Investigating How Concentration Affects Osmosis

Investigating How Concentration Affects Osmosis
Aim Looking at how concentration affects the movement of water particles through a partially permeable membrane Prediction I predict that when the potato is submerged into highly concentrated sugar solution the mass will decrease. However, when the potato is submerged into plain water the mass will increase. I predict that relationship between the pure water will be linear however the relationship between the sugar solution will be a smooth curve. Scientific background Diffusion Diffusion is the process by which a substance moves from a region of high concentration of that substance to a region of low concentration of the same substance. Diffusion occurs because the molecules of which substance s are made in random motion (kinetic theory). The rate of diffusion depends upon: 1. The concentration Gradient ? The greater the difference in concentration between two regions of a substance the greater the rate of diffusion. Organisms must therefore maintain a fresh supply of a substance to be absorbed by creating a stream over the diffusion transported away. 2. The distance over which diffusion takes place ? The shorter the distance between two regions of different concentration the greater the rate of diffusion. The rate is proportional to the reciprocal of the square of the distance (inverse square law).
Any structure in an organism across which diffusion regularly takes place must therefore be thin. Cell membranes for example are only 7.5nm thick and even epithelial layers such as those lining the alveoli of the lungs are as thin as 0.3μm across. 3. The area over which diffusion takes place -The larger the surface area the greater the rate of diffusion. Diffusion surfaces frequently have structures for increasing their surface area and hence the rate at which they exchange materials. These structures include villi and microvilli. 4. The nature of any structure across which diffusion occurs - Diffusion frequently takes place across epithelial layers or cell membranes. Variations in their structure may affect diffusion. For example, the greater the number and size of pores in cell membranes the greater the rate of diffusion. 5. The size and nature of the diffusing molecule ? Smaller molecules diffuse faster than large ones. Fat ? soluble ones diffuse more rapidly through cell membranes than water ? soluble ones. Facilitated diffusion is a special form of diffusion which allows more rapid exchange. It appears to involve channels within a membrane, which make diffusion of specific substances easier. Carrier molecules may also be involved. The process is passive, not involving any energy expenditure. Osmosis Osmosis is a special form of diffusion, which involves the movement of solvent molecules. The solvent in biological systems is invariably water. Most cell membranes are permeable to water and certain solutes only. Such membranes are termed partially permeable. Osmosis in living organisms can therefore be defined as : the passage of water from a region where it is highly concentrated to a region where its concentration is lower, through a partially permeable membrane. If a solution is separated from its pure solvent, the pressure which must be applied to stop water entering that solution and so prevent osmosis, is called the osmotic pressure. The more concentrated a solution the greater is its osmotic pressure. This is a hypothetical situation and, as a solution does not actually exert a pressure under normal circumstances, the term ?osmotic potential? is preferred. As the osmotic potential is in effect the potential of a solution to pull water into it, it always has a negative value. A more concentrated solution therefore has a more positive osmotic pressure but a more negative osmotic potential. Osmosis not only occurs when a solution when a solution is separated from its pure solvent by a partially permeable membrane but also arises when such a membrane separates two solutions of different hypotonic, solution, to the more concentrated, or hypotonic, solution. When a dynamic equilibrium is established and both solutions are of equal concentration they are said to be isotonic. The above terms should only be applied to animal cells. The osmotic relationship of plant cells should be described in terms of water potential. The water potential of a system is the difference between the chemical potential of water in the system and the chemical potential of pure water, under standard conditions, the water potential of pure water is zero. In effect, water potential is a measure of the tendency of water to leave a solution. This tendency is increased of solutes. As solute molecules in a solution tend to prevent water molecules leaving it, the solution will have a lower water potential than pure water, i.e. its value will be negative. The more concentrated the solution the more negative will be its water potential. Water will diffuse from a region of higher to a region of lower potential. Water potential values (Ψ) [image]Higher Lower Direction Zero 0 of water Less negative ? 