Diffusion and Osmosis Lab

Name/s Block: Diffusion and Osmosis lab. Data and Answer Sheet Pre-Lab: 1. What environmental factors affect diffusion? What are their effects? Temperature of solute and solvent, concentration gradient, and size of molecules affect diffusion. ● Higher temperatures increase the rate of diffusion, lower temperatures decrease the rate of diffusion ● The greater the difference in concentration, the more rapid diffusion. The closer material is to equilibrium, the slower the diffusion rate ● Heavier molecules diffuse slowly, lighter molecules diffuse quickly 2. Cells lining the small intestines are folded into microvilli. How does this help the cells absorb nutrients? Microvilli are tiny hairs that help with the absorption of nutrients; they are located in the small intestines. They help the cells absorb nutrients because they increase surface area which increases the efficiency of absorption. 3. Will water move into or out of a plant cell if the cell has a higher water potential than the surrounding environment? Water will move out of a plant cell if the cell has a higher water potential than the surrounding environment. 4. What happens if a plant cell is exposed to a hypertonic salt solution? The plant cell will shrink or shrivel. It will not burst, so the cell membrane may shrink from the cell wall. 5. How does a plant cell control its internal (turgor) pressure? Plants have cellulose, which are flexible and stronger. Therefore, the turgor pressure will be controlled because the plant cells will not break or burst. 6. Calculate the surface area: volume ratio of the following three “cells” (divide to get a single decimal answer): (a) 2cm on each side (show calculations for all parts) SA= 6(2)2 SA= 24 V= 23 V= 8 24:8→ 3:1 (b) 4cm on each side SA= 6(4) 2 SA=96 V= 43 V= 64 96:64→ 3:2 (c) 1cm x 2cm x 4cm SA=2lw+2lh+2wh SA= 28 V=(1)(2)(4) V=8 28:8→ 7:2 7. Which “cell” from #6 will allow a diffused substance to reach the center fastest? Cell A will allow for a faster diffusion because its surface area to volume ratio is the largest, which allows the substance to reach the center faster. Osmosis in Living Plant Cell Pieces of white potato will be placed in various concentrations of sucrose to determine the solute potential of the potato. The pieces will be massed before and after being in sucrose for 20 minutes. Change of mass will be calculated, and solute potential will be determined. 1. Write a Purpose/summary/background information of the lab The purpose of this lab is to determine water potential and solute potential of the potato. Make prediction of the following experiments and put in your lab notebook a. What will happen to a potato core if placed in 0% sucrose solution (distilled water)? The potato core will swell and fill with water. b. What will happen to a potato core if placed in 1% sucrose solution (highly concentrated)? The potato core will shrink and shrivel. c. What will happen to a potato core if placed in an isotonic sucrose solution? Isotonic solutions are at dynamic equilibrium and have no net movement. Therefore, nothing will happen to the potato core. Add Experimental Design a. Research Question: How does the mass of a potato core change when placed in molarities from 0.0M to 1.0 M of sucrose solution? b. Hypothesis i. Null: As potato cores are placed in various concentrations of solutions, the mass of the potato will not change. ii. Alternative: As potato cores are placed in 0.0 M to 1.0 M concentrations of sucrose, the mass of the potato cores will decrease. c. IV: the concentration of solutions that the potato will be placed in d. DV: the mass of the potato e. Control: a potato’s normal mass, when not placed in any solutions f. Constants: the distilled/ionized water, and the type of potato 4. Write out the procedures a. Get the initial mass of each potato core and record the initial temperature b. Place each potato core in each of the beaker containing the sucrose solution (label each w/ 40 mL of different molarity solutions) c. Wait for 20 minutes. After 20 minute, get the final mass of each potato core. Before weighing the potato, blot it using a paper towel. d. Record all data. While waiting for the final data, make sure to answer questions that do not need any of your data. (use your time wisely). 5. Data Collection: Water Potential Contents in Cup 0.0 M sucrose (distilled water) 0.2 M sucrose 0.4 M sucrose 0.6 M sucrose 0.8 M sucrose 1.0 M sucrose Temperature Initial Mass (°C) 21.5 °C 6.1 g Final Mass 21.5 °C 21.5 °C 21.5 °C 21.5 °C 21.5 °C 5.3 g 7.2 g 6.9 g 8.2 g 5.4 g 5.9 g Change in Mass 0.2 g % Change in Mass -3.3 % 5.0 g 6.9 g 6.5 g 7.8 g 5.0 g 0.3 g 0.3 g 0.4 g 0.4 g 0.4 g -5.7 % -4.2 % -5.8 % -4.9 % -7.4 % Formulas for this lab (refer to cheat sheet for lab): ● % Change in mass = (final mass-initial mass)/initial mass x 100 ● Water potential(ψ): = Pressure potential (ψP) + Solute Potential (ψS) ● Solute potential using the formula: (ψS) = – iCRT i= 1, C=molar concentration for which there was no change in mass, R=0.0831,and T=temperature in K (273 + °C). 