Class IX Biology - Chapter: Cell - Topic: Osmosis Numerical Problems

Welcome to the advanced problem set on Osmosis. This section will test your ability to apply concepts of water potential, osmolarity, and tonicity to quantitative and analytical scenarios.

Problem 1: The U-Tube Experiment (Osmolarity and Dissociation)

Scenario: A U-shaped tube is divided into two halves, Side A and Side B, by a semi-permeable membrane at the bottom. The membrane is permeable to water but impermeable to solutes.

Side A is filled with $0.15\text{ M}$ Sodium Chloride ( $\text{NaCl}$ ) solution.
Side B is filled with $0.20\text{ M}$ Glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) solution.

Questions:

Calculate the effective osmotic concentration (osmolarity) of the solutions on both sides.
Determine the direction of the net movement of water.
What will be the physical result on the water levels in Side A and Side B at equilibrium?

Step-by-Step Solution

Step 1: Calculate Osmolarity Osmolarity depends on the number of particles a solute forms in solution.

Side A (NaCl): Sodium chloride is an ionic compound that dissociates into two ions ( $\text{Na}^+$ $Na^{+}$ and $\text{Cl}^-$ $Cl^{-}$ ) in water.
- $\text{Osmolarity of A} = \text{Molarity} \times \text{Number of particles}$
- $\text{Osmolarity of A} = 0.15\text{ M} \times 2 = \mathbf{0.30\text{ Osmolar}}$
Side B (Glucose): Glucose is a covalent compound and does not dissociate in water (it remains as 1 particle).
- $\text{Osmolarity of B} = 0.20\text{ M} \times 1 = \mathbf{0.20\text{ Osmolar}}$

Step 2: Determine Net Water Movement Water always moves from a region of lower osmolarity (higher water concentration/hypotonic) to a region of higher osmolarity (lower water concentration/hypertonic).

Side A is hypertonic ( $0.30\text{ Osmolar}$ ) compared to Side B ( $0.20\text{ Osmolar}$ ).
Result: Net movement of water is from Side B to Side A.

Step 3: Physical Result Because water moves into Side A, the fluid level in Side A will rise, and the fluid level in Side B will fall, until the hydrostatic pressure equals the osmotic pressure.

U-Tube Net Water Movement:

Side B: 0.20 M Glucose ➔ 0.20 Osmolar (Hypotonic)
Flow: Net water flows from Side B, across the Semi-Permeable Membrane, into Side A.
Side A: 0.15 M NaCl ➔ 0.30 Osmolar (Hypertonic)

[!WARNING] Trap Question / Common Pitfall Students often look solely at the molarity (0.15 vs 0.20) and conclude water moves from A to B. Osmosis is driven by the number of particles, not the type. Always check if the solute is ionic (dissociates) or covalent (does not dissociate) when determining osmotic pressure.

Problem 2: Sequential Tonicity and Plant Cells

Scenario: A student extracts a cylindrical core of potato tissue. The internal cell sap of the potato cells has a solute concentration equivalent to $0.4\text{ M}$ sucrose. The potato cylinder is placed sequentially into three beakers for 30 minutes each, with a brief pat-dry between transfers:

Beaker X: $0.1\text{ M}$ Sucrose
Beaker Y: $0.4\text{ M}$ Sucrose
Beaker Z: $0.7\text{ M}$ Sucrose

Questions:

Identify the tonicity of each beaker relative to the original potato cells.
Predict the biological state (turgid, flaccid, or plasmolyzed) of the cells at the end of their time in each beaker.

Step-by-Step Solution

Step 1: Beaker X Analysis

Concentration: External ( $0.1\text{ M}$ ) < Internal ( $0.4\text{ M}$ ).
Tonicity: The solution is Hypotonic.
Process: Endosmosis occurs. Water enters the cells.
State: The cell vacuoles expand, pushing the membrane against the cell wall. The cells become Turgid.

Step 2: Beaker Y Analysis

Note: Assuming the cells return roughly to normal, or comparing strictly to original concentration.
Concentration: External ( $0.4\text{ M}$ ) = Original Internal ( $0.4\text{ M}$ ).
Tonicity: The solution is Isotonic.
Process: Water enters and leaves at the same rate (dynamic equilibrium). No net movement.
State: The cells are not fully swollen nor shrunken. They are Flaccid.

Step 3: Beaker Z Analysis

Concentration: External ( $0.7\text{ M}$ ) > Internal ( $0.4\text{ M}$ ).
Tonicity: The solution is Hypertonic.
Process: Exosmosis occurs. Water leaves the cells.
State: The cytoplasm and cell membrane shrink away from the rigid cell wall. The cells become Plasmolyzed.

