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Osmosis Lab Purpose The purpose of this lab is to determine the concentration of solutes inside of a potato s cell. We also want to know the composition of potato water and stuff. Also the scientific Phenomenon studied of diffusion/ Osmosis. Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration. Osmosis is process by which water molecules diffuse across a cell membrane from an area of higher concentration to an area of lower concentration. Hypothesis In the potato I think that there will be 8. 05 % solute.
I think that the potato inside the test tube that is labeled # 1, that has no sugar inside will gain mass. I believe this because of osmosis, I think that the water will diffuse into the potato because the water it is in has a higher concentration of solute than the potato. In test tube # 2 I also think that the potato will gain mass. In the # 3 test tube the factor of osmosis will come in and the potato will gain water. In the # 4 test tube it will gain even more mass because of the sugar factor.
In # 5 and # 6 they will lose mass, because there is more than 8. 05 % of sugar in the water. My entire hypothesis is based on the laws of diffusion. Diffusion: The moving of molecules from an area of high concentration to an area of low concentration. In my hypothesis I said that the potato in test tube # 1 without any sugars in it would gain mass. This will occur because the potato has a lower concentration of water than water in the test tube, (High Concentration) to inside the potato, (Low concentration) thus the diffusion. So in test tube # 4 with 8. 05 % of sugar will become equilibrium (same amount of diffusion between the two) with the potato.
Materials: 1. 5 test tube 2. 12 grams of sugar 3. Gradated Cylinder 4. Aluminum foil 5 1 x 1 in. 5. ! beaker 6. Scoop 7. Balance 8. 10 Paper towels 9.
Masking tape 10 in. 10. Potato Slice 11. Forceps 12. Pen 13.
Pen 14. Knife 15. Distilled water 1 cup 16. Weighing paper 2 1 x 1 in. 17. Cork Born 18.
Stirring Rod Procedure: (Day 1) 1) Put a small piece of tape on each test tube and label them as shown: Group 1 0 1. 37 2. 73 4. 12. 5. 63 Group 2 0 1. 15 2. 42 2. 78 5. 21 Group 3 0 1. 03 2. 14 3. 42 4. 85 Group 4 0. 75 1. 61 3. 04 4. 52 2) Put 20 ml of water in each test tube. 3) Into each test tube put the correct amount of sugar indicated by the test tube s label. (Refer to # 1) Calculate the concentration of sugar in each test tube by dividing the amount of sugar by 20 ml. Record this. 4) Now cut 5 pieces of potato with the cork borer for consistent potato pieces. Measure and cut the pieces of potato 6 -cm in length. Be sure to cut off the pieces with the skin. 5) Use the stirring rod on the bottom to get all the sugar dissolved. (Some of the concentrations of sugar very high and may not dissolve) 6) One at a time find the mass of potato pieces: put a piece of paper on the balance and find the mass. Put the potato on the paper and find the mass of them together. Subtract the mass of the paper.
Record the mass of the potato on your data table in the row for the test tubes. It will go into, and then put the potato into its test tube. 7) Put an aluminum foil hat on the test tubes and seal it as well as you can. 8) Place the test tubes into a beaker. Using masking tape label the beaker with your names and period. 9) Place test beaker on the assigned place on the countertop. Clean up. 10) Set up your rough draft of the lab write up. Complete the purpose through procedure. 11) Make your predictions about how the mass of the potato spears will change in each sugar solution. Explain why you think certain changes will occur (Look back at the section of diffusion). (Day 2) 12) One at a time, using forceps, remove your potato spears from the test tubes. 13) Carefully blot it with paper; roll the potato on paper towel to remove excess water from the outside. 14) Find the mass of each potato spear using the same method you used on Day 1 and record it. 15) Clean your test tubes, and then rinse the forceps.
