The Effect of Solute Concentration on the Rate of Osmosis

The Effect of Solute Concentration on the Rate of Osmosis
Aim: To test and observe how the concentration gradient between a potato and water & sugar solution will affect the rate of osmosis. Introduction: Osmosis is defined as, diffusion, or net movement, of free water molecules from high to low concentration through a semi-permeable membrane. When a substance, such as sugar (which we will be using in the experiment we are about to analyse), dissolves in water, it attracts free water molecules to itself, and in doing so, stops them from moving freely. The effect of this, is that the concentration of (free) water molecules in that environment goes down. There are less free water molecules, and therefore less water molecules to pass across a semi-permeable membrane, through which sugar molecules and other molecules attached to them are too big to diffuse across with ease. In the diagram below on the right, we see two solutions divided by a partially/selectively permeable membrane (i.e. one that is porous, but allows water molecules through faster than dissolved substances).
Text Box: Originally, the two solutions were; pure water, on the left, and sugar solution with a high sugar concentration, on the right. The pure water solution is said to have higher water potential than the concentrated sugar solution, because the water will flow from the area of high concentration of free water molecules (the dilute solution), to the area of low concentration of free water molecules (the concentrated solution). In other words, to the sugar solution. However, in this diagram we see that osmosis has been taking place for a short while, because water molecules have started to diffuse to the right, across the membrane, so that there are now many present on the right side of the membrane, and a few sugar molecules are starting to diffuse across the membrane in the opposite direction, to the left side of the membrane as we see it. Through moving from an area of lots of free water molecules, to an area where there are less, the sugar concentration (or, ratio of sugar to water molecules) on the left side will increase, and that on the right side will decrease, until they come to a point where the concentrations on either side of the membrane are equal. At this point of equilibrium, the diffusion across the membrane will stop, due to there not being an area of higher or lower concentration. Factors affecting Rate of Osmosis: Before I start thinking about what I am going to test in my experiment, and how I am going to carry it out, I have to know the range of factors that might influence these choices. There are several factors that could change the rate of osmosis, but I have to identify, and thenceforth concentrate, on the most important: ? Temperature affects osmosis, because the greater the temperature, the greater the energy (which must be heat), the faster the molecules under the influence of the heat will move, and the faster they will diffuse across the membrane. This isn?t the only worry, because heat can have a physically detrimental affect on this experiment as well. There is a possibility that the cells? membranes will be denatured by the heat and behave differently to the norm. The water molecules in the solution and in the potatoes might also evaporate, a process I wouldn?t be able to completely control. ? Type of cell is vital in osmosis, because of the fact that the permeability of the membrane and the surface area of the membrane can be different in potatoes of exactly the same type, even from the same region. If the membranes of potato cells I use in the experiment have different levels of permeability, water will flow more easily through some than through others, changing the rate of osmosis. Again, if the surface areas of the membranes are vastly different between potato samples, diffusion will occur faster in cells where there is a greater surface area, as it means there are more places to cross the membrane. ? Concentration gradient between two solutions is a factor that affects the rate of osmosis a lot, and can be controlled. The greater the difference in free water molecules between two areas, the faster water will pass across the membrane, from an area where there is more water, to an area where there is less. Solutions can be made to have a fixed, specific molarity of your choice, which can be varied easily to see the effect it has. ? Pressure is the final factor affecting the rate of osmosis that I know of. If there is a greater pressure on the outside of a potato, for example, than in, water will be forced into the potato from the outside, to even the two pressures up. It is my conclusion that the concentration gradient is the key factor, because it is easy to monitor in an experiment and, as I will show in the hypothesis, has a definite, measurable effect on potatoes, which can be easily backed up by the osmosis theory. Prediction: I predict that, if the molarity of the solution is greater inside the potato, than out, the potato will gain mass over time. I predict that, if the molarity of the solution is the same inside the potato, as it is out, the potato?s mass will not change at all. I predict that, if the molarity of the solution is less inside the potato, than out, the potato will lose mass over time. I also predict that, the greater the concentration gradient, the faster the rate of osmosis. Hypothesis: In plant cells, which are important to us as our experiment involves potato chunks, osmosis in similar conditions to those shown in the previous diagram, has noticeable physical effects on the potatoes, which tell us precisely what is happening with the water molecules. Potatoes contain sugar, and any potato will have a given molarity. M.B.V Roberts (the author of a respected biology textbook) tells us that he found it to be 0.27M for the potato he experimented upon. Text Box: So, if a chunk of this potato were put in a sugar solution of 0.1M, there would be a higher concentration of free water molecules outside of the potato, than in. Due to osmosis, water molecules would move from the outside solution into the potato?s cells, and this would cause both an increase in mass in the potato, but the each individual cell would become turgid. If a similar chunk were put into a 0.27M solution, the amount of free water molecules inside and outside the potato would be equal, so no diffusion would occur, and no mass would be gained or lost. Finally, if another chunk were in a sugar solution of concentration 0.8M, there would be more free water molecules inside the potato than in the solution, and due to osmosis, water would move from the potato into the sugar solution. This would result in the potato losing mass as it lost water molecules, and the individual cells would become plasmolysed (flaccid), but not completely lose their form, due to the strong cell walls holding the structure in place. The above statements are not designed to imply that the rate at which the potato?s mass would change, would stay the same for different concentrations. The greater the concentration gradient (or, difference between the two concentrations), the faster the water molecules would transfer across the membrane, in an effort to fill the bigger water-free area up in the same time as any other, to create an equal molarity on both sides. This theory is shown in this hypothetical graph: Text Box: Method: ? Using a corer of fixed diameter, I extracted a few cylinders of potato tissue from one potato (so that the cells would have similar characteristics, including permeability). ? With a ruler and scalpel, I measured the cylinders into smaller lengths and then cut them up to form equal blocks (same surface area, so that given similar conditions, every block would end up with the same amount of water diffused in/out of it), so we had 18 to work with. ? Six clean beakers were filled with even amounts of sugar solution: 0.0M, 0.2M, 0.4M, 0.6M, 0.8M & 1M respectively. This range of solutions was chosen to go with the potato, because we knew that the molarity of a potato was around 0.3M and we wanted solutions that would be above, below and of similar molarity to the potato, to gain a good range of results. As well as this, because we didn?t go particularly high (e.g. 8M), the osmosis would happen slowly enough for us to note the gradual change in mass over as long period of time, hopefully. ? Three exact blocks of potato were placed in each of the beakers (so we could take an average of results, reducing the effect of anomalies or slightly different surface areas) at periods of 2 minutes or so (to give us time to weigh each beaker?s contents individually) and a stopwatch was started. ? All of the above occurred in the same part of the same room, and all of the beakers were exposed to the same conditions, so that there was no question about our results being affected by temperature of pressure. ? Every beaker?s potato blocks were weighed in periods of 15 minutes, for an hour, and exactly the same technique was used. Each potato block was dabbed once on each of its sides with a paper towel (so that we weren?t weighing the water itself), and was out of its solution for the same time as all the other blocks had been before them, as it was being weighed. To tell the pieces of potato apart, we made a different mark, in pen, on each of the three chunks. The effect of the pen, on the permeability of the potato block, was minimal. ? Once all the results had been written down in the table (see Results), we got rid of the actual values, and changed them to ?