When a cell is kept in a 0.5 M solution of Sucrose, its volume does not alter if the same cell is placed in 0.5 M solution of sodium chloride its volume

Options

A. increases
B. decreases
C. does not show any change
D. none of the above

Answer

To solve the question regarding the behavior of a cell when placed in different solutions, we need to analyze the effects of the two solutions on the cell's volume.

### Step-by-Step Solution:

1. **Understanding the Solutions**:
- The first solution is a 0.5 M sucrose solution.
- The second solution is a 0.5 M sodium chloride (NaCl) solution.

2. **Behavior of Sucrose**:
- Sucrose is a non-ionic solute, meaning it does not dissociate into ions when dissolved in water.
- A 0.5 M sucrose solution is osmotically isotonic to the internal environment of the cell. This means that the concentration of solutes inside the cell is similar to that of the sucrose solution.
- As a result, when the cell is placed in the 0.5 M sucrose solution, there is no net movement of water into or out of the cell, leading to no change in the cell's volume.

3. **Behavior of Sodium Chloride (NaCl)**:
- Sodium chloride is an ionic compound that dissociates into sodium ions (Na⁺) and chloride ions (Cl⁻) when dissolved in water.
- A 0.5 M NaCl solution is considered hypertonic compared to the internal environment of the cell because it has a higher effective solute concentration due to the presence of ions.
- Water moves from an area of lower solute concentration (inside the cell) to an area of higher solute concentration (the NaCl solution) in order to balance the concentrations.
- As water exits the cell, the volume of the cell decreases.

4. **Conclusion**:
- In summary, when the cell is placed in a 0.5 M sucrose solution, its volume remains unchanged. However, when placed in a 0.5 M sodium chloride solution, the cell's volume decreases due to the movement of water out of the cell.

### Final Answer:
- The volume of the cell decreases when placed in a 0.5 M solution of sodium chloride.

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