500Pa diffusion More negative -1000 Pa The water is in fact diffusing from: the weaker solution (high solution) into the stronger solution (low concentration). The tiny holes in the membrane allow small water molecules to pass through. But the large solute molecules are too big to pass through the partially permeable membrane. Before Osmosis After Osmosis Osmosis in plant cells Osmosis is the way in which many living things take up water. Water will move into plant cells by osmosis. � The cell membrane of the plant cell acts as a partially permeable membrane. � The cell sap inside the vacuole is a strong solution. � Water passes into the plant cell by osmosis. � The concentration of the sap in the vacuole is now weaker. � Water passes from the weak solution into the strong solution in the next cell by osmosis. Turgidity When plant cells are placed in water, the water enters the cells. This is because their cell sap contains a strong solution. So water passes into the cells by osmosis. The cell membrane is the partially permeable membrane. As water enters it makes the cell swell up. The water pushed against the cell wall. Eventually the cell contains as much water as it can hold. It?s like a blown-up balloon. The strong cell wall stops the cell bursting. We say that the cell is turgid. Plasmolysis When plant cells are placed into a strong sugar or salt solution water passes out of the cells by osmosis. As water passes out, the sap vacuole starts to shrink. These cells are no longer firm, they are limp. We say that they are flaccid. As more water leaves the cells the cytoplasm starts to peel away from the cell wall. These cells are now plasmolysed. Osmosis in animal cells The cell membrane is partially permeable in animal cells. The red blood cells in this, pictures have been placed in distilled water. Their cytoplasm is a strong solution. Water passes into the cells by osmosis. But animal cells have no cell wall to stop them swelling too much ? so they burst. We call this haemolysis. A constant blood concentration The red blood cells in this diagram have been placed in a concentrated solution Wtaer has moved out of their cytoplasm by osmosis. The cells have shrunk. I can see that it is important to keep the concentration of our blood constant. The red blood cells in this diagram are in a solution that is the same concentration as their cytoplasm. The solutions on each side of their cell membranes are the same concentration. So the cells neither gain or lose water. Wtaer enters Amoeba by osmosis. The water goes into a contractile vacuole. Eventually the contractile vacuole becomes so full of water that it moves to the cell membrane and bursts. The water has been removed from the cell. Active Transport Diffusion and osmosis are passive processes, i.e. they occur without the expenditure of energy. Some molecules are transported in and out of cells by active means, i.e. energy is required to drive the process The energy is necessary because molecules are transported against a concentration gradient, i.e. from a region of low concentration to one of a high concentration. It is thought that the process occurs through the proteins that span the membrane. These accept the molecule to be transported on one side of the membrane and, by a change in the structure of the protein, convey it to the other side. Due to the energy expenditure necessary to move molecules against a concentration gradient, cells and tissues carrying out active transport are characterized by: 1. The presence of numerous mitochondria. 2. A high concentration of atp. 3. A high respiratory rate. As a consequence of 3, any factor, which increases the rate of respiration, e.g. a higher temperature or increased concentration of oxygen, will increase the rate of active transport. Any factor reducing the rate of respiration or causing it to cease, e.g. the presence of cyanide, will cause active transport to slow or stop altogether. Cell can take up some particles and keep them in high concentration. Sometimes cells can keep hold of particles and not let them diffuse out. What?s more they can even take in more particles against a concentration gradient. Look at the histograms: It shows the concentrations of some salts inside the cells of a water plant and outside in the water. These salts cannot have been taken in by diffusion. They are taken in against a concentration gradient. They are taken in by active transport. Active transport needs energy from respiration to make it happen. Active transport is the uptake of particles by cells against a concentration gradient. Apparatus � 9 test-tubes � 2 test tube racks � Pure distilled water � 1M of sugar solution � Potato � Ruler � Knife � 1 Cutting tile � 2 Measuring cylinders � Potato borer � Stop-watch � Tweezers � Tissuepaper This is how my experiment is going to be set: Method The following is a summary of my experimental method: 1) I will collect all my apparatus and assemble them according to the above. 