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Lab / experiment Grade 11 Engineering & Design

Modeling Osmosis: A Dialysis-Tubing Investigation for Desalination

A grade 11 lab where students track osmosis into dialysis tubing holding 20% and 40% sucrose, estimate osmotic pressure, and weigh reverse-osmosis trade-offs.

Objectives

  1. Students will measure and compare the total mass change, percent mass change and interval rates of osmotic water gain in dialysis-tubing cells holding 20% and 40% sucrose solutions.
  2. Students will calculate the ideal osmotic pressure of each solution and explain why reverse osmosis requires an applied pressure greater than the osmotic pressure of the water being treated.
  3. Students will evaluate energy-cost and other engineering trade-offs between treating brackish water and treating seawater by reverse osmosis.

Vocabulary

Reverse osmosis
Reverse osmosis cleans water by pushing it through a membrane with holes so tiny that salt and most other dissolved things cannot fit through. It can even turn seawater into fresh water.
Desalination
Turning salty water, like ocean water, into fresh water people can use. It works, but it takes a lot of energy and leaves behind very salty water that must be handled carefully.
Brine
The extra-salty water left over after desalination pulls fresh water out of seawater. Plants must dispose of it carefully so it does not harm sea life.

Teacher overview

Reverse osmosis, the process behind large desalination plants such as the one in Carlsbad, California, works against a natural process: osmosis, the movement of water across a selectively permeable membrane from a less concentrated solution toward a more concentrated one. In this lab, students fill dialysis tubing with sugar solutions at two concentrations, place each in fresh water, and measure how fast water moves in by weighing the tubes every 10 minutes. They then estimate the osmotic pressure of each solution and use those results to reason about why treating seawater by reverse osmosis takes more pressure and more energy than treating less salty water.

Sugar (sucrose) stands in for salt because dialysis tubing does not hold salt back. Its pores are small enough to retain large molecules such as starch and proteins, yet sodium and chloride ions pass through them easily, so a salt solution sealed in the tubing leaks salt into the surrounding water, the concentration difference collapses, and the mass gain is small and unreliable. Sucrose molecules are larger than those ions and cross the tubing only slowly, so over a 40-minute run the tubing behaves much like a membrane that rejects the dissolved substance. A reverse-osmosis membrane is engineered to do what dialysis tubing cannot: let water through while rejecting dissolved salt.

The lab also addresses a common misconception. Many students expect reverse osmosis to move water in the same direction as ordinary osmosis, when in fact it requires applied pressure to force water the opposite way, out of the concentrated solution and against the concentration difference this lab lets them observe directly.

Teacher preparation. Pre-soak the tubing for at least 5 minutes before class (or have groups do it in Step 3). The 40% solution dissolves slowly; groups can stir it in lukewarm water and let it cool, or the teacher can prepare class stocks of both solutions in advance using the same proportions (for example, 100 g of sugar made up to 500 mL for the 20% solution and 200 g made up to 500 mL for the 40% solution). Students work the reference-data questions during the 10-minute waits between weighings, which keeps the lab within a single class period.

Modeling Osmosis: A Dialysis-Tubing Investigation for Desalination — Grade 11 Engineering & Design
Name: ____________________________ Date: ______________

Procedure

Step 1. Safety: wear goggles while preparing and handling solutions, do not taste or drink any solution used in this lab, and wipe up spills promptly, since sugar solutions make floors and benches sticky and slippery. Wash hands at the end of the lab.

Step 2. Prepare two sucrose solutions. Solution A (20% sucrose): stir 10.0 g of table sugar into about 30 mL of distilled water in a small cup until the liquid is completely clear, pour it into the 100 mL graduated cylinder, rinse the cup with a few milliliters of distilled water into the cylinder, and add distilled water until the total volume reads 50 mL. Solution B (40% sucrose): repeat with 20.0 g of sugar, again making the total volume 50 mL. These are weight-per-volume percentages: 20 g of sugar in every 100 mL of solution for A and 40 g per 100 mL for B. Skip this step if your teacher has prepared the solutions for you.

Step 3. Soak two dialysis tubing segments in distilled water until soft, then rub one end of each between your fingers to open it. Seal the other end of each tube by folding it over and tying it tightly with string, or with a clip, about 2 cm from the end. Label one 600 mL beaker "A" and the other "B" with the marker.

Step 4. Measure 20 mL of Solution A in the 25 mL graduated cylinder and pour it into one tube using the funnel or pipette. Squeeze out as much air as you can and seal the open end the same way. Fill the second tube with 20 mL of Solution B and seal it. Leave some slack in each tube rather than stretching it tight.