Tonicity Effects on Potato Cells (0.4M Sucrose):

Beaker X (0.1M - Hypotonic): Endosmosis (Water Enters) ➔ Turgid State (Cell swells against wall)
Beaker Y (0.4M - Isotonic): Dynamic Equilibrium ➔ Flaccid State (No net change)
Beaker Z (0.7M - Hypertonic): Exosmosis (Water Leaves) ➔ Plasmolyzed State (Membrane pulls from wall)

[!CAUTION] Trap Question / Common Pitfall Believing that in an isotonic solution (Beaker Y), osmosis stops completely. Osmosis never stops as long as molecules have kinetic energy. In an isotonic state, there is a dynamic equilibrium where the rate of water entering equals the rate of water leaving, resulting in zero net movement.

Problem 3: Clinical Application - Intravenous Fluids

Scenario: A patient arrives at the hospital needing intravenous (IV) fluids. A normal red blood cell (RBC) has an internal solute concentration equivalent to $0.9\%$ $\text{NaCl}$ . By mistake, a trainee nurse prepares to administer a $0.05\%$ $\text{NaCl}$ drip.

Questions:

Quantify the concentration difference (ratio) between the RBC interior and the IV drip.
Explain what will happen to the patient's red blood cells if this drip is administered, and name the specific biological term for this event.

Step-by-Step Solution

Step 1: Calculate the Concentration Ratio

Internal RBC concentration: $0.90\%$
External IV fluid concentration: $0.05\%$
Ratio = $0.90 / 0.05 = \mathbf{18}$ .
The RBC has an internal solute concentration that is 18 times higher than the surrounding fluid. The IV fluid is severely hypotonic.

Step 2: Cytological Consequences

Because the external fluid is extremely hypotonic, water will rush into the red blood cells via endosmosis to equalize the concentration gradient.
Unlike plant cells, animal cells (RBCs) lack a rigid cell wall to exert turgor pressure and prevent over-expansion.
Result: The RBCs will swell continuously until their delicate cell membranes rupture. This specific bursting of red blood cells is called Hemolysis.

Clinical Scenario: RBC in Hypotonic IV Fluid RBC (0.9% NaCl) ➔ Placed in IV Fluid (0.05% NaCl) ➔ Massive Endosmosis ➔ RBC Swells ➔ Lacks Cell Wall ➔ Cell Membrane Ruptures ➔ Hemolysis (Cell Death)

[!TIP] Trap Question / Common Pitfall Applying plant cell logic to animal cells and stating the RBC will simply become "turgid". Turgidity requires a cell wall. Without it, hypotonic environments lead to cytolysis (cell bursting). In RBCs specifically, it is termed hemolysis.

Problem 4: The Potato Osmometer (Quantitative Rate Calculation)

Scenario: A potato osmometer is set up. The cavity in the potato contains $25\text{ mL}$ of a $20\%$ sucrose solution, and the initial level of the solution is marked with a pin at $2.0\text{ cm}$ from the base of the cavity. The potato is placed in a beaker of distilled water. After exactly $150\text{ minutes}$ , the fluid level in the cavity has risen to $6.5\text{ cm}$ .

Questions:

Calculate the rate of osmosis in terms of change in height per hour ( $\text{cm/hr}$ ).
Suppose the experiment is repeated, but the potato is boiled for 15 minutes before setting up the osmometer. What will be the fluid level after 150 minutes? Justify your answer.

Step-by-Step Solution

Step 1: Calculate the Rate of Osmosis

Initial Level ( $h_1$ ) = $2.0\text{ cm}$
Final Level ( $h_2$ ) = $6.5\text{ cm}$
Change in Height ( $\Delta h$ ) = $6.5\text{ cm} - 2.0\text{ cm} = 4.5\text{ cm}$
Time taken ( $t$ ) = $150\text{ minutes}$ = $150 / 60\text{ hours} = 2.5\text{ hours}$
Rate of Osmosis = $\Delta h / t = 4.5\text{ cm} / 2.5\text{ hrs} = \mathbf{1.8\text{ cm/hr}}$

Step 2: The Boiled Potato Scenario

Result: The fluid level will remain at exactly $2.0\text{ cm}$ (no change).
Justification: Boiling kills the plant cells and denatures the proteins in the cell membrane. This destroys the selective semi-permeability of the membrane, turning it into a freely permeable boundary or completely destroying its functional integrity. Osmosis specifically requires a living (or physically intact functional) semi-permeable membrane. Without it, the sucrose will simply diffuse out into the water, and no osmotic pressure gradient will be established to draw water up into the cavity.