2) Place the potato on the cutting tile. 3) Using a potato borer, bore out 9 identical strips of potato which are about 4cm long (make sure that you use the same borer to bore each strip of potato otherwise you may end up with some thick and the others thin which is not a fair test). 4) Cut off all the potato skins on the potato. (this is so that the potatoes are clean before you begin your experiment) 5) Line up all the potato strips together and place them alongside a ruler. 6) Cut the potatoes so that they are 3cm long. 7) Measure the weights of each of the potato strips using a weigher and put the results in a table. 8) Draw up a results chart like the one below to store all your results in. Experiments Volume (ml) Time left for (minutes) Mass of the potato before (g) Mass of the potato after (g) Percentage change in mass (%) Concentrated sugar solution of 1M Pure Water 1 20 0 10 2 18 2 10 3 15 5 10 4 13 7 10 Note: In this table to show that the mass has lost I will us ? ? ? and to show that the mass has increased I will use ?+?. Also the reason the concentrated sugar solution is placed next to the pure distilled water column with the same volume is to show a clear comparison of the results collected. 9) Now get your test-tube racks and place the test-tubes inside it. 10) Allocate one measuring cylinder for the sugar solution and the other for pure distilled water. 11) Decide that your most concentrated solution is going to be the furthest away from you and the more solvent, less concentrated one closer to you (this will stop you from getting confused). 12) Using a measuring cylinder pour in each solution into each test tube according to the given ratio (e.g. for the first one 20:0, meaning 20ml of concentrated sugar solution to 0 ml pure water solution. For the second one 18:2, meaning 18 ml of concentrated sugar solution and 2 ml of pure water). 13) When you pour in the solutions using a measuring cylinder, make sure that you use one cylinder for the concentrated sugar solution and another measuring cylinder for the pure water (this will prevent you from mixing more than the given ratio). 14) Allocate one person to be in charge of the time and tell them to set the stop-watch to 00?00?00. 15) The rest of your group will place the potatoes into each test tube and leave it in there for 10 minutes (make sure that the potatoes are put in at the same time, i.e. not putting one potato in after another potato has been in the solution or solvent for 1 minute). 16) After 10 minutes, take out the potatoes using tweezers and place the potato strips on some tissue paper (make sure that you now which potato was in which test tube otherwise your results may be wrong). 17) Once again weigh your potato to see if its mass has gained or lost. 18) Keep a note of this result. 19) Draw up a results chart like the one above to store all your results in. 20) Fill in the necessary details. 21) After the experiment is over you will need to do some calculations to find out whether the mass has gained or lost. To do this get your calculator and take away the mass of the potato after from the mass of the potato before. 22) Afterwards you will need to change this value into a percentage. You can do this by using this formula: %change in mass = (mass of the potato before) ? (mass of the potato after) / (Mass of the potato before) x 100% 23) Then round you answer to nearest whole number. 24) If it has decreased put a minus in front of the value however if it has increased put a plus in front of the value. Fair Test I will ensure that my investigation is a fair test by: 1. Leaving my potato strips in the solution for only 10 minutes, use the same two solutions expect different volumes, and I will finally make sure that the lengths of each potato strip is the same (i.e. The lengths of each potato strip is 3cm long). 2. Ensure that the same potato was used to take every strip. 3. Use the same apparatus with the same configuration and position. 4. Doing the experiment on the same day therefore, constant environment (i.e. not one day cold, second day hot) 5. Allowed each strip of potato the same amount of time. 6. I will use the same cork borer to take each piece of potato, otherwise the investigation will not be a fair test. The factors, which will affect my investigation, are: 1. Potato weights (i.e. the weights may not be precisely the same). 2. Diameter of the potato. 3. Thickness of the potato. 4. Type of potato (whether the type of potato absorbs well or doesn?t absorb well). I will control them by checking to see that the same potato is used therefore the weights will be almost the same and notify what type of potato it is. I will also use the same cork borer therefore the thickness of the potato will be the same. The factors, which I am unable to control, are: 1. Fluctuation in the potato strips. 