6. Graph: Percent Change in Mass of Potato for each sucrose solution record the initial mass, the mass after 20 minutes, and calculate the percent change in mass) Insert your graph here: *Excel sheets won’t let me put the 0 in the middle for %change in mass, so I had to put it at the top. 7. Calculation: (Osmosis in Living Plant Cell) using data from the graph: calculate solute potential using the formula: (ψS) = – iCRT Solute potential calculation for 0.0 M sucrose (distilled water): (ψS) = -1(0)(0.0831)(294.5) (ψS) = 0 bars Solute potential calculation for 0.2 M sucrose: (ψS) = -1(0.2)(0.0831)(294.5) (ψS) = -4.89 bars Solute potential calculation for 0.4 M sucrose: (ψS) = -1(0.4)(0.0831)(294.5) (ψS) = -9.79 bars Solute potential calculation for 0.6 M sucrose: (ψS) = -1(0.6)(0.0831)(294.5) (ψS) = -14.68 bars Solute potential calculation for 0.8 M sucrose: (ψS) = -1(0.8)(0.0831)(294.5) (ψS) = -19.58 bars Solute potential calculation for 1.0 M sucrose: (ψS) = -1(1.0)(0.0831)(294.5) (ψS) = -24.47 bars 8. Write a discussion section to include analysis of data; accept or reject null Hypothesis; error analysis; and conclusion According to the data, the change in mass of 0.6 M, 0.8 M, and 1.0 M concentrations of sucrose had the highest change in mass. However, 0.0 M of sucrose, or distilled water, had the highest percent change in mass because the changes in mass were all negative (final mass was lower than initial mass). The data shows that the potatoes had a higher change in mass as the concentration of the sucrose solutions increased. Our null hypothesis was that as potato cores are placed in various concentrations of solutions, the mass of the potato will not change. This null hypothesis was rejected because the mass of the potatoes did change as the potatoes were placed in various concentrations of sucrose solutions. Our alternative hypothesis was that as potato cores are placed in 0.0 M to 1.0 M concentrations of sucrose, the mass of the potato cores will decrease. This hypothesis was accepted because the potato cores’ masses decreased from the initial mass as the concentration of the sucrose solution that the potato cores were placed in increased. Our results of this experiment were skewed, so there are some factors to understand through error analysis. One reason that our results could have been skewed is that we did not dry the potato before placing it on the scale to measure the mass. This means that the mass was not of just the potato because it was actually the mass of the potato and the water together. This would have made the mass larger than it should have been because the water would have been calculated with the mass of the potato. Another error is that we did not wait for exactly 20 minutes before getting the final mass of the potato core. We waited for more time because we started the timer late. This would give the potato cores more time for osmosis, so the final mass may not be accurate. These errors in our experiment could have been the cause of our skewed results for the experiment. In conclusion, through this experiment, we learned that the solute potential decreases and the sucrose concentration increases. We also learned that the increase in concentration of sucrose leads to a decrease in the final mass of potato cores. This can illustrate the concept of osmosis in the potato, as the water is traveling from higher to lower concentrations. Our null hypothesis was rejected, but our alternative hypothesis was accepted. As the concentration of the sucrose solution that the potato cores will be placed in increases, the mass of the potato cores will decrease. Post Lab Questions: Osmosis Living Plant Cell: 1. How did you measure the plant pieces to determine the rate of osmosis? We began by weighing the potato and recording the results. After the 20-,minute soak, we removed the potatoes and weighed and recorded the new weight of each potato. We were able to calculate the percentage change in mass of all the potatoes after comparing the two mases. . 2. Which solution had a water potential equal to that of the plant cells? How do you know? 3. Why are most cells small, and why do some have cell membranes with many convolutions? Smaller cells have a higher surface area to volume ratio, and convolutions increase surface area due to folds. A larger surface area improves material transport efficiency. 4. What organelles inside the cell have membranes with many convolutions? Why? Endoplasmic reticulum and mitochondria are two organelles with many convolutions in their membranes. The greater the number of folds on an endoplasmic reticulum, the more space for ribosomes. Ribosomes are responsible for protein synthesis. The folds of mitochondria provide more surface area, allowing for more ATP synthesis. 5. Do you think osmosis occurs when a cell is in an isotonic solution? Explain your reasoning. Yes, I believe osmosis occurs in an isotonic solution, but it occurs in such a way that water moves in and out of the cell at a constant rate. 6. What is the water potential of pure water at standard temperature and pressure? The water potential of pure water at standard temperature and pressure is zero. 7. A chef chops vegetables into a bowl of water. Would you expect the vegetable slices to gain or lose water? Explain your answer in terms of water potential. The vegetable slices would gain water because water has a higher water potential than the vegetable slices meaning water will move into the vegetable slices. 8. A marine clam is mistakenly added to a freshwater aquarium. What will happen to the clam and why? The clams cells will swell/lyse and possibly burst causing the clam to die. (hypotonic)