Practice questions · Topic 2.7b

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

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 tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences and show any calculation. 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 is placed 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 0.3 M sucrose solution. 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 on the plunger 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, because the two solutions have the same concentration B. Out of the syringe into the beaker, because pressure has raised the water potential inside C. Into the syringe from the beaker, because pressure lowers the water potential inside D. Sucrose moves out of the syringe, pushed through the membrane by the plunger

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: toward the soil, where water enters the root B. From C to B to A: toward the bigger number, ignoring the sign C. No net movement: all three values are negative D. From A to B to C: toward the lower water potential

Question 7

A walled cell has a solute potential of −12.0 bar; assume this does not change as water moves. It is placed in an open beaker of 0.10 M sucrose at 27 °C and left 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. 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. 0.363 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 20 °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 parsnip, weighs them, leaves each in a different sucrose solution at 20 °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 0.28 M. R = 0.0831 L·bar/(mol·K).

What is the water potential of the parsnip tissue?

A. −13.6 bar B. −6.82 bar C. −0.465 bar D. −4.87 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. Six cores are cut from one apple with the same cutter, blotted dry and weighed. Each core is placed in a beaker of sucrose solution at 25 °C: 0.0, 0.2, 0.4, 0.6, 0.8 or 1.0 M. After two hours each core is blotted and weighed 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; the percent change 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 percent change in mass of 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 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{final - initial}{initial} \times 100

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

percent change = \frac{0.6}{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, and estimate 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 and then turns into a loss. The change is +3.9% at 0.4 M and −1.3% at 0.6 M, so zero falls between them and nearer to 0.6 M: about 0.55 M is isotonic to the tissue.

Check the box for each point your answer earns

as the sucrose concentration rises, the gain in mass shrinks and becomes a loss, AND the isotonic concentration lies between 0.4 M (+3.9%) and 0.6 M (−1.3%), about 0.55 M (accept 0.5 to 0.6 M).

Accept: any estimate from 0.45 M to 0.60 M with the two rows named. 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.

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 cores lost mass because water left the apple cells by osmosis, from the cells toward the solution, which held more solute per liter (a lower water potential) than the cells. Sucrose cannot cross the membranes, so only water moved, and the water lost is the mass lost.

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 is at about −13.6 bar, the solute potential of the 0.55 M sucrose solution that is isotonic to it.

i = 1

C = 0.55 mol/L

R = 0.0831 L·bar/(mol·K)

T = 25 + 273 = 298 K

Ψp = 0 bar (open beaker), so Ψ = Ψs

Ψ_{s} = - i C R T

Ψ_{s} = - i C R T

Ψ_{s} = −1 \times 0.55 \times 0.0831 \times 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; for C = 0.55 M this is −13.6 bar (accept −12.4 to −14.9 bar for estimates from 0.50 to 0.60 M; 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. Predict whether it gains or loses mass, and justify your prediction 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. The 0.70 M solution has Ψ = −17.3 bar, which is lower, more negative, than the tissue's water potential of about −13.6 bar, so net water movement is from the cells into the solution.

i = 1

C = 0.70 mol/L

R = 0.0831 L·bar/(mol·K)

T = 25 + 273 = 298 K

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

Ψ_{s} = - i C R T

Ψ_{s} = - i C R T

Ψ_{s} = −1 \times 0.70 \times 0.0831 \times 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 core loses mass, because 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 about −13.6 bar, so net water movement is from the cells into the solution.

Accept a justification 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 point for the right prediction with no comparison of water potentials or table rows.

Common slip: Giving the right prediction 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. The cell is placed in an open beaker of 0.10 M sucrose at 25 °C. At the moment it is put in, its contents rest against the 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 a turgid cell pressed against its wall has a positive Ψp. The solution in the open beaker has Ψp = 0 bar, because nothing presses on it.

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 wall), AND Ψp = 0 bar for the solution in the open beaker, because nothing presses on it.

Accept: 'the push of the 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 = 25 + 273 = 298 K

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} = - i C R T

Ψ_{s} = −1 \times 0.10 \times 0.0831 \times 298

Ψ_{s} = −2.48 bar

Ψ = Ψ_{p} + Ψ_{s} = 0 + (−2.48) = −2.48 bar

A full-credit answer: 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 into the cell, because the solution at −2.48 bar has the higher water potential. As water enters, the contents press against the wall and the pressure potential rises until the cell's water potential equals −2.48 bar, at Ψp = +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} = 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 wall (Ψp = 0 bar), and predict what the cell looks like in a solution more concentrated than that. (1 point)

A full-credit answer: The matching concentration is 0.26 M sucrose. In anything more concentrated, water leaves the cell, the contents shrink away from the wall while the wall keeps its shape (plasmolysis), and 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 = 25 + 273 = 298 K

Ψ_{s} = - i C R T

C = \frac{- Ψ_{s}}{i R T}

C = \frac{- Ψ_{s}}{i R T}

C = \frac{6.5}{1 \times 0.0831 \times 298}

C = \frac{6.5}{24.8}

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 wall (plasmolysis) while the 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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