2. The uniformity of the potato strip diameter. 3. Holes on the potato. 4. The condition of the potato throughout the strip (I.e. one end may be quite flaccid whilst the middle and the other end could be turgid). The apparatus or source of evidence will make my measurements and evidence accurate as when I take the readings from the weigher, the readings will be digital, therefore I will have a precise digit, however the digits fluctuated a bit but settled within a few seconds. The measurements of the potato strip length were precise and accurate as the calculator showed whether the number was positive or negative allowing me to say whether the mass increased or decreased. Safety The safety precautions which we might have to be aware of is: � Not to touch the potato strips when the experiment is in process otherwise it may cause a differ in the results. � To make sure that the equipment used is not placed at the end of the table, otherwise there is a chance of it being knocked over. � To make sure that all the equipment used is clean before you use them otherwise there could be something wrong in the results. � Make sure that the work surface you use is clean otherwise if it is wet then your equipment may slip and fall on the floor causing the test tubes to break and hurt someone. � Make sure to set your experiment in an area where there is a lot of space otherwise you may find it hard working as it is overcrowded and there will be accidents due to collisions. � Not to keep the power pack on too long otherwise the wire might burn. � Be careful when using the knife because if you are not concentrating you may cut your fingers. � Concentrate on what you are doing otherwise if you day-dream or don?t concentrate you may hurt yourself badly. I am planning to do a trial experiment to help decide what evidence I am going to collect, how I am going to collect it and between what ranges I am going to take my readings from. These were the results from my trial experiment: Experiments Volume (ml) Time left for (minutes) Mass of the potato before (g) Mass of the potato after (g) Percentage change in mass (%) Concentrated sugar solution of 1M Pure Water 1 30 0 10 1.05 0.93 - 11.4 2 25 5 10 1.02 0.92 - 9.8 3 20 10 10 1.03 0.94 - 8.7 4 15 15 10 1.02 0.97 - 4.9 5 12 18 10 0.95 0.94 - 1.1 6 9 19 10 1.01 1.03 1.9 7 6 24 10 0.96 1.06 10.4 8 4 26 10 0.97 1.08 11.3 9 2 28 10 1.06 1.18 11.3 10 0 30 10 0.99 1.11 12.1 The results of this trial experiment have determined what ranges I am going to use for my main experiment. It has helped me finalize my plan, as now I am able to say why I have chosen these ranges and why it helps me. On the other page, I have drawn up a few graphs to present my trial results. Results The results from my investigation is summarised in the following table: Experiments Volume (ml) Time left for (minutes) Mass of the potato before (g) Mass of the potato after (g) Percentage change in mass (%) Concentrated sugar solution of 1M Pure Water 1 20 0 10 1.11 0.98 - 11.7 2 18 2 10 1.05 0.99 - 5.7 3 15 5 10 1.07 1.01 - 5.60 4 13 7 10 1.07 1.03 - 3.7 5 10 10 10 1.09 1.10 - 0.9 6 6 14 10 1.08 1.10 1.9 7 4 16 10 1.12 1.13 3.7 8 2 18 10 1.08 1.14 5.6 9 0 20 10 1.06 1.16 9.4 I have decided that I don?t need to repeat the whole of my investigation or any part of my results because I have gathered enough evidence to support my conclusion. On the other page, I have drawn a few graphs to present my result. Conclusions Now that I have collected all my results and evidence, I will now present in the form of a ?Smooth Curve graph?. This will therefore help me to read off my results clearly, precisely and quickly. I have found out that my prediction was correct. That when the potato is submerged into highly concentrated sugar solution of 1M, the mass did decrease. However, when the potato is submerged into plain water the mass did increase. I found out that the reason for this was because of Osmosis. This is a special form of diffusion, which involves the movement of solvent molecules. The solvent in biological systems is invariably water. Most cell membranes are permeable to water and certain solutes only. Through the evidence I have gathered, such membranes are termed partially permeable or semi permeable. Due to my investigation I can say and prove that Osmosis in living organisms is the passage of water from one region where it is highly concentrated to a region where its concentration is lower, through a partially permeable membrane. I have developed my results to say that if a solution is separated from its pure solvent, the pressure, which must be applied to stop water entering that solution, and so prevent osmosis, is called the osmotic pressure. In my results I found out that the cut out potato strips placed