below is animal cells placed in beakers of various concentrations.
1. draw an arrow to show which way the water should move by osmosis
2. fill in any missing percentages (water or solute)
3. identify the type of solution (isotonic, hypertonic, or hypotonic) by circling your choice

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Accepted Answer

### Solving for Missing Percentages (Water or Solute) For any solution, the sum of water percentage and solute percentage is $ 100\% $ (since $ \text{Water}\% + \text{Solute}\% = 100\% $). We use this to find missing values. #### 1. Fourth Beaker (Animal Cell: $ 55\% \, \text{H}_2\text{O}, 45\% \, \text{Solute} $; Beaker: $ 55\% \, \text{H}_2\text{O} $) ## Step 1: Use $ \text{Water}\% + \text{Solute}\% = 100\% $ Let $ x = \text{Solute}\% $ in the beaker. $ 55\% + x = 100\% $ ## Step 2: Solve for $ x $ $ x = 100\% - 55\% = 45\% $ #### 2. Fifth Beaker (Animal Cell: $ 15\% \, \text{H}_2\text{O}, 85\% \, \text{Solute} $; Beaker: $ 35\% \, \text{H}_2\text{O} $) ## Step 1: Use $ \text{Water}\% + \text{Solute}\% = 100\% $ Let $ x = \text{Solute}\% $ in the beaker. $ 35\% + x = 100\% $ ## Step 2: Solve for $ x $ $ x = 100\% - 35\% = 65\% $ #### 3. Sixth Beaker (Animal Cell: $ \_ \% \, \text{H}_2\text{O}, 95\% \, \text{Solute} $; Beaker: $ 25\% \, \text{H}_2\text{O}, 75\% \, \text{Solute} $) ## Step 1: Find animal cell’s $ \text{H}_2\text{O}\% $ (since $ \text{Water}\% + \text{Solute}\% = 100\% $) $ \text{H}_2\text{O}\% = 100\% - 95\% = 5\% $ #### 4. Seventh Beaker (Animal Cell: $ \_ \% \, \text{H}_2\text{O}, 55\% \, \text{Solute} $; Beaker: $ \_ \% \, \text{H}_2\text{O}, 25\% \, \text{Solute} $) ## Step 1: Animal cell’s $ \text{H}_2\text{O}\% $ $ \text{H}_2\text{O}\% = 100\% - 55\% = 45\% $ ## Step 2: Beaker’s $ \text{H}_2\text{O}\% $ (since $ \text{Water}\% + \text{Solute}\% = 100\% $) $ \text{H}_2\text{O}\% = 100\% - 25\% = 75\% $ #### 5. Eighth Beaker (Animal Cell: $ 80\% \, \text{H}_2\text{O}, \_ \% \, \text{Solute} $; Beaker: $ \_ \% \, \text{H}_2\text{O}, 90\% \, \text{Solute} $) ## Step 1: Animal cell’s $ \text{Solute}\% $ $ \text{Solute}\% = 100\% - 80\% = 20\% $ ## Step 2: Beaker’s $ \text{H}_2\text{O}\% $ $ \text{H}_2\text{O}\% = 100\% - 90\% = 10\% $ #### 6. Ninth Beaker (Animal Cell: $ 5\% \, \text{H}_2\text{O}, \_ \% \, \text{Solute} $; Beaker: $ 65\% \, \text{H}_2\text{O}, \_ \% \, \text{Solute} $) ## Step 1: Animal cell’s $ \text{Solute}\% $ $ \text{Solute}\% = 100\% - 5\% = 95\% $ ## Step 2: Beaker’s $ \text{Solute}\% $ $ \text{Solute}\% = 100\% - 65\% = 35\% $ ### Identifying Solution Types (Isotonic, Hypertonic, Hypotonic) - **Isotonic**: Equal solute concentration (water moves equally). - **Hypertonic**: Beaker has higher solute (water moves out of cell). - **Hypotonic**: Beaker has lower solute (water moves into cell). #### 1. First Beaker (Cell: $ 10\% \, \text{Solute} $; Beaker: $ 60\% \, \text{Solute} $) Beaker solute > Cell solute → **Hypertonic** #### 2. Second Beaker (Cell: $ 60\% \, \text{Solute} $; Beaker: $ 30\% \, \text{Solute} $) Beaker solute < Cell solute → **Hypotonic** #### 3. Third Beaker (Cell: $ 80\% \, \text{Solute} $; Beaker: $ 80\% \, \text{Solute} $) Equal solute → **Isotonic** #### 4. Fourth Beaker (Cell: $ 45\% \, \text{Solute} $; Beaker: $ 45\% \, \text{Solute} $) Equal solute → **Isotonic** #### 5. Fifth Beaker (Cell: $ 85\% \, \text{Solute} $; Beaker: $ 65\% \, \text{Solute} $) Beaker solute < Cell solute → **Hypotonic** #### 6. Sixth Beaker (Cell: $ 95\% \, \text{Solute} $; Beaker: $ 75\% \, \text{Solute} $) Beaker solute < Cell solute → **Hypotonic** #### 7. Seventh Beaker (Cell: $ 55\% \, \text{Solute} $; Beaker: $ 25\% \, \text{Solute} $) Beaker solute < Cell solute → **Hypotonic** #### 8. Eighth Beaker (Cell: $ 20\% \, \text{Solute} $; Beaker: $ 90\% \, \text{Solute} $) Beaker solute > Cell solute → **Hypertonic** #### 9. Ninth Beaker (Cell: $ 95\% \, \text{Solute} $; Beaker: $ 35\% \, \text{Solute} $) Beaker solute < Cell solute → **Hypotonic** ### Water Movement (Osmosis) - **Hypertonic (beaker > cell solute)**: Water moves **out of the cell** (arrow from cell to beaker). - **Hypotonic (beaker < cell solute)**: Water moves **into the cell** (arrow from beaker to cell). - **Isotonic**: No net movement (arrows equal or none). ### Final Missing Percentages: - Fourth Beaker Solute: $ 45\% $ - Fifth Beaker Solute: $ 65\% $ - Sixth Beaker H₂O: $ 5\% $ - Seventh Beaker H₂O (cell): $ 45\% $; H₂O (beaker): $ 75\% $ - Eighth Beaker Solute (cell): $ 20\% $; H₂O (beaker): $ 10\% $ - Ninth Beaker Solute (cell): $ 95\% $; Solute (beaker): $ 35\% $ ### Answer (Key Values): - Missing Percentages: - 4th: $ 45\% $ Solute - 5th: $ 65\% $ Solute - 6th: $ 5\% $ H₂O - 7th: $ 45\% $ H₂O (cell), $ 75\% $ H₂O (beaker) - 8th: $ 20\% $ Solute (cell), $ 10\% $ H₂O (beaker) - 9th: $ 95\% $ Solute (cell), $ 35\% $ Solute (beaker) - Solution Types: 1. Hypertonic 2. Hypotonic 3. Isotonic 4. Isotonic 5. Hypotonic 6. Hypotonic 7. Hypotonic 8. Hypertonic 9. Hypotonic - Water Movement: - Hypertonic: Out of cell - Hypotonic: Into cell - Isotonic: No net movement

