Practice questions · Topic 2.7

Unit 2 · Practice for the Topic 2.7 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 hen's egg has its shell dissolved away, leaving the thin membrane beneath, which lets water through but not sugar. The egg is weighed and left overnight in corn syrup, a concentrated sugar solution. In the morning it is lighter and shrunken.

What moved, and which way?

A. Water left the egg for the syrup, toward the side with more solute B. Sugar entered the egg from the syrup, toward the side with less sugar C. Water entered the egg from the syrup, toward the side with more water D. Sugar left the egg for the syrup, and water followed it out

Question 2

Sea urchin eggs, each 270 pL (picoliters) in volume, were placed in three solutions of a solute that cannot cross their membranes. After ten minutes: in 0.60 mol/L the eggs were 350 pL; in 1.00 mol/L, 270 pL; in 1.40 mol/L, 210 pL.

How should the 0.60 mol/L solution be classified relative to the eggs, and what did water do?

A. Hypotonic: water entered the eggs B. Hypertonic: water left the eggs C. Isotonic: water crossed both ways at equal rates D. Hypertonic: water entered the eggs

Question 3

An animal cell holds 0.30 mol/L of solute that cannot cross its membrane. It is placed in 0.30 mol/L glycerol, a small molecule that crosses the membrane freely. At first the cell's volume holds. Over the next half hour the glycerol spreads until it is equal inside and outside the cell.

Predict what happens to the cell's volume over that half hour.

A. The volume holds: the total solute concentration is 0.30 mol/L on both sides B. The volume grows: only the cell's own solute now counts, so water enters C. The volume shrinks: the glycerol inside adds to the cell's solute and draws water out D. The volume holds: a solute that crosses the membrane carries water along with it

Question 4

Red blood cells from a marine fish hold about 0.35 mol/L of solute that cannot cross their membranes. A few drops of the fish's blood fall into a freshwater stream, whose water holds almost no solute.

Predict what happens to the red blood cells.

A. The cells shrink as water leaves them for the stream B. The cells hold their volume: the stream is isotonic to them C. The cells swell until their membranes tear: osmotic lysis D. The cells shrink as their solute leaks out into the stream

Question 5

An Amoeba, a single-celled organism with no cell wall, lives in pond water that holds far less solute than its cytosol. In pond water its contractile vacuole fills and empties every 30 seconds. Moved into water holding 0.08 mol/L of a solute its membrane blocks, it empties its vacuole every 4 minutes.

Explain the change in the emptying rate.

A. The vacuole now keeps water out, so it rarely needs to empty B. Less water enters in the new solution, so the vacuole fills more slowly C. Solute enters the Amoeba, and the vacuole must remove the solute instead D. More water enters in the new solution, and the vacuole stores it

Question 6

A seabird drinks seawater, which holds far more solute than its blood. A gland above each eye pumps salt out of the blood into a very salty fluid that drips from the bird's beak, and the bird's blood stays near 0.30 mol/L solute all day.

Which term names what the bird is doing?

A. Osmotic lysis B. Facilitated diffusion C. Plasmolysis D. Osmoregulation

Question 7

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 8

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 9

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 10

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%

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%.

Write down the values in the question:

initial mass = 6.2 g
final mass = 6.8 g

Write down the equation:

                 final − initial
percent change = ─────────────── × 100
                     initial

Substitute in the values, and calculate:

percent change = (6.8 − 6.2) / 6.2 × 100
percent change = 0.6 / 6.2 × 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.

Write down the values in the question:

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

Write down the equation:

Ψs = −iCRT

Substitute in the values, and calculate:

Ψs = −1 × 0.55 × 0.0831 × 298
Ψs = −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.

Write down the values in the question:

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)

Write down the equation:

Ψs = −iCRT

Substitute in the values, and calculate:

Ψs = −1 × 0.70 × 0.0831 × 298
Ψs = −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)

A full-credit answer: The solution's water potential is −2.48 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:

Ψs = −iCRT

Substitute in the values, and calculate:

Ψs = −1 × 0.10 × 0.0831 × 298
Ψs = −2.48 bar
Ψ = Ψp + Ψs = 0 + (−2.48) = −2.48 bar

Check the box for each point your answer earns

Ψ = Ψs = −1 × 0.10 × 0.0831 × 298 = −2.48 bar (accept −2.4 to −2.5 bar), with Ψp = 0 bar in the open beaker.

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, which gives −0.21 bar. Convert to kelvin first.

(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.

Write down the values in the question:

Ψ of the cell at equilibrium = Ψ of the solution = −2.48 bar
Ψs of the cell = −6.5 bar

Write down the equation:

Ψ = Ψp + Ψs

Make Ψp the subject:

Ψp = Ψ − Ψs

Substitute in the values, and calculate:

Ψp = (−2.48) − (−6.5)
Ψp = +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.

Write down the values in the question:

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

Write down the equation:

Ψs = −iCRT

Make C the subject:

      −Ψs
C = ─────
      iRT

Substitute in the values, and calculate:

C = 6.5 / (1 × 0.0831 × 298)
C = 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 ÷ (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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