Diffusion and osmosis
Grade 9 Science Worksheet - Challenge Me

Circle the best answer.
Ari reads a graph of particles on two sides of a membrane they cross. What does the flat part of both lines, after 30 minutes, show?
- The membrane ran out of energy for moving particles at 30 minutes
- Particles stopped crossing once the sides reached 50
- The sides are equal near 50, with particles still crossing both ways
- Particles crossed from side 1 to side 2, and no particle crossed back
Bo reads a graph of one substance crossing a membrane in warm water and in cold water. Which conclusion does the graph support?
Graph (model data). Particles that crossed a membrane, in warm and in cold water. 100 particles started on one side. Time (minutes) across, 0 to 40, a mark every 10. Particles crossed (count) up, 0 to 60, a mark every 10. Warm line joins (0, 0), (10, 30), (20, 45), (30, 50), (40, 50) with straight pieces. Cold line joins (0, 0), (10, 15), (20, 28), (30, 40), (40, 50) with straight pieces. Graph (model data). Particles that crossed a membrane, in warm and in cold water. 100 particles started on one side. Time (minutes) across, 0 to 40, a mark every 10. Particles crossed (count) up, 0 to 60, a mark every 10. Warm line joins (0, 0), (10, 30), (20, 45), (30, 50), (40, 50) with straight pieces. Cold line joins (0, 0), (10, 15), (20, 28), (30, 40), (40, 50) with straight pieces. Particles that crossed a membrane 0 10 20 30 40 0 10 20 30 40 50 60 Time (minutes) Particles crossed (count) Warm Cold 100 particles started on one side. - Cold water is slower because the membrane spends less energy there
- Warm particles stop moving after 30 minutes, when the line goes flat
- Warmth raises the rate but not the end point: both reach 50
- Warmth changes the end point: warm water ends higher than cold
Esme reads a graph of one substance moving into a lab chamber. The line climbs steeply, then flattens. Why does the climb slow down?
Graph (model data). Concentration in a lab chamber that a substance crosses into from a second chamber of the same size, which started at 80. Time (minutes) across, 0 to 30, a mark every 5. Concentration (units) up, 0 to 40, a mark every 10. The line joins (0, 0), (5, 20), (10, 30), (15, 35), (20, 38), (25, 39), (30, 40) with straight pieces. Graph (model data). Concentration in a lab chamber that a substance crosses into from a second chamber of the same size, which started at 80. Time (minutes) across, 0 to 30, a mark every 5. Concentration (units) up, 0 to 40, a mark every 10. The line joins (0, 0), (5, 20), (10, 30), (15, 35), (20, 38), (25, 39), (30, 40) with straight pieces. Graph (model data). Concentration in a lab chamber that a substance crosses into from a second chamber of the same size, which started at 80. 0 5 10 15 20 25 30 0 10 20 30 40 Time (minutes) Concentration (units) - The chamber is running low on energy to power diffusion
- The gradient shrinks as the chambers near equal, so diffusion slows
- Particles cross into the chamber and do not cross back, so the first empties
- Particles slow down and stop moving as the line flattens
Read this first
Use this
Lab card (model data). A bag of dialysis tubing, a kind of lab membrane, held starch molecules and sat in a beaker with iodine molecules. The lab membrane lets iodine molecules cross but not starch molecules. Table with 2 rows. Columns: Particle | In the bag at 0 minutes | In the bag at 30 minutes | In the beaker at 30 minutes. Rows: starch | 10 | 10 | 0; iodine | 0 | 7 | 3. The beaker held 10 iodine molecules and no starch at 0 minutes. A teacher ran the test; read about only.
Jun reads a teacher's card about a bag of dialysis tubing, a lab membrane, set in a beaker. Which conclusion does the card support?
- Iodine moved into the bag because the membrane spent energy moving it
- Iodine moved into the bag, and no iodine molecule moved back out again
- Iodine moved into the bag down its gradient; starch stayed inside
- Starch moved out of the bag down its gradient as iodine moved in
Use this
Lab card (model data). A lab-made bilayer held 10 carrier proteins for glucose. The card gives the glucose outside and the rate of crossing. Table with 4 rows. Columns: Glucose outside (units) | Rate (molecules per second). Rows: 2 | 5; 4 | 9; 8 | 12; 16 | 12. A teacher ran the test; read about only.