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Explanation:

Solving for Missing Percentages (Water or Solute)

1. Fourth Beaker (Animal Cell: \( 55\% \, \text{H}_2\text{O}, 45\% \, \text{Solute} \); Beaker: \( 55\% \, \text{H}_2\text{O} \))

Step 1: Use \( \text{Water}\% + \text{Solute}\% = 100\% \)

Step 2: Solve for \( x \)

2. Fifth Beaker (Animal Cell: \( 15\% \, \text{H}_2\text{O}, 85\% \, \text{Solute} \); Beaker: \( 35\% \, \text{H}_2\text{O} \))

Step 1: Use \( \text{Water}\% + \text{Solute}\% = 100\% \)

Step 2: Solve for \( x \)

3. Sixth Beaker (Animal Cell: \( \_ \% \, \text{H}_2\text{O}, 95\% \, \text{Solute} \); Beaker: \( 25\% \, \text{H}_2\text{O}, 75\% \, \text{Solute} \))

Step 1: Find animal cell’s \( \text{H}_2\text{O}\% \) (since \( \text{Water}\% + \text{Solute}\% = 100\% \))

4. Seventh Beaker (Animal Cell: \( \_ \% \, \text{H}_2\text{O}, 55\% \, \text{Solute} \); Beaker: \( \_ \% \, \text{H}_2\text{O}, 25\% \, \text{Solute} \))

Step 1: Animal cell’s \( \text{H}_2\text{O}\% \)

Step 2: Beaker’s \( \text{H}_2\text{O}\% \) (since \( \text{Water}\% + \text{Solute}\% = 100\% \))

5. Eighth Beaker (Animal Cell: \( 80\% \, \text{H}_2\text{O}, \_ \% \, \text{Solute} \); Beaker: \( \_ \% \, \text{H}_2\text{O}, 90\% \, \text{Solute} \))

Step 1: Animal cell’s \( \text{Solute}\% \)

Step 2: Beaker’s \( \text{H}_2\text{O}\% \)

6. Ninth Beaker (Animal Cell: \( 5\% \, \text{H}_2\text{O}, \_ \% \, \text{Solute} \); Beaker: \( 65\% \, \text{H}_2\text{O}, \_ \% \, \text{Solute} \))

Step 1: Animal cell’s \( \text{Solute}\% \)

Step 2: Beaker’s \( \text{Solute}\% \)

Identifying Solution Types (Isotonic, Hypertonic, Hypotonic)

1. First Beaker (Cell: \( 10\% \, \text{Solute} \); Beaker: \( 60\% \, \text{Solute} \))

2. Second Beaker (Cell: \( 60\% \, \text{Solute} \); Beaker: \( 30\% \, \text{Solute} \))

3. Third Beaker (Cell: \( 80\% \, \text{Solute} \); Beaker: \( 80\% \, \text{Solute} \))

4. Fourth Beaker (Cell: \( 45\% \, \text{Solute} \); Beaker: \( 45\% \, \text{Solute} \))

5. Fifth Beaker (Cell: \( 85\% \, \text{Solute} \); Beaker: \( 65\% \, \text{Solute} \))

6. Sixth Beaker (Cell: \( 95\% \, \text{Solute} \); Beaker: \( 75\% \, \text{Solute} \))

7. Seventh Beaker (Cell: \( 55\% \, \text{Solute} \); Beaker: \( 25\% \, \text{Solute} \))

8. Eighth Beaker (Cell: \( 20\% \, \text{Solute} \); Beaker: \( 90\% \, \text{Solute}…

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