Question
The line cross concentration (M) 3 times one at from positive percentage to 0.3M and reach -10.04% and goes or rise Up from negative to positive percentage crossing at 0.49M and then comes down from positive to negative percentage at 0.67M to 14.53%. how can i determine the molarity for potato cell?
Answers
Answered by
GPT-4o mini
To determine the molarity of the solution in which potato cells are placed, based on the information you've provided, it appears that you're describing a graphical representation of how the concentration of a solution affects the percentage change (likely in mass or length) of the potato cells through osmosis.
Here's a breakdown of how you might interpret and analyze the data you've given:
1. **Understanding the Data Points**:
- **0.3 M**: The percentage change is -10.04%, indicating that the potato cells are likely losing water and becoming flaccid in this solution (hypotonic).
- **0.49 M**: The percentage change is 0%, which suggests that this concentration is the isotonic point; the concentration of the solution is the same as the internal concentration of the potato cells, resulting in no net movement of water.
- **0.67 M**: The percentage change is +14.53%, indicating that the potato cells are gaining water and becoming turgid, which means the solution is hypertonic relative to the cell's internal environment.
2. **Determining Molarity**:
- The molarity where the percentage change is 0% (0.49 M) represents the point at which the concentration of solutes inside the potato cells is equal to the concentration of solutes in the surrounding solution. This is the isotonic point, which means that the molarity of the solution surrounding the potato cells, where there is no net movement of water, can be inferred to be approximately **0.49 M**.
3. **Conclusion**:
- Based on this analysis, if you are looking for the molarity of the solution at which the potato cells experience no change in mass, the answer would be around **0.49 M**.
This value indicates the osmotic balance point for potato cells under the conditions you described. To conduct a more formal experiment, you would test various concentrations of the solution, measure the change in mass, and plot this data to clearly see the trend and calculate the isotonic point more accurately.
Here's a breakdown of how you might interpret and analyze the data you've given:
1. **Understanding the Data Points**:
- **0.3 M**: The percentage change is -10.04%, indicating that the potato cells are likely losing water and becoming flaccid in this solution (hypotonic).
- **0.49 M**: The percentage change is 0%, which suggests that this concentration is the isotonic point; the concentration of the solution is the same as the internal concentration of the potato cells, resulting in no net movement of water.
- **0.67 M**: The percentage change is +14.53%, indicating that the potato cells are gaining water and becoming turgid, which means the solution is hypertonic relative to the cell's internal environment.
2. **Determining Molarity**:
- The molarity where the percentage change is 0% (0.49 M) represents the point at which the concentration of solutes inside the potato cells is equal to the concentration of solutes in the surrounding solution. This is the isotonic point, which means that the molarity of the solution surrounding the potato cells, where there is no net movement of water, can be inferred to be approximately **0.49 M**.
3. **Conclusion**:
- Based on this analysis, if you are looking for the molarity of the solution at which the potato cells experience no change in mass, the answer would be around **0.49 M**.
This value indicates the osmotic balance point for potato cells under the conditions you described. To conduct a more formal experiment, you would test various concentrations of the solution, measure the change in mass, and plot this data to clearly see the trend and calculate the isotonic point more accurately.
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