Make your lab space clean. Remove masking tape from test tubes and beakers. Data collection and observations-Formulas: Volume = 3. 14 r^ 2 L Averages = The sum of all the numbers / how many numbers you added Mass = whatever the balance indicates Percent change in volume = volume of day 2 day 1 Day 1 100 Percent change in mass = mass of day 2 day 1 Day 1 100 To calculate the change % of concentration of solute in each test tube: Equation: Amount of sugar x 100 Amount of water Test tube with 0 g of sugar: 0 g x 100 = 0 % concentration of solute 20 g Test tube with. 75 g of sugar: . 75 g x 100 = 3. 75 % 20 g Test tube with 1. 61 g of sugar: 1. 61 g x 100 = 8. 05 % 20 g Test tube with 3. 04 g of sugar: 3. 04 g x 100 = 15. 2 % 20 g Test tube with 4. 52 g of sugar: 4. 52 g x 100 = 22. 6 % 20 g Weigh in grams: Potato Paper Total Potato went into of solute 1 1. 12 2. 45 1. 33 0 % 2 1. 12 2. 41 1. 29 3. 75 % 3 1. 12 2. 40 1. 28 8. 05 % 4 1. 12 2. 35 1. 23 15. 2 % 5 1. 12 2. 35 1. 23 22. 6 %Fig. 1 - this chart shows the weight of the potato and the patter used to weigh it, as well as where each potato went into of solute. (Day 2) Potato mass in grams: Potato in Mass No sugar 1. 94. 75 1. 651. 61 1. 543. 04 1. 114. 52 1. 00 Fig. 1 - this chart shows the potatoes mass gathered in day two. To calculate percent change in mass Equation: Mass Day 2 Mass Day 1 x 100 Mass Day 1 Potato in 0 % solute: 1. 94 1. 33 x 100 = 45. 9 % 1. 33 Potato in 3. 75 % solute: 1. 65 1. 29 x 100 = 27. 9 % 1. 29 Potato in 8. 05 % solute: 1. 45 1. 28 x 100 = 20. 9 % 1. 28 Potato in 15. 2 % solute: 1. 11 1. 23 x 100 = - 9. 76 % 1. 23 Potato in 22. 6 % solute: 1. 00 1. 23 x 100 = - 21. 6 % 1. 23 Observations for day two Potato in test tube with no sugar: This potato spear is stiff, and is bigger on one side than the other. There seems to be a foul odor coming from this test tube. Potato in. 75 g test tube: The potato is a little stiff to our surprise.
One side appears to be thicker than the other is. The potato looks fuzzy. Potato in 1. 61 g test tube: The potato is a little stiff. One side appears to be thicker than the other is. There seems to be little oxygen bubbles found in the water. Potato in 3. 04 g test tube: The potato is surprisingly limp.
Both sides appear to be the same thickness. Potato 4. 52 g test tube: The potato is also very limp. At the top of the water there appears to be a white cloud of some sort. Interpretation of data: Both living and nonliving things are composed of molecules made from chemical elements such as carbon, hydrogen, oxygen, and nitrogen. The organization of these molecules into cells is one feature that distinguishes living things from all other matter. A cell is the smallest unit of matter that can carry on all of the process of life.
A cell cannot survive if it is totally isolated from its environment. All cells must take in nutrients and other materials, and they must also dispose of the wastes they produce. Therefore, both nutrients and wastes must pass through the cell membrane. The cell membrane controls the ease with which substances pass into and out of the cell-some substances easily cross the membrane, while others cannot cross at all. This is the reason behind diffusion. A potato cell has what we call starch granules, which are too big to go through the cell membrane.
The only thing that can go through is water or osmosis. Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration. Osmosis is process by which water molecules diffuse across a cell membrane from an area of higher concentration to an area of lower concentration. The more glucose put in with the potato the more mass the potato will lose. According to graph # 2 when the solute percents went up the mass of the potato went down.
According to my data (see graph # 1) where from the equation 0 = - 3. 0151 x + 42. 458. The percent solute in the potato is 14. 08 %. According to class data graph (see graph # 2) with the equation 0 = - 1. 4973 x + 23. 953. The percent solute in the potato is 15. 99 % in the potato.
Equilibrium is when the concentration the molecules of a substance is the same throughout a space. It is significant to answering the question in this lab, because equilibrium (isotonic) is what the cell is trying to get too. In order to make this lab easier to read for you the reader, there is a few words that I should describe for you. A solution is a mixture in which one or more substances are uniformly distributes in another substance. Solute is the substance in which the solute is dissolved. When the concentration of solute molecules outside the cell is lower than the concentration in the cytosol, the solution outside is Hypotonic.
When the concentration of solute molecules outside the cell is higher the concentration in the cytosol, the solution outside is hypertonic. Isotonic is when it is all equal in and outside the cell. When the concentration is different from that of the potato then it would be a time for osmosis to occur. After the potato has reached the equilibrium point the mass of the potato would have changed from the original mass. Anything with 10 % solute or more will lose mass. There are a few things that may have cause data error.
Such as when reading the balance there could have been some gunk left over from past experiments and since were using such small measurements. This could effect the weight and make the reading of the weight a little different, and this would change some things around. Such as at what rate diffusion occurs. Another data error could be human error, such as partners giving each other the wrong information. And last but not least, is that all the potato spears werent cut exactly alike.
So if we could of cut them more alike then the data would be more precise. Conclusion The purpose of this lab was to determine the percent of concentration of solute in a potato cell by placing the potato spears in differently labeled test tubes, with different amounts of sugar of sugar solution. This lab also helped with the study of osmosis. The data showed that in general, a potato does contain a certain percent of water and a certain percent of solutes. According to # 3.
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