in pure water soon went firm. This is because they have taken in water by osmosis and are now turgid. Turgid cells are useful to plants as the turgid cells gives the plant support and they keep the stems of many plants upright, however if plants lose water the cells are no longer firm and turgid and the plant stems that have lost water will therefore wilt. Using this evidence the potato strips in the sugar of 1M almost began to wilt because less water was taken in through osmosis. There were more solute molecules than solvent particles. My result also clearly explains diffusion. Using my results I can say and prove that diffusion is the process where which a substance moves from a region of high concentration of that substance to a region of low concentration of the same substance. Both solvent (water) and solute (glucose) molecules are in random motion, but only solvent (water) molecules are able to pass the partially permeable membrane. This they do until their concentration is equal on both sides of the membrane. Solute molecules Solvent molecules e.g. sugar or glucose e.g. water Partially permeable membrane Once the water molecules are evenly distributes, in theory a dynamic equilibrium should be established. However, the water molecules on the left of the membrane are impeded to some extent but the glucose molecules form passing the membrane. With no glucose present on the right of the membrane, water molecules move more easily to the left than in the reverse direction. I have found out that a situation is reached whereby additional water molecules build up on the left of the membrane, until their greater concentration offsets the blocking effect of the glucose. The probability of water molecules moving in either direction is the same, and a dynamic equilibrium is established. N.B. Solution = solute + solvent e.g. Glucose solution = glucose powder + water I have also found out that my results were accurate as when I drew my Line of Best-Fit, I could easily see that most of the points were already in line. I did carry out a few calculations so I could firstly work out the average of the masses and lengths and secondly find out whether the mass of the potato strip increased or decreased. After I did these calculations I plotted them onto my graph to draw a perfect and accurate Line of Best-Fit and a smooth curve. I have drawn a smooth curve to represent my results as I also found it convenient to read off and say what the change in weight was. I also found it easier to summarise and analyse my results. I have decided that the line on my graph is going to be a curve. This is because the more water added into each test tube, allowed more water to cross through the partially permeable membrane making the potato strip heavier, whereas if the solution is concentrated there are fewer water molecules. This then means that only a specific amount of water molecules pass the partially permeable membrane making the potato lighter. The data is represented graphically. In my result there is a pattern. Using the graph I have drawn I can clearly see that the pattern is a smooth curve which shows the relationship of the change in weight against the pure water and the change in weight of the concentrated sugar solution. There is a definite relationship between the two variables in my investigation as I can say that when the volume is low the amount gained or lost is higher compared to when the volume is higher the amount gained or lost is lower. This is because of osmosis. That water particles move form a highly concentrated region to a lower concentrated region through a partially permeable membrane. From my graph I can tell that there is a negative correlation for the pure water and a negative correlation for the concentrated sugar solution. Water potential values (Ψ) [image]Higher Lower Direction Zero 0 of water Less negative ? 500Pa diffusion More negative -1000 Pa My conclusion compared with my prediction because I predicted that if the potato was submerged into highly concentrated solution the mass would decrease. In my conclusion I have proved that the mass did decrease however now I am able to say that it did decrease because the amount of water molecules in the solution was less compared to the amount of solute particles. This therefore meant that fewer solvent molecules were entering through the partially permeable membrane and more water molecules were coming out of the chip rather than going into the chip making the weight of the potato chip lighter. I also predicted that if the potato chip was submerged into plain water the mass will increase. In my conclusion I have proved that the mass did increase however now I am able to say that it did because the amount of water molecules in the solution was greater compared to the amount of solute particles in the solution. This therefore meant that more water molecules were entering through the partially permeable