Step 5. Rinse the outside of each filled tube with distilled water, blot it dry with a paper towel, and record its starting mass to the nearest 0.1 g in the 0-minute row of the data table.

Step 6. Pour 400 mL of distilled water into each beaker, submerge each tube in its labeled beaker so it is fully covered, and start the timer.

Step 7. Every 10 minutes for 40 minutes, remove each tube, blot the outside dry the same way each time, record its mass, and return it to its beaker within about 30 seconds. While you wait between weighings, work on the reference-data parts of questions 1 through 5, and finish the parts that use your own data after the last weighing.

Step 8. Handle the tubing gently throughout. If a tube tears or drips at a seal, or its mass suddenly drops, that trial is no longer valid; start it over with a fresh segment and note the restart in your data table.

Step 9. Clean up: pour the solutions and beaker water down the drain with running water, discard the tubing and string in the trash, and wipe down the bench and balance.

Data & Analysis

Your group's data

Time (min) Mass of Tube A, 20% sucrose (g) Mass of Tube B, 40% sucrose (g)
0
10
20
30
40

Reference data set. Suppose another grade 11 class ran this lab with the same solutions, the same 20 mL fill volume and the same 400 mL of distilled water, at a room temperature of 22 °C (295 K). One group recorded the masses below.

Time (min) Mass of Tube A, 20% sucrose (g) Mass of Tube B, 40% sucrose (g)
0 23.0 24.2
10 23.7 25.6
20 24.3 26.8
30 24.8 27.8
40 25.2 28.6

Useful values: the molar mass of sucrose is 342.3 g/mol; the gas constant is R = 0.0821 L·atm/(mol·K); and for an ideal solution of a substance that does not split into ions, the van 't Hoff equation gives osmotic pressure as π = MRT, where M is molarity in mol/L and T is temperature in kelvins. Use T = 295 K throughout.

  1. For the reference data set, calculate the total mass change (40-minute mass minus starting mass) and the percent mass change (total change divided by starting mass, times 100) for Tube A and Tube B. Then calculate the same values for your own group's tubes. Explain, in terms of osmosis, why both tubes gained mass rather than lost it, and why percent change is a fairer comparison than total change when starting masses differ.
  2. For the reference data set, calculate each tube's average rate of mass gain over the full 40 minutes, and its rate during the first 10-minute interval (0 to 10 minutes) and the last 10-minute interval (30 to 40 minutes), in grams per minute. Describe how the rate changed over time, then check whether your own data show the same pattern.
  3. Calculate the molarity of Solution A and of Solution B (round each to three significant figures), then use the van 't Hoff equation and your rounded molarities to estimate the ideal osmotic pressure of each solution in atmospheres, to one decimal place.
  4. Divide Solution B's osmotic pressure by Solution A's, and divide Tube B's total mass gain in the reference data by Tube A's. Compare the two ratios, then compare them with the same mass-gain ratio from your own data. What do these ratios suggest about how the size of the osmotic pressure difference affects the rate at which water crosses a membrane?
  5. Tube B started with 20 mL of 40% sucrose, which contains 8.0 g of sugar. Assume no sugar left the tube and that each gram of water gained added 1 mL to the volume inside. Estimate the sucrose concentration inside Tube B at 40 minutes, as a weight-per-volume percent to one decimal place, and the percent by which that concentration fell. Compare that drop with the percent by which Tube B's interval rate fell between the first and last 10-minute intervals. Does dilution alone explain the slowdown? Propose at least one other cause.
  6. A reverse-osmosis unit must do the opposite of what you observed: it pushes water out of a concentrated solution through a membrane, which requires an applied pressure greater than the solution's osmotic pressure. Using your answers to question 3, what is the minimum ideal pressure needed to begin pushing pure water out of Solution A, and out of Solution B? Then suppose a unit pulled enough pure water out of Solution A to cut its volume in half. What would the ideal osmotic pressure of the remaining concentrated solution (the brine) be, and what does that imply about the pressure a plant needs as it recovers more fresh water from each gallon of feed?
  7. The Carlsbad, California desalination plant produces about 50 million gallons of drinking water per day by seawater reverse osmosis, which uses roughly 11 to 15 kilowatt-hours (kWh) of energy per 1,000 gallons, compared with about 1 to 2 kWh per 1,000 gallons for conventional treatment and delivery of local surface water or groundwater. Calculate the range of daily energy use for 50 million gallons by seawater reverse osmosis and by conventional treatment. Then, using your answers to questions 3 and 6, explain why reverse osmosis of a lower-salinity brackish source uses less energy per gallon than reverse osmosis of seawater.
  8. A town is choosing between a reverse-osmosis plant that treats brackish groundwater and one that treats seawater. Name two criteria other than energy cost that should be weighed, say which you would rank higher and why, and support your reasoning with evidence from this investigation, including what it showed about the membrane itself.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: Tube A gained 2.2 g (9.6%) and Tube B gained 4.4 g (18.2%) in the reference data; own-data values vary by group.
    Tube A: 25.2 − 23.0 = 2.2 g, and 2.2 ÷ 23.0 × 100 = 9.565, or 9.6%. Tube B: 28.6 − 24.2 = 4.4 g, and 4.4 ÷ 24.2 × 100 = 18.18, or 18.2%. Each group's own values should show both tubes gaining mass, with Tube B gaining more. Both tubes gained mass because the solution inside had a higher concentration of dissolved sugar than the distilled water outside, so water crossed the membrane inward, from lower to higher solute concentration, while the sugar was largely held inside. Percent change is fairer because it scales the gain to each tube's starting mass, so a tube that started heavier does not appear to have gained more simply because it held more.

