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Worksheet Grade 7 Science

Quarry Bend's Reverse Osmosis Unit: Percent Removal and Recovery

A grade 7 science worksheet on a fictional town's under-sink reverse osmosis unit: students calculate percent of dissolved solids removed and the unit's water recovery.

Objectives

  1. Students will calculate the percent of dissolved solids a reverse osmosis membrane removes from feed water.
  2. Students will calculate a reverse osmosis unit's percent water recovery and the volume it sends to drain.

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.
Total dissolved solids (TDS)
Total dissolved solids is a measure of everything dissolved in water that you cannot see or filter out, like minerals and salts. Water with very high TDS can taste salty or bitter, even though it might look perfectly clear.

Also: membrane, permeate, recovery

Teacher overview

Reverse osmosis pushes water through a thin membrane that blocks most dissolved salts and minerals, so the water that passes through, called permeate, is much cleaner than the water that went in. The membrane cannot recover every drop, though, because some water must carry the rejected salts away to the drain. This worksheet is set at a community center in the fictional town of Quarry Bend, where students track one under-sink unit and turn its readings into percent problems.

Quarry Bend's Reverse Osmosis Unit: Percent Removal and Recovery — Grade 7 Science
Name: ____________________________ Date: ______________

Student Worksheet

The fictional town of Quarry Bend runs an under-sink reverse osmosis unit at its community center. Use these readings for the problems below.

  • The feed water going in has a total dissolved solids (TDS) level of 600 mg/L.
  • The permeate coming out has a TDS level of 30 mg/L.
  • In one hour the unit takes in 80 gallons of feed water, produces 60 gallons of permeate, and sends 20 gallons to the drain.
  1. What percent of the dissolved solids does the membrane remove? Use (feed TDS minus permeate TDS) divided by feed TDS, times 100. Round to the nearest whole percent.
  2. The recovery of a unit is the permeate volume divided by the feed volume, times 100. What is this unit's percent recovery for the hour?
  3. What percent of the feed water is sent to the drain? Explain how your answer to this item and your answer to item 2 are related.
  4. In one or two sentences, explain why the permeate has a lower TDS level than the feed water.
  5. A second unit is rated to reject 98 percent of the dissolved solids. Starting from the same 600 mg/L feed, what would its permeate TDS be? Use feed TDS times (100 percent minus 98 percent).
  6. How many mg/L cleaner is the second unit's permeate than the first unit's permeate?
  7. The community center needs 300 gallons of permeate in a day. At the first unit's recovery rate of 75 percent, how many gallons of feed water must the unit take in?
  8. Using your answer to item 7, how many gallons go to the drain that day?
  9. Name one trade-off of using reverse osmosis, based on what you found in items 7 and 8.
  10. A sticker on the first unit claims it "reduces dissolved solids by about 95 percent." Using your answer to item 1, is that claim accurate? Explain.

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: 95%.
    (600 − 30) ÷ 600 × 100 = 570 ÷ 600 × 100 = 95 percent removed.
  2. Answer: 75%.
    60 ÷ 80 × 100 = 75 percent recovery.
  3. Answer: 25%.
    20 ÷ 80 × 100 = 25 percent to the drain. This plus the recovery from item 2 equals 100 percent, because every gallon of feed water becomes either permeate or drain water.
  4. Answer: The membrane blocks most of the dissolved salts.
    Only water molecules pass through easily, so the salts stay behind and the permeate ends up with a much lower TDS level than the feed.
  5. Answer: 12 mg/L.
    600 × (100 percent − 98 percent) = 600 × 0.02 = 12 mg/L in the permeate.
  6. Answer: 18 mg/L.
    30 − 12 = 18 mg/L cleaner for the second unit.
  7. Answer: 400 gallons.
    300 ÷ 0.75 = 400 gallons of feed water are needed to make 300 gallons of permeate at 75 percent recovery.
  8. Answer: 100 gallons.
    400 − 300 = 100 gallons sent to the drain that day.
  9. Answer: It wastes some water.
    The unit sends 100 gallons to the drain to make 300 gallons of clean permeate, so a quarter of the feed water is not delivered to the tap. Accept any answer that names the drain water or the extra energy the pump uses.
  10. Answer: Yes, the claim is accurate.
    Item 1 shows the unit removes 95 percent of the dissolved solids, which matches the sticker's claim of about 95 percent.

Differentiation

Support

Give students the removal and recovery formulas already written out on a reference strip, and work item 1 together as a class before they continue. Allow the recovery percent (75 percent) to be reused in item 7 without recomputing it.

Extension

Ask students to find what recovery rate the community center would need so that only 60 gallons of drain water are produced when making 300 gallons of permeate, and to show the calculation that supports their answer.

ELL supports

Pre-teach "feed," "permeate," "drain," and "recovery" with a simple labeled diagram showing water entering the membrane and splitting into a clean stream and a drain stream. Allow item 4 and item 9 to be answered with a labeled sketch plus a short phrase.