Luz reads a card about glucose crossing through carrier proteins. The rate stays at 12 when glucose outside rises from 8 to 16. Why?
- The carriers ran out of cell energy once glucose reached 8 units
- Every carrier was busy, so more glucose could not raise the rate
- Glucose stopped moving at 8 units, because the sides reached equilibrium
- Glucose crossed the bilayer alone at 16 units, so carriers were not needed
Use this
Notes card written by a student. Line 1: Diffusion moves a substance from higher to lower concentration. Line 2: The cell spends ATP to run diffusion. Line 3: Warmer conditions speed diffusion up.
Milo's notes on diffusion have three lines. One line is wrong. Which correction fixes the notes?
- Line 2 is wrong: diffusion is passive, so the cell spends no ATP
- Line 3 is wrong: warmer conditions slow diffusion down
- Line 1 is wrong: substances move from lower to higher concentration
- Line 2 is right: diffusion through proteins also spends ATP
Use this
Notes card written by a student. The left arm holds 2 sugar dots and the right arm holds 8. Line 1: Sugar cannot cross the membrane. Line 2: Water crosses, so water moves toward the left arm, where there is less sugar. Line 3: The left level will rise.
Noah's notes explain a U-tube with a membrane that passes water but not sugar. One line is wrong. Which correction fixes the notes?
- Sugar moves to the left arm, which has less sugar, so its level rises
- Water moves to the right arm, which has more sugar, so its level falls
- Water moves to the right arm, which has more sugar, so its level rises
- Water moves to the left arm, which has less sugar, so its level rises
Pia's notes say a molecule in diffusion only moves from high to low concentration, so it never moves back. Which correction fixes the notes?
Notes card and membrane drawing (model, not to scale). A bilayer of phospholipids with watery fluid outside the cell above and watery fluid inside the cell below. No proteins are drawn. Outside the cell: 12 oxygen molecules. Inside the cell: 4 oxygen molecules. A long arrow points inward and a short arrow points outward. Notes card and membrane drawing (model, not to scale). A bilayer of phospholipids with watery fluid outside the cell above and watery fluid inside the cell below. No proteins are drawn. Outside the cell: 12 oxygen molecules. Inside the cell: 4 oxygen molecules. A long arrow points inward and a short arrow points outward. Notes watery fluid outside the cell watery fluid inside the cell model, not to scale No proteins are drawn. 12 oxygen molecules 4 oxygen molecules O O O O O O O O O O O O O O O O inward outward - No correction: molecules move from high to low, and a move back needs cell energy
- Molecules move both ways until equilibrium, and then they stop moving
- Molecules move both ways, but more move outward, so the net movement is outward
- Molecules move at random both ways; the net movement is from high to low
Ren compares claims about glucose and carrier protein C when both sides hold 8 glucose molecules. Which claim, with its reason, fits?
Membrane drawing (model, not to scale). A bilayer of phospholipids with watery fluid outside the cell above and watery fluid inside the cell below. Letter C points to a protein that reaches all the way across the bilayer and changes shape to carry glucose. Outside the cell: 8 glucose molecules. Inside the cell: 8 glucose molecules. Membrane drawing (model, not to scale). A bilayer of phospholipids with watery fluid outside the cell above and watery fluid inside the cell below. Letter C points to a protein that reaches all the way across the bilayer and changes shape to carry glucose. Outside the cell: 8 glucose molecules. Inside the cell: 8 glucose molecules. model, not to scale watery fluid outside the cell watery fluid inside the cell glucose phospholipids C - No net movement of glucose, because the two sides match and the gradient is gone
- No glucose crosses, because the carrier stops working when the sides match
- Net movement inward, because the carrier uses cell energy to pull glucose in
- Net movement inward, because the carrier points toward the inside
Use the table
Model data (not to scale)
Cell Width (units) Cell 1 10 Cell 2 40 Both cells start with the same oxygen concentration at the surface and none at the center. Oxygen enters by diffusion at the same temperature.
Tala compares two round model cells at the same temperature. Which claim, with its reason, fits for oxygen reaching the center?
- Both cells get oxygen to the center together, because diffusion is passive
- Cell 2 gets oxygen to its center sooner, because a larger cell has a steeper gradient
- Cell 1 gets oxygen to its center sooner, because the distance is shorter
- Cell 2 gets oxygen to its center sooner, because a larger cell spends more energy