  2. Answer: Tube A averaged 0.055 g/min and Tube B 0.11 g/min; first and last intervals were 0.07 and 0.04 g/min for A and 0.14 and 0.08 g/min for B.
    Averages: 2.2 g ÷ 40 min = 0.055 g/min for A and 4.4 g ÷ 40 min = 0.11 g/min for B. Tube A gained 23.7 − 23.0 = 0.7 g in the first interval (0.7 ÷ 10 = 0.07 g/min) and 25.2 − 24.8 = 0.4 g in the last (0.04 g/min). Tube B gained 25.6 − 24.2 = 1.4 g in the first interval (0.14 g/min) and 28.6 − 27.8 = 0.8 g in the last (0.08 g/min). In both tubes the rate fell steadily with each interval. Most groups' own data show the same slowing; accept an own-data description that is consistent with the student's table.

  3. Answer: Solution A is 0.584 M with an ideal osmotic pressure of 14.1 atm; Solution B is 1.17 M with 28.3 atm.
    Solution A: 10.0 g ÷ 342.3 g/mol = 0.02921 mol in 0.0500 L, which is 0.5843 M, or 0.584 M. (Using 20 g in 100 mL gives the same result.) π = 0.584 × 0.0821 × 295 = 0.584 × 24.22 = 14.14, or 14.1 atm. Solution B: 20.0 g ÷ 342.3 g/mol = 0.05843 mol in 0.0500 L, which is 1.169 M, or 1.17 M. π = 1.17 × 24.22 = 28.34, or 28.3 atm. These are ideal estimates; real concentrated sugar solutions deviate somewhat from the van 't Hoff equation, so the true values are not exactly these, yet the comparison between the two solutions still holds.

  4. Answer: Both ratios are about 2.0 in the reference data; own-data ratios vary by group.
    Osmotic pressure ratio: 28.3 ÷ 14.1 = 2.007, or about 2.0. Mass-gain ratio: 4.4 ÷ 2.2 = 2.0. Doubling the solute concentration doubled the predicted osmotic pressure and, in the reference data, doubled the amount of water that crossed in the same time. This suggests the rate of water flow across a membrane is roughly proportional to the osmotic pressure difference that drives it, the same relationship engineers use when they size reverse-osmosis systems, where water flow depends on how far the applied pressure exceeds the osmotic pressure. Accept an own-data ratio that differs from 2.0 if the student explains a likely reason, such as a leaking seal, uneven blotting or a tube stretched tight at the start.

  5. Answer: About 32.8% at 40 minutes, a drop of about 18%, while the interval rate fell by about 43%; dilution alone does not explain the slowdown.
    Volume inside at 40 minutes: 20 mL + 4.4 mL = 24.4 mL. Concentration: 8.0 g ÷ 24.4 mL × 100 = 32.79, or 32.8% weight per volume. Drop: (40 − 32.8) ÷ 40 × 100 = 18%. Rate drop: (0.14 − 0.08) ÷ 0.14 × 100 = 42.9, or about 43%. Because the driving concentration fell by much less than the rate did, other causes must contribute. Accept any of these with a reasonable explanation: as the tube fills and its wall becomes taut, the stretched tubing presses on the solution inside, creating a pressure that pushes back against incoming water, which is a small-scale version of the applied pressure used in reverse osmosis; some sucrose slowly leaks through the tubing, lowering the concentration difference further; or water entering the tube dilutes the layer of solution right next to the membrane faster than that layer mixes with the rest, so the concentration at the membrane surface is lower than the average (engineers call this concentration polarization, and it also reduces performance in reverse-osmosis plants).