% Mass Change?, to eliminate any bias caused by the sizes and masses of the blocks being slightly different to start off with. Diagram of the Apparatus: [image] Results: Molarity of solution: 0.0M Time Mass (1) % Mass Change (1) Mass (2) %Mass Change (2) Mass (3) % Mass Change (3) Average % Mass Change 0 1.23 0.00 1.20 0.00 1.22 0.00 0.00 15 1.26 2.44 1.27 5.83 1.30 6.56 4.94 30 1.35 9.76 1.35 12.50 1.35 10.66 10.97 45 1.37 11.38 1.37 14.17 1.39 13.93 13.16 60 1.38 12.20 1.39 15.83 1.42 16.39 14.81 Molarity of solution: 0.2M Time Mass (1) % Mass Change (1) Mass (2) %Mass Change (2) Mass (3) % Mass Change (3) Average % Mass Change 0 1.27 0.00 1.24 0.00 1.11 0.00 0.00 15 1.28 0.79 1.26 1.61 1.20 8.11 3.50 30 1.33 4.72 1.32 6.45 1.30 17.12 9.43 45 1.35 6.30 1.33 7.26 1.33 19.82 11.13 60 1.37 7.87 1.33 7.26 1.35 21.62 12.25 Molarity of solution: 0.4M Time Mass (1) % Mass Change (1) Mass (2) %Mass Change (2) Mass (3) % Mass Change (3) Average % Mass Change 0 1.07 0.00 1.21 0.00 1.31 0.00 0.00 15 1.05 -1.87 1.19 -1.65 1.27 -3.05 -2.19 30 1.03 -3.74 1.18 -2.48 1.25 -4.58 -3.60 45 0.98 -8.41 1.14 -5.79 1.25 -4.58 -6.26 60 0.96 -10.28 1.09 -9.92 1.21 -7.63 -9.28 Molarity of solution: 0.6M Time Mass (1) % Mass Change (1) Mass (2) %Mass Change (2) Mass (3) % Mass Change (3) Average % Mass Change 0 1.00 0.00 1.15 0.00 1.22 0.00 0.00 15 0.95 -5.00 1.10 -4.35 1.19 -2.46 -3.94 30 0.94 -6.00 1.04 -9.57 1.15 -5.74 -7.10 45 0.88 -12.00 0.99 -13.91 1.09 -10.66 -12.19 60 0.86 -14.00 0.96 -16.52 1.07 -12.30 -14.27 Molarity of solution: 0.8M Time Mass (1) % Mass Change (1) Mass (2) %Mass Change (2) Mass (3) % Mass Change (3) Average % Mass Change 0 1.28 0.00 1.15 0.00 1.27 0.00 0.00 15 1.21 -5.47 1.10 -4.35 1.23 -3.15 -4.32 30 1.12 -12.50 1.03 -10.43 1.15 -9.45 -10.79 45 1.08 -15.63 1.01 -12.17 1.08 -14.96 -14.25 60 1.05 -17.97 0.96 -16.52 1.05 -17.32 -17.27 Molarity of solution: 1.0M Time Mass (1) % Mass Change (1) Mass (2) %Mass Change (2) Mass (3) % Mass Change (3) Average % Mass Change 0 1.11 0.00 1.17 0.00 1.11 0.00 0.00 15 1.04 -6.31 1.13 -3.42 1.05 -5.41 -5.04 30 0.96 -13.51 1.06 -9.40 1.00 -9.91 -10.94 45 0.87 -21.62 0.96 -17.95 0.91 -18.02 -19.20 60 0.73 -34.23 0.84 -28.21 0.83 -25.23 -29.22 Analysis: To make my results relevant to the hypotheses, I need to change the data I have into ?rate of osmosis? As the units for this measure are ?% mass change / minutes?, I will divide the average % mass change for each concentration by the minutes, at a particular point on the line of best fit, on a scattergraph of ?average % mass change vs. time?. I will do this for each concentration, and then hopefully be able to analyse my data properly. Text Box: Text Box: Text Box: Text Box: Text Box: Text Box: For each of the graphs, I found the equation of the line. The method I used to do this, was to make a right-angled triangle with the line at any given point (as the best-fit line is consistent the whole time) and divided the change in y-axis value by the change in x-axis value. Because the line compared Average % mass change and Time (the units of measure for ?rate of osmosis?), I could define the rate of osmosis for each concentration from the gradient of each line: 0.0M y = 0.2522x + 1.21 0.2M y = 0.2142x + 0.8371 0.4M y = -0.1508x + 0.2588 0.6M y = -0.2453x ? 0.14 0.8M y = -0.2965x ? 0.4335 1.0M y = -0.484x + 1.6386 There were other methods through which I could have worked out the rate of osmosis for each concentration. One of these was to take the 1st and last values and divide them by 2, getting an average. However, although this displayed the general trend of the data, it didn?t show you what happened in the interim period. Also, by having a line of best fit, you take all the values into account and negate the effect o any anomalies. Now that I have the data that I need in correct units, I have to see whether my predictions were correct. To do this, I have to bear in mind that I made the predictions according to mass gain. This is now incorporated into my rate of osmosis data, through % mass change, and therefore, an increase in rate of osmosis is the same as an increase in mass. To prove my predictions best, I will use a graph of concentration vs. rate of osmosis, which will clearly show any increases or decreases in rate of osmosis and will tell me approximately what concentration my potato was. [image] I can see from the graph, that at the point where the rate of osmosis is 0% mass change per minute and therefore where the molarity would be the same inside and outside the potato, there is not a data point, meaning I can?t work out exactly what the molarity of my potato is, and I can?t say for sure whether the mass would change or not. However, I can estimate the molarity of the potato to be around 0.35M judging from the line produced by the other results, which sounds feasible as mvb Roberts? found his to be 0.27M. Using this as the basis, I can see that for the concentrations that were lower than 0.35M, the rate of osmosis increased over time, and therefore the weight did increase. For those concentrations higher than 0.35M, the rate of osmosis has been a negative number, and therefore weight has been lost. Also, as you can see from the fairly consistent slope of the data, the further away from 0.35M you