membrane and water was coming out and going into the potato chip making the weight of the potato chip heavier. . My prediction is supported by my result as demonstrated by the smooth curve relationship of my data on my graph. My results are supported by my prediction as now I can prove that the passage of water molecules from a region where it is highly concentrated to a region where its concentration is lower, through a partially permeable membrane. This therefore allows me to conclude by saying that my prediction was correct. Evaluation Using the evidence from my experiment, I found that when the amount of water is greater compared to the amount of concentrated sugar solution there is a greater chance of osmosis taking place whereas if the amount of concentrated sugar solution is greater compared to the amount of water, there is a less chance of osmosis taking place. This is mainly because there are more solute particles in the pure water, so the water molecules can cross freely plus there are more water molecules, which can cross through the partially permeable membrane. However if there are fewer solute particles and more solvent molecules in a lower concentration more water molecules will cross the partially permeable membrane making the weight of the chip heavier. Using the evidence I have obtained I can tell that my graphs have a negative correlation whereby as the volume increases the amount lost or gained decreases. Here is a sketch to show that there is a negative correlation between the pure water and the sugar solution. My actual results fitted with my expected results as I predicted that the mass of the potatoes put in the highly concentrated sugar solution will decrease because there are more solute particles than solvent molecules passing through the partially permeable membrane and therefore the potato chip becomes flaccid whereas the mass of the potatoes put in the pure water will increase because there are more solvent molecules than solute particles passing through the partially permeable membrane and therefore the potato chip become turgid. I had two anomalous results. I had one anomalous result at 18ml and I think the reason for this was because the % change in mass was greater than the concentration, so I got a reading of ? 5.7%. I had another anomalous result at 6 ml and I think the reason for this was because the % change in mass was lower than the concentration, therefore I got a reading of 1.85%. I would say that my investigation was a fair test because: 1. I made sure the each potato strip was only in the solvent or solution for 10 minutes. 2. I made sure that I used the same cork borer therefore I got the same thickness of potato all the way round. 3. I made sure that there weren?t any holes in each potato strip.(making the weight lighter than the other strips) 4. I made sure I used the same type of potato and the same potato to take my strips of potato out. 5. I made sure I did my experiment on the same day, constant environment (i.e. not one day cold, and the other day hot) 6. I made sure that the weights of the potatoes were reasonable the same. 7. I also made sure that the total volume in each test tube was 20ml therefore the volume will be same through out the whole experiment. 8. I also made sure that each test tube had 20ml of solution so that firstly the potato will be completely covered, (that one wont have a larger surface area covered than another) and secondly there will be a ratio of showing how many parts were pure water and how many parts were the concentrated sugar solution. I think I have collected enough evidence to support my conclusion from my trial and actual experiment however if I had to improve on my results or find more evidence I would do another trial experiment to find more precise length ranges. To do this I would use smaller increments in length. This would help me to verify my final result with more precision and accuracy. The difficulties I had whilst carrying out my investigation was ensuring that I did not run out of time and, as group coordinator that everyone in the group understood, and knew what to do. Typically I ensured that everything was weighed correctly, recorded and a drawing a suitable table (i.e. concentration of solution (1M), Time left for (minutes), Mass of the potato before, Mass of the potato after and Mass gained or lost). If I were to repeat this investigation again the things I would change in my method would be to use a better weigher that would give more accurate readings; use better pieces of potato strips that was more in a straight line and had no holes because a small hole could represent 3mm making my investigation invalid and therefore presenting that my results were incorrect. Overall I would do more experiments, if I had more time, and so obtain more evidence to support my conclusion.

Investigating How Concentration Affects Osmosis 8.6 of 10 on the basis of 3721 Review.