[!NOTE] Fundamental Principle Osmosis is a physical process, but in biological systems, it relies on the biological integrity of the cell membrane. Dead tissues lose their selective permeability, allowing free diffusion rather than osmosis.

Class IX Biology - Chapter: Cell - Topic: Osmosis Numerical Problems

Welcome to the advanced problem set on Osmosis. This section will test your ability to apply concepts of water potential, osmolarity, and tonicity to quantitative and analytical scenarios.

Problem 1: The U-Tube Experiment (Osmolarity and Dissociation)

Scenario: A U-shaped tube is divided into two halves, Side A and Side B, by a semi-permeable membrane at the bottom. The membrane is permeable to water but impermeable to solutes.

Side A is filled with $0.15\text{ M}$ Sodium Chloride ( $\text{NaCl}$ ) solution.
Side B is filled with $0.20\text{ M}$ Glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) solution.

Questions:

Calculate the effective osmotic concentration (osmolarity) of the solutions on both sides.
Determine the direction of the net movement of water.
What will be the physical result on the water levels in Side A and Side B at equilibrium?

Step-by-Step Solution

Step 1: Calculate Osmolarity Osmolarity depends on the number of particles a solute forms in solution.

Side A (NaCl): Sodium chloride is an ionic compound that dissociates into two ions ( $\text{Na}^+$ $Na^{+}$ and $\text{Cl}^-$ $Cl^{-}$ ) in water.
- $\text{Osmolarity of A} = \text{Molarity} \times \text{Number of particles}$
- $\text{Osmolarity of A} = 0.15\text{ M} \times 2 = \mathbf{0.30\text{ Osmolar}}$
Side B (Glucose): Glucose is a covalent compound and does not dissociate in water (it remains as 1 particle).
- $\text{Osmolarity of B} = 0.20\text{ M} \times 1 = \mathbf{0.20\text{ Osmolar}}$

Side A is hypertonic ( $0.30\text{ Osmolar}$ ) compared to Side B ( $0.20\text{ Osmolar}$ ).
Result: Net movement of water is from Side B to Side A.

U-Tube Net Water Movement:

Side B: 0.20 M Glucose ➔ 0.20 Osmolar (Hypotonic)
Flow: Net water flows from Side B, across the Semi-Permeable Membrane, into Side A.
Side A: 0.15 M NaCl ➔ 0.30 Osmolar (Hypertonic)

[!WARNING] Trap Question / Common Pitfall Students often look solely at the molarity (0.15 vs 0.20) and conclude water moves from A to B. Osmosis is driven by the number of particles, not the type. Always check if the solute is ionic (dissociates) or covalent (does not dissociate) when determining osmotic pressure.

Problem 2: Sequential Tonicity and Plant Cells

Beaker X: $0.1\text{ M}$ Sucrose
Beaker Y: $0.4\text{ M}$ Sucrose
Beaker Z: $0.7\text{ M}$ Sucrose

Questions:

Identify the tonicity of each beaker relative to the original potato cells.
Predict the biological state (turgid, flaccid, or plasmolyzed) of the cells at the end of their time in each beaker.

Step-by-Step Solution

Step 1: Beaker X Analysis

Concentration: External ( $0.1\text{ M}$ ) < Internal ( $0.4\text{ M}$ ).
Tonicity: The solution is Hypotonic.
Process: Endosmosis occurs. Water enters the cells.
State: The cell vacuoles expand, pushing the membrane against the cell wall. The cells become Turgid.

Step 2: Beaker Y Analysis

Note: Assuming the cells return roughly to normal, or comparing strictly to original concentration.
Concentration: External ( $0.4\text{ M}$ ) = Original Internal ( $0.4\text{ M}$ ).
Tonicity: The solution is Isotonic.
Process: Water enters and leaves at the same rate (dynamic equilibrium). No net movement.
State: The cells are not fully swollen nor shrunken. They are Flaccid.

Step 3: Beaker Z Analysis

Concentration: External ( $0.7\text{ M}$ ) > Internal ( $0.4\text{ M}$ ).
Tonicity: The solution is Hypertonic.
Process: Exosmosis occurs. Water leaves the cells.
State: The cytoplasm and cell membrane shrink away from the rigid cell wall. The cells become Plasmolyzed.

