Osmosis Worksheet: Analyze water movement in animal cells placed in beakers with different solute concentrations.

Explanation:
We are given 9 beakers, each with an animal cell inside. Each beaker has two solutions:
- One inside the cell (inner circle)
- One outside the cell (outer ring)

For osmosis:
- Water moves from where it’s more concentrated (higher % H₂O / lower solute) → to where it’s less concentrated (lower % H₂O / higher solute).
- So: compare % water inside vs. outside.
- If inside has more water than outside → water leaves cell → hypertonic solution outside
- If inside has less water than outside → water enters cell → hypotonic solution outside
- If equal % water → no net movement → isotonic

Also, since % H₂O + % Solute = 100%, we can fill missing percentages easily.

Let’s go one beaker at a time:

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Beaker 1 (top-left)
Inside: 90% H₂O, 10% Solute
Outside: 40% H₂O, 60% Solute
→ Inside has more water → water moves out of cell
→ Outside is hypertonic
→ Circle: Hyper

Beaker 2 (top-middle)
Inside: 40% H₂O, 60% Solute
Outside: 70% H₂O, 30% Solute
→ Inside has less water → water moves in
→ Outside is hypotonic
→ Circle: Hypo

Beaker 3 (top-right)
Inside: 20% H₂O, 80% Solute
Outside: 20% H₂O, 80% Solute
→ Equal → no net movement
→ Isotonic
→ Circle: Iso

Beaker 4 (middle-left)
Inside: 55% H₂O, 45% Solute
Outside: 55% H₂O, ___% Solute
→ Since 55% H₂O, solute = 100 − 55 = 45%
→ Equal → isotonic
→ Circle: Iso

Beaker 5 (middle-center)
Inside: 15% H₂O, 85% Solute
Outside: 35% H₂O, ___% Solute
→ Outside solute = 100 − 35 = 65%
Compare water: inside 15% < outside 35% → water moves in
→ Outside is hypotonic
→ Circle: Hypo

Beaker 6 (middle-right)
Inside: ___% H₂O, 95% Solute
→ So H₂O = 100 − 95 = 5%
Outside: 25% H₂O, 75% Solute
Inside 5% H₂O < outside 25% H₂O → water moves in
→ Outside is hypotonic
→ Circle: Hypo

Beaker 7 (bottom-left)
Inside: ___% H₂O, 55% Solute
→ H₂O = 100 − 55 = 45%
Outside: ___% H₂O, 25% Solute
→ Outside H₂O = 100 − 25 = 75%
Inside 45% < outside 75% → water moves in
→ Outside is hypotonic
→ Circle: Hypo

Beaker 8 (bottom-center)
Inside: 80% H₂O, ___% Solute
→ Solute = 100 − 80 = 20%
Outside: ___% H₂O, 90% Solute
→ Outside H₂O = 100 − 90 = 10%
Inside 80% > outside 10% → water moves out
→ Outside is hypertonic
→ Circle: Hyper

Beaker 9 (bottom-right)
Inside: 5% H₂O, ___% Solute
→ Solute = 100 − 5 = 95%
Outside: 65% H₂O, ___% Solute
→ Solute = 100 − 65 = 35%
Inside 5% < outside 65% → water moves in
→ Outside is hypotonic
→ Circle: Hypo

Now summarize all missing percentages and circled choices:

1. No missing % — Hyper
2. No missing % — Hypo
3. No missing % — Iso
4. Outside solute = 45% — Iso
5. Outside solute = 65% — Hypo
6. Inside H₂O = 5% — Hypo
7. Inside H₂O = 45%, Outside H₂O = 75% — Hypo
8. Inside solute = 20%, Outside H₂O = 10% — Hyper
9. Inside solute = 95%, Outside solute = 35% — Hypo

Let me double-check each calculation:

- Beaker 4: 55% H₂O inside, 55% H₂O outside → solute both 45% ✔
- Beaker 5: inside 15% H₂O → 85% solute; outside 35% H₂O → 65% solute → water in ✔
- Beaker 6: inside 95% solute → 5% H₂O; outside 25% H₂O → 75% solute → water in ✔
- Beaker 7: inside 55% solute → 45% H₂O; outside 25% solute → 75% H₂O → water in ✔
- Beaker 8: inside 80% H₂O → 20% solute; outside 90% solute → 10% H₂O → water out ✔
- Beaker 9: inside 5% H₂O → 95% solute; outside 65% H₂O → 35% solute → water in ✔

All correct.

Final Answer:
Beaker 1: Hyper
Beaker 2: Hypo
Beaker 3: Iso
Beaker 4: 45% Solute, Iso
Beaker 5: 65% Solute, Hypo
Beaker 6: 5% H₂O, Hypo
Beaker 7: 45% H₂O (inside), 75% H₂O (outside), Hypo
Beaker 8: 20% Solute (inside), 10% H₂O (outside), Hyper
Beaker 9: 95% Solute (inside), 35% Solute (outside), Hypo