A cell when dipped in 0.5 M sucrose solution has no effect but when the same cell will be dipped in 0.5 M NaCl solution the cell will

Options

A. Decrease in volume
B. Increase in volume
C. No change in volume
D. None of the above

Explanation

To solve the question regarding the effect of dipping a cell in 0.5 M sucrose and 0.5 M NaCl solutions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Solutions**:
- Sucrose is a non-ionic solute, meaning it does not dissociate into ions in solution.
- NaCl is an ionic solute, which dissociates into sodium (Na⁺) and chloride (Cl⁻) ions when dissolved in water.

2. **Osmotic Pressure**:
- The osmotic pressure of a solution is influenced by the number of solute particles in the solution. Since NaCl dissociates into two ions (Na⁺ and Cl⁻), a 0.5 M NaCl solution effectively has an osmotic pressure equivalent to 1 M of solute particles (0.5 M Na⁺ + 0.5 M Cl⁻).
- In contrast, the 0.5 M sucrose solution has an osmotic pressure equivalent to only 0.5 M of solute particles.

3. **Effect on the Cell**:
- When the cell is placed in the 0.5 M sucrose solution, the osmotic pressure is balanced, and there is no net movement of water into or out of the cell. Therefore, there is no effect on the cell's volume.
- When the same cell is placed in the 0.5 M NaCl solution, the osmotic pressure is higher due to the dissociation of NaCl into ions. This causes water to move out of the cell to balance the concentration of solutes outside the cell.

4. **Resulting Cell Condition**:
- As water leaves the cell, the cell will lose volume and may undergo plasmolysis, where the cell membrane pulls away from the cell wall due to loss of turgor pressure.

5. **Conclusion**:
- The correct answer to the question is that when the cell is dipped in 0.5 M NaCl solution, the cell will **decrease in volume**.

### Final Answer:
- The cell will **decrease in volume** (Option A).