Tonicity Effects on Potato Cells (0.4M Sucrose):

Beaker X (0.1M - Hypotonic): Endosmosis (Water Enters) ➔ Turgid State (Cell swells against wall)
Beaker Y (0.4M - Isotonic): Dynamic Equilibrium ➔ Flaccid State (No net change)
Beaker Z (0.7M - Hypertonic): Exosmosis (Water Leaves) ➔ Plasmolyzed State (Membrane pulls from wall)

[!CAUTION] Trap Question / Common Pitfall Believing that in an isotonic solution (Beaker Y), osmosis stops completely. Osmosis never stops as long as molecules have kinetic energy. In an isotonic state, there is a dynamic equilibrium where the rate of water entering equals the rate of water leaving, resulting in zero net movement.

Problem 3: Clinical Application - Intravenous Fluids

Questions:

Quantify the concentration difference (ratio) between the RBC interior and the IV drip.
Explain what will happen to the patient's red blood cells if this drip is administered, and name the specific biological term for this event.

Step-by-Step Solution

Step 1: Calculate the Concentration Ratio

Internal RBC concentration: $0.90\%$
External IV fluid concentration: $0.05\%$
Ratio = $0.90 / 0.05 = \mathbf{18}$ .
The RBC has an internal solute concentration that is 18 times higher than the surrounding fluid. The IV fluid is severely hypotonic.

Step 2: Cytological Consequences

Because the external fluid is extremely hypotonic, water will rush into the red blood cells via endosmosis to equalize the concentration gradient.
Unlike plant cells, animal cells (RBCs) lack a rigid cell wall to exert turgor pressure and prevent over-expansion.
Result: The RBCs will swell continuously until their delicate cell membranes rupture. This specific bursting of red blood cells is called Hemolysis.

[!TIP] Trap Question / Common Pitfall Applying plant cell logic to animal cells and stating the RBC will simply become "turgid". Turgidity requires a cell wall. Without it, hypotonic environments lead to cytolysis (cell bursting). In RBCs specifically, it is termed hemolysis.

Problem 4: The Potato Osmometer (Quantitative Rate Calculation)

Questions:

Calculate the rate of osmosis in terms of change in height per hour ( $\text{cm/hr}$ ).
Suppose the experiment is repeated, but the potato is boiled for 15 minutes before setting up the osmometer. What will be the fluid level after 150 minutes? Justify your answer.

Step-by-Step Solution

Step 1: Calculate the Rate of Osmosis

Initial Level ( $h_1$ ) = $2.0\text{ cm}$
Final Level ( $h_2$ ) = $6.5\text{ cm}$
Change in Height ( $\Delta h$ ) = $6.5\text{ cm} - 2.0\text{ cm} = 4.5\text{ cm}$
Time taken ( $t$ ) = $150\text{ minutes}$ = $150 / 60\text{ hours} = 2.5\text{ hours}$
Rate of Osmosis = $\Delta h / t = 4.5\text{ cm} / 2.5\text{ hrs} = \mathbf{1.8\text{ cm/hr}}$

Step 2: The Boiled Potato Scenario

Result: The fluid level will remain at exactly $2.0\text{ cm}$ (no change).
Justification: Boiling kills the plant cells and denatures the proteins in the cell membrane. This destroys the selective semi-permeability of the membrane, turning it into a freely permeable boundary or completely destroying its functional integrity. Osmosis specifically requires a living (or physically intact functional) semi-permeable membrane. Without it, the sucrose will simply diffuse out into the water, and no osmotic pressure gradient will be established to draw water up into the cavity.

[!NOTE] Fundamental Principle Osmosis is a physical process, but in biological systems, it relies on the biological integrity of the cell membrane. Dead tissues lose their selective permeability, allowing free diffusion rather than osmosis.

Numerical Problems - Osmosis

Class IX Biology - Chapter: Cell - Topic: Osmosis Numerical Problems

Problem 1: The U-Tube Experiment (Osmolarity and Dissociation)

Problem 2: Sequential Tonicity and Plant Cells

Problem 3: Clinical Application - Intravenous Fluids

Problem 4: The Potato Osmometer (Quantitative Rate Calculation)

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Numerical Problems - Osmosis

Class IX Biology - Chapter: Cell - Topic: Osmosis Numerical Problems

Problem 1: The U-Tube Experiment (Osmolarity and Dissociation)

Problem 2: Sequential Tonicity and Plant Cells

Problem 3: Clinical Application - Intravenous Fluids

Problem 4: The Potato Osmometer (Quantitative Rate Calculation)

On this page