  6. Answer: More than 14.1 atm for Solution A and more than 28.3 atm for Solution B; at half volume, Solution A's brine reaches about 28.3 atm.
    Pressure must exceed the osmotic pressure before water flows outward through the membrane, so the minimum ideal pressures are just above 14.1 atm and 28.3 atm. Halving the volume of Solution A while keeping all the sugar doubles its molarity to 2 × 0.584 = 1.168 M, and 1.168 × 24.22 = 28.29, or 28.3 atm, the same as Solution B. As a plant recovers more fresh water from each gallon of feed, the brine left behind becomes more concentrated and its osmotic pressure rises, so the pumps must push harder at the concentrated end of the system. Strong answers note that a higher-salinity feed starts at a higher osmotic pressure and climbs from there, and that dissolved salt splits into two ions per formula unit, so a salt solution has roughly twice the osmotic effect of a sugar solution of the same molarity.

  7. Answer: About 550,000 to 750,000 kWh per day by seawater reverse osmosis, compared with about 50,000 to 100,000 kWh per day by conventional treatment.
    50,000,000 gallons ÷ 1,000 = 50,000 units of 1,000 gallons. Seawater reverse osmosis: 50,000 × 11 = 550,000 kWh and 50,000 × 15 = 750,000 kWh. Conventional: 50,000 × 1 = 50,000 kWh and 50,000 × 2 = 100,000 kWh. These are estimates from the industry ranges, not the plant's metered use. Brackish water has a much lower dissolved-salt concentration than seawater, so, as questions 3 and 6 showed for the sugar solutions, its osmotic pressure is lower, and the brine it produces at a given recovery is also lower in osmotic pressure. The pumps therefore need less pressure to push each gallon through the membrane, and pumping pressure is where most of the energy goes.

  8. Answer: Answers vary; accept two relevant criteria ranked with a justification supported by evidence from the lab.
    Relevant criteria include: brine disposal, since seawater brine is more concentrated, while an inland brackish plant may have no ocean outlet and must find another way to dispose of its brine; recovery rate, since the question 6 result shows that pushing recovery higher raises the required pressure, so a seawater plant typically recovers a smaller share of its feed; membrane selectivity and replacement, since the lab showed that a membrane is only useful if it holds back the dissolved substance, and higher-pressure seawater membranes and pumps cost more to buy and maintain; and long-term reliability of the source, since brackish aquifers can be depleted or become saltier over time while the ocean supply is steady. A strong answer prioritizes one criterion over another and explains the ranking for the town's situation rather than listing criteria without weighing them.

Differentiation

Support (below grade level)

Record the 0-minute masses alongside each group as a model for the later weighings, and give students the reference-data subtractions for question 1 already set up (25.2 − 23.0 and 28.6 − 24.2). Work question 3 for Solution A together as a class before students do Solution B on their own. Provide a sentence frame for the explanation in question 1: "Water moved into the tube because the sugar solution had a ___ concentration of dissolved particles than the water outside, so osmosis moved water toward the ___ concentration."

Extension (above grade level)

Have students add a third tube at an intermediate concentration, 30% sucrose (15.0 g made up to 50 mL, about 15 g of additional sugar per group), predict its ideal osmotic pressure before running it, and graph total mass gain against predicted osmotic pressure for all three tubes to test whether the relationship is close to proportional. As a second option, students can run a salt control to see why this lab uses sugar: seal a dilute salt solution (for example, 5 g of salt made up to 100 mL) in a third tube, place it in distilled water, and read the outside water with a conductivity or TDS meter every 5 minutes. A rising reading shows salt crossing the tubing, and students can explain in writing how that leak would undermine a desalination membrane and why reverse-osmosis membranes are built as dense films rather than porous sieves.

ELL supports

Provide a labeled diagram showing a membrane with small water molecules passing through and larger dissolved sugar particles held on one side, with the terms osmosis, membrane, concentration, osmotic pressure and brine marked directly on the diagram. Allow data recording and the calculations in questions 1 through 3 in numbers and short phrases before requiring full-sentence explanations for questions 4 through 8.