go in either direction, the further away from 0% is the rate of osmosis, highlighting the fact that the greater the concentration gradient, the faster the rate of osmosis. There is one strange result ? that at 1.0M concentration. The results at 0.4M, 0.6M & 0.8M would suggest that the rate of osmosis was levelling out. This makes sense, as there has to be a limit on how fast water can travel from a cell, because there is a limited membrane surface area through which water may pass. I thought that the rate of osmosis would have levelled out further at 1.0M, possibly even being at that very limit, but it actually just goes right down! I think that this could be because, during the experiment, just as I was picking up the solution, my elbow was knocked and a bit of the solution was spilt. Whether the concentration of the solution changed or the volume of solution affected the potato chunks, I don?t know, but that is the only incident that could have caused the anomaly. Evaluation: There were several inaccuracies in the experiment that could have resulted in this anomaly and the fact that the results weren?t quite perfect: ?When we were drying the potatoes before weighing them, we didn?t time how long we dried each chunk, so there is a chance we either soaked some of the potato material out of the cells themselves, but we might also have not completely dried the water off the potato. ?Also when we were weighing, if we left some of the potatoes outside their solution for longer than others, different amounts of evaporation would have occurred, resulting in different masses. ?The fact that we used an ink pen to mark the three separate chunks of potato in each solution could have affected the chunks by either adding extra solute to the solution or by blocking exchange over the membrane, in both cases changing the amount of water passing in/out of the potato. ?Another problem that occurred during weighing was the water left by other groups, on the scales, meaning that the chunks picked up liquid from solutions of different concentrations or, simply, a greater weight was recorded, than that of the chunk on its own. ?An inaccuracy that was very hard to stop, since we were performing the experiment manually, without the aid of mechanical cutters, was making sure that each chunk had exactly the same surface area. Obviously, they all probably had slightly different ones, which would affect the rate of osmosis, although with the care we took, the difference would be minimal. ?There was a chance that, because we were handling the potato chunks with sharp tweezers, that every time we picked a chunk out of a solution, or put it back in, we were either damaging potato tissue or making some fall off altogether. ?Despite the fact that there were so many little inaccuracies, I know that all the main factors affecting the rate of osmosis were controlled: 1. Temperature ? we did the whole experiment in the same laboratory, so there wouldn?t have been a temperature change between one point and another. 2. Type of cell ? all of the chunks came from just two potatoes, both of which had come from the same field, and were in the same packaging. 3. Pressure ? again, because we did all aspects of the experiment in the same room, there wouldn?t have been any differences in air pressure. As well as inaccuracies, as I was doing the experiment, I saw clearly where improvements could be made to make it more successful and accurate: ?We could have further mitigated the effects of anomalies (such as the anomaly at 1.0M) by using even more potato chunks in each solution, and taking an average of them. ?The inaccuracies to do with evaporation, drying potatoes and water on the scales were all to do with our being rushed, careless and suffering, because of what other people had done. If we had had more scales, we could have taken our time over things and perhaps erased these inaccuracies. ?Instead of having to rely on a rough line to estimate the molarity of the potato, we could have used more concentrations, all around the 0.25M, 0.3M, 0.35M mark, and then have a more concrete foundation from which to analyse the results. ?On the topic of concentrations, we could have maybe gone above 1.0M, and then the level-off point at which the rate of osmosis can be no faster would have been more clearly illustrated, and anomalies would also be easier to spot. There are two ways I think we could extend this coursework so that there is more to talk about, and we can learn more about the conditions osmosis thrives in: ?We could investigate the other variables and see if their effects on the rate of osmosis are lesser or greater than concentration gradient factors, and if so, why? ?The other option would be to investigate rate of osmosis in other plants, and see maybe if there are specific cell structures etc. under which the rate of osmosis changes.

The Effect of Solute Concentration on the Rate of Osmosis 9 of 10 on the basis of 1758 Review.