Results
%Δ mass = (mass final – mass initial) / mass initial
Beaker containing Initial mass Final Mass Mass difference %change in mass
Distill water 1.38g 1.60g 0.22g 15.94%
0.2M 1.40g 1.40g 0g 0 %
0.4M 1.36g 1.18g -0.18g -13.24%
0.6M 1.31g 1.06g -0.25g -19.08%
0.8M 1.40g 1.14g -0.26g -18.57%
1.0M 1.29g 0.83g -0.46g -35.66%
The results were presented on a graph against the percentage change in mass against the concentration of sucrose molarity
The molar concentration of the potato cell is 0.2M
The solute potential can be find using the formula given below
Ψ = -iCRT where
i = the ionization
C = molar sucrose concentration at equilibrium
R = pressure constant (0.0831litres bar per mole
T = temperature K (273 + ℃)
Givens
i= (-1) (0.2M) (0.0831) (25℃ +273 = 298℃)
= -4.9527
Ψ = -4.94 bars
Explain how you would determine the molarity of a potato
In the given data, the molarity of the potato cell is determined to be 0.2M based on the percentage change in mass. The concentration where the percentage change in mass matches that of the potato cell is the molarity of the potato.
Complete the post lab questions
a.if all the potatoes cores allowed to dehydrated, by sitting in the open air would the water potential of the potato cell decrease or increase. Why?
If the potato cores are allowed to dehydrate by sitting in the open air, the water potential of the potato cell would decrease. This is because the water potential is a measure of the potential of water to move from one place to another, and as the potato cores lose water and dehydrate, there would be a decrease in the amount of free water molecules available in the cell. This would result in a decrease in the water potential of the potato cell.
b.If the plant cell has the lower water potential than its surrounding environment and if pressure is equal to zero, is the cell hypertonic (in terms of solute concentration) or hypotonic to its environment. Will the cell loss or gain water?
The cell would be hypertonic to its environment in terms of solute concentration. Since water moves from areas of higher water potential to lower water potential, the cell would lose water to its environment.
c.if the water potential for the sucrose solution is dialysis bag is -6.25 bars and it is immerse in the cup of sucrose solution having a water potential of -3.36 bars and if the water potential inside and outside the bag is zero, will the bag gain or lose mass. Explain your answers
The bag will lose mass. This is because water will move from the area of higher water potential (-3.36 bars) to the area of lower water potential (-6.25 bars) in an attempt to reach equilibrium. Since the water potential inside and outside the bag is zero, water will move out of the bag into the sucrose solution, causing the bag to lose mass.
Discussion
The results from the experiment show a clear trend in the percentage change in mass of potato cores as the molarity of the sucrose solution increases. As the concentration of sucrose solution increases, there is a corresponding decrease in the mass of the potato cores. This is indicative of the process of osmosis, where water moves from an area of higher water potential (inside the potato) to an area of lower water potential (the sucrose solution) through a semi-permeable membrane.
The data presented in the table and graph clearly demonstrate that the molarity of the potato cell is approximately 0.2M based on the percentage change in mass. The cell is at equilibrium when the percentage change in mass matches that of the 0.2M sucrose solution. This can be further confirmed by calculating the solute potential of the potato cell, which was determined to be -4.94 bars.
In terms of practical implications, these findings have significance in understanding the osmotic processes that occur within plant cells. The movement of water in and out of cells is crucial for maintaining cell turgidity and overall function. The experiment highlights how changes in solute concentration can impact water movement and ultimately affect the overall mass of the potato cores.
Overall, the experiment successfully demonstrates the principles of osmosis and provides valuable insights into the molarity and water potential of plant cells. Further studies can be conducted to explore the effects of different solutes on osmotic processes and cellular function. The data obtained from this experiment can serve as a basis for further research and understanding of plant physiology
write a good conclusion using the information given above
In conclusion, the experiment on osmosis using potato cores immersed in different concentrations of sucrose solutions has provided valuable insights into the principles of water movement in plant cells. The results clearly show a direct relationship between the molarity of the sucrose solution and the percentage change in mass of the potato cores, indicating the process of osmosis in action.
By determining that the molarity of the potato cell is approximately 0.2M based on the percentage change in mass, and calculating the solute potential of the potato cell to be -4.94 bars, we have gained a better understanding of the equilibrium state and water potential within the cell. These findings have practical implications in understanding the osmotic processes that are vital for maintaining cell turgidity and function in plants.
Overall, the experiment has successfully demonstrated how changes in solute concentration can affect water movement and ultimately impact the mass of plant cells. This study provides a foundation for further research on osmotic processes and cellular physiology. By exploring the effects of different solutes on osmosis, researchers can deepen their understanding of plant physiology and potentially discover new applications in agriculture and biotechnology.