A cell placed in 0.5 M sugar solution was found to be with no change in cell volume. If the same cell is placed in 0.5 M of sodium chloride solution , then there will be

Options

A. increase in cell volume
B. decrease in cell volume
C. no change in volume
D. bursting of the cell

Explanation

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The cell is placed in a 0.5 M sugar solution. Since there is no change in cell volume, we can conclude that the solution is isotonic with respect to the cell. This means the concentration of solutes inside the cell is equal to that of the surrounding sugar solution.

2. **Analyzing the Sugar Solution**:
- Sugar (such as glucose) does not dissociate into ions in solution. Therefore, the 0.5 M sugar solution has a certain concentration of solute molecules, which does not change the osmotic balance with the cell.

3. **Introducing the Sodium Chloride Solution**:
- Now, when the same cell is placed in a 0.5 M sodium chloride (NaCl) solution, we need to consider that NaCl dissociates into two ions: Na⁺ and Cl⁻. This means that a 0.5 M NaCl solution effectively has a concentration of 1.0 M of solute particles (0.5 M Na⁺ + 0.5 M Cl⁻).

4. **Determining the Effect on the Cell**:
- Since the NaCl solution has a higher concentration of solute particles compared to the sugar solution, it creates a hypertonic environment for the cell. In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside the cell.

5. **Predicting Water Movement**:
- Water will move out of the cell (exosmosis) to balance the solute concentration between the inside and outside of the cell. As water leaves the cell, the cell will lose volume.

6. **Conclusion**:
- Therefore, when the cell is placed in a 0.5 M sodium chloride solution, the cell volume will decrease due to the loss of water.

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

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