−17.3 bar is lower than −13.6 bar, so water leaves the core

Check the box for each point your answer earns

the claim that the core loses mass, supported by evidence AND reasoning: 0.70 M sucrose at 25 °C has Ψ = −(1)(0.70)(0.0831)(298) = −17.3 bar (accept −17.2 to −17.4 bar), which is lower (more negative) than the tissue’s value from the student’s own part (e) (anywhere from −12.4 to −14.9 bar; about −13.6 bar for 0.55 M), so net water movement is from the cells into the solution.

Accept support from the table: 0.70 M lies between 0.6 M (−1.3%) and 0.8 M (−6.3%), both losses, so the core loses. Do not award the claim with no comparison of water potentials or table rows.

Common slip: Making the claim with no comparison of the two water potentials. The point needs the two values side by side.

Free-response score: 0 of 6

Free response 2 · Conceptual Analysis · 4 points

A cell from the leaf of a pondweed has a rigid cell wall. Its solute potential is −6.5 bar; assume this does not change as the cell gains or loses water. A student puts the cell into an open beaker of 0.10 M sucrose at 25 °C. At the moment the cell goes in, its contents rest against the cell wall without pushing on it. R = 0.0831 L·bar/(mol·K).

(a) Describe what the pressure potential of a cell is, and state the pressure potential of the sucrose solution in the open beaker. (1 point)

A full-credit answer: Pressure potential is the part of water potential that comes from pressure on the water.
Pressure pushing on water raises its water potential.
So the contents of a turgid cell, pressed against the cell wall, have a positive Ψp.
Nothing presses on the solution in the open beaker.
So the solution has Ψp = 0 bar.

Check the box for each point your answer earns

pressure potential is the part of water potential due to pressure on the water (pressure pushing on water raises its water potential; it is positive for the contents of a turgid cell pressed against its cell wall), AND Ψp = 0 bar for the solution in the open beaker, because nothing presses on it.

Accept: 'the push of the cell wall on the contents' as the description. Both the description and the 0 bar are needed for the point.

Common slip: Giving Ψp = 0 bar with no statement of what pressure potential is, or the other way around. Both parts are needed.

(b) Calculate the water potential of the 0.10 M sucrose solution. (1 point)

bar

Write down the values in the question:

i = 1

C = 0.10 mol/L

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψp = 0 bar (open beaker), so Ψ = Ψs

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.10) (0.0831) (298)

Calculate:

Ψ_{s} = −2.48 bar

Ψ = 0 + (−2.48) = −2.48 bar

A full-credit answer: In an open beaker Ψp = 0 bar, so Ψ = Ψs.
Sucrose has i = 1, and T = 25 + 273 = 298 K.
Ψs = −(1)(0.10)(0.0831)(298) = −2.48 bar.
So the solution’s water potential is −2.48 bar.

Do not award the point for a positive value, for 25 used in place of 298 K, or for i = 2.

(c) Predict which way water moves at first, and calculate the pressure potential the cell reaches once net water movement has stopped. (1 point)

A full-credit answer: Water moves from higher water potential to lower.
The solution is at −2.48 bar.
The cell is at −6.5 bar.
So water moves into the cell.
As water enters, the contents press against the cell wall.
The cell wall presses back, so the pressure potential rises.
Net movement stops when the cell's water potential equals −2.48 bar. Ψp = Ψ − Ψs = (−2.48) − (−6.5) = +4.0 bar.

Ψ of the cell at equilibrium = Ψ of the solution = −2.48 bar

Ψs of the cell = −6.5 bar

Ψ = Ψ_{p} + Ψ_{s}

Ψ_{p} = Ψ - Ψ_{s}

Ψ_{p} = (−2.48) - (−6.5)

Ψ_{p} = +4.0 bar

Check the box for each point your answer earns

water moves into the cell, because the solution's −2.48 bar is higher than the cell's −6.5 bar, AND the cell's Ψ rises to match −2.48 bar, so Ψp = Ψ − Ψs = (−2.48) − (−6.5) = +4.0 bar (accept +3.9 to +4.1 bar).

Accept 'the cell becomes turgid' for the effect. Do not award the point for water leaving the cell, or for Ψp = +6.5 bar (the pure-water value).

Common slip: Giving +6.5 bar, the pressure the cell would reach in pure water. This solution is at −2.48 bar, so the cell's Ψ only has to rise that far.

(d) Calculate the sucrose concentration at 25 °C in which the cell's contents would just stop pressing on the cell wall (Ψp = 0 bar), and predict what the cell looks like in a solution more concentrated than that. (1 point)

A full-credit answer: At Ψp = 0 bar the cell's water potential is its solute potential, −6.5 bar.
So the matching solution must be at −6.5 bar.
C = −Ψs ÷ (iRT) = (6.5) ÷ ((1)(0.0831)(298)) = 0.26 M sucrose.
In a more concentrated solution, the solution's water potential is lower than the cell's.
So water leaves the cell.
The contents shrink away from the cell wall (plasmolysis).
The cell wall keeps its shape.
The tissue goes limp.

Ψp = 0 bar, so Ψ of the cell = Ψs = −6.5 bar

the solution must have Ψs = −6.5 bar

i = 1

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψ_{s} = - i C R T

C = \frac{- Ψ_{s}}{i R T}

C = \frac{(6.5)}{(1) (0.0831) (298)}

C = 0.26 mol/L

Check the box for each point your answer earns

at Ψp = 0 bar the cell's Ψ is its Ψs, −6.5 bar, so the solution must be at −6.5 bar: C = (6.5) divided by ((1)(0.0831)(298)) = 0.26 M (accept 0.25 to 0.27 M), AND in a more concentrated solution water leaves, the contents shrink away from the cell wall (plasmolysis) while the cell wall keeps its shape, and the tissue goes limp.

Accept 'the cell loses turgor' or 'goes limp' for the effect. Do not award the point for a negative concentration, for 25 used in place of 298 K, or for a cell that bursts (walled cells do not).

Common slip: Answering −0.26 M. A concentration is never negative; the two minus signs cancel when Ψs is negative.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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