Question
if the line cross concentration (M) twice one at 0.3M and reach -10.04% and goes or rise Up crossing at 0.49M and then comes down at 0.67M to 14.53%. how can i calculate and determine the molarity for potato cell?
Answers
Answered by
GPT-4o mini
To determine the molarity at which potato cells are in an isotonic solution using the provided data, we need to examine the points where the concentration of the solution and the percentage change in mass of the potato cells intersect, specifically where the change in mass is close to zero (indicative of no net movement of water).
Given the information:
1. The potato cells have a mass change of -10.04% at a concentration of 0.3 M (indicating that the cells are losing mass).
2. The cells are in equilibrium and exhibit no mass change around 0.49 M, as they are starting to gain mass.
3. At 0.67 M, the percentage change is 14.53%, indicating that the cells are gaining mass significantly.
### Determining Isotonic Molarity
The isotonic point—that is, the molarity of the solution at which there is no net movement of water (and thus no mass change)—is usually found between the values given. Based on your data:
1. At 0.3 M, there is a -10.04% mass change (cells lose mass).
2. At 0.49 M, it crosses to a higher percentage, starting to gain mass.
3. At 0.67 M, there is a 14.53% change (cells gain mass).
From this, we can infer that the isotonic point is between 0.3 M and 0.67 M concentrations. Specifically, it is likely close to 0.49 M where the mass change shifts from negative to positive.
### Conclusion
The isotonic molarity for the potato cells can be estimated to be around **0.49 M** based on the provided data, as this is where the concentration is closer to zero change in mass. To get a more precise value, a broader range of concentrations and more data points would be helpful to pinpoint where the mass change equals zero.
### Final Note
In a lab, you would typically plot the concentration (M) on the x-axis and the percentage mass change on the y-axis, and look for the concentration at which your line crosses the x-axis (i.e., where mass change is 0%). This would give you a more accurate value for the isotonic point. In the absence of a complete dataset, estimating visually between your values is a good approach.
Given the information:
1. The potato cells have a mass change of -10.04% at a concentration of 0.3 M (indicating that the cells are losing mass).
2. The cells are in equilibrium and exhibit no mass change around 0.49 M, as they are starting to gain mass.
3. At 0.67 M, the percentage change is 14.53%, indicating that the cells are gaining mass significantly.
### Determining Isotonic Molarity
The isotonic point—that is, the molarity of the solution at which there is no net movement of water (and thus no mass change)—is usually found between the values given. Based on your data:
1. At 0.3 M, there is a -10.04% mass change (cells lose mass).
2. At 0.49 M, it crosses to a higher percentage, starting to gain mass.
3. At 0.67 M, there is a 14.53% change (cells gain mass).
From this, we can infer that the isotonic point is between 0.3 M and 0.67 M concentrations. Specifically, it is likely close to 0.49 M where the mass change shifts from negative to positive.
### Conclusion
The isotonic molarity for the potato cells can be estimated to be around **0.49 M** based on the provided data, as this is where the concentration is closer to zero change in mass. To get a more precise value, a broader range of concentrations and more data points would be helpful to pinpoint where the mass change equals zero.
### Final Note
In a lab, you would typically plot the concentration (M) on the x-axis and the percentage mass change on the y-axis, and look for the concentration at which your line crosses the x-axis (i.e., where mass change is 0%). This would give you a more accurate value for the isotonic point. In the absence of a complete dataset, estimating visually between your values is a good approach.
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