The aim of this experiment is to find the concentration inside a potato chip. To find it, the potato and surrounding solution must reach equilibrium.

Osmosis

Aim

Introduction

Some membranes in plant and animal cells allow certain particles to pass through them and not others. They are selectively permeable. The diffusion of water through a selectively permeable membrane is called osmosis. Osmosis is one of the most important processes that take place inside living organisms. The rate at which osmosis takes place is affected by the concentration of water in the two solutions on each side of the membrane. If a selectively permeable membrane separates the two solutions, water moves through it in both directions at the same time. However, more water leaves a dilute solution (high water concentration) and passes into a more concentrated solution (low water concentration) than enters it. Although the water appears to move across the membrane in one direction, it is in fact moving in both directions but more one way than the other. When the concentration of water is the same on both sides of the membrane, the movement of water will be the same in both directions. At this point, the net exchange of water is zero and the system is in equilibrium.

This diagram shows a semi permeable membrane with pores that only let water molecules through because they are smaller. The water mainly diffuses from right to left on this diagram, from the weak solution to the strong sugar solution.

Salt molecule Water molecule

Animals have kidneys to resolve our water troubles when we change our concentrations by sweating, breathing and drinking. It is of course, different in plants. Plants need water for photosynthesis and to keep them upright. A plant that loses too much water wilts. The reason for this is that plants are blown up with water in the vacuole. The vacuole has a fairly strong solution and water around the cell passes into the vacuole by osmosis, making it swell up. This makes the whole cell swell until it is full and because the cell walls are strong and will not burst, the whole cell becomes hard and rigid. A cell like this is called turgid. When the vacuole loses water it shrinks, pulling away the cytoplasm from the cell wall making it weak.

I will use these theories to help me investigate the effect concentration has on a plant cell. I will carry out preliminary tests to give me a clearer idea of what I must do and what to expect for my final tests.

Preliminary tests

These are the preliminary test results. These help because they give me a clearer idea of what to expect during the final experiments and they tell me if I need to change anything, and I do. I will make my final results more accurate and I will take procedures to stop any other factors affecting the experiment.

The apparatus included 3 75ml beakers each with 3 chips in the same size, all from the same potato. Each potato chip was weighed with an electronic scale so that they were all the same mass to begin with. The beginning mass of the potato chips left in the salt solution for 60 minutes was 1 gram. Cling film was wrapped over the top of the beakers to stop any outside interference.

These results aren’t very accurate because they show that the biggest decrease was in the 1.5 molar solution where there is less salt than the 2M but more than the 1M solution. So there should have been a smaller decrease with the 1M solution because there is a higher water concentration and osmosis goes from a high water concentration to a lower concentration (the potato chip), but a bigger decrease with the 2M solution. There wasn’t so it must mean that my results were inaccurate. From these results one can see that the concentration of the potato lies somewhere between ...

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Prediction

I predict that as the salt concentration increases the mass of the potato will decrease and as the water concentration increases (therefore the salt concentration decreasing) the potato mass will increase. I believe this because if there is a low water concentration in the solution the water from the potato will diffuse to it and therefore lose mass. I think that as the concentration halves the mass gained will double, this is known as inversely proportional. From the preliminary testing it can be seen that the concentration of the potato chip lies between 0 molar and 1 molar so I will use the concentrations amid them.

Final Method

A cork barer will be used The cylinders of potato They will then be placed

to push into the potato will be cut into small in salt solutions for

and release smaller cubes and weighed so the same amount of time

cylinders of potato. they’re all the same and re-weighed to see

weight. any changes in mass.

For each different experiment a different number of containers with different solutions in will be used. The first experiment apparatus will include 6 containers each enclosing different concentrations of water and salt. The first will be distilled water hence 0 molar, then 0.1 M, 0.25 M, 0.5 M, 0.75 M and finally 1 M. To achieve these solutions different concentrations will be mixed e.g. to achieve 0.5 M, distilled water and 1 M will be mixed. Each of the six containers will contain 3 potato chips all approximately the same size, shape and of course mass.

A cork barer will be inserted into a large potato and from the potato take a sharp knife will be used to cut the potato chips into cubes. An electronic scale will be used to make sure they are all the same mass. Cling film will be wrapped around the top to prevent any changes in the air affecting the outcome. To make sure the temperature doesn’t change the results and is constant throughout a thermometer will be placed through the cling film and into each container. The amount they are left in the containers for will be timed to 60 minutes.

Once the time is up the potato chips shall be re-weighed so any changes in mass can be recorded. Whilst doing this the potato chips waiting to be weighed shall be wrapped in cling film to stop any evaporation and therefore anomalies with the results.

A second set of results shall then be taken from a second experiment involving a different mass and different concentrations. The mass used will be 2 grams as opposed to 1g used before. This is to see if there are any changes. The concentrations used in this experiment will be 0.12 M, 0.14 M, 0.16 M, and finally 0.18 M. these will be achieved the same way as before and the same method will be repeated.

From the three results for each chip taken from each concentration an average will be recorded to help make comparisons.

Factors

There are many factors that can affect the outcome of this experiment so they must be kept constant throughout.

Temperature of air must be kept the same otherwise it could cause more or less water to be evaporated. On certain days the air could be quite humid so less water would be evaporated because of the water already in the air. It could affect the actual solutions as well because on a warm day the heat could give the particles more energy and the rate of osmosis will be increased. But on a colder day it could slow down the rate of osmosis. This is very unlikely because the solutions aren’t out long enough for the temperature of the air to affect it.

The concentration obviously affects the outcome directly because that is what the investigation is about. If there is a lower water concentration outside of the chip in the solution then water will diffuse from the chip and into the solution. At the same time water will also diffuse into the chip at a slower rate until equilibrium is met, when the concentrations are the same. The rate at which osmosis happens depends on the concentrations in each medium. The process of osmosis always goes from a higher concentration to a lower concentration so if they are highly unbalanced water will diffuse at a faster rate. I will control this by rinsing the containers after each use aswell as wrapping cling film over the containers to stop the temperature of the air affecting the result. It also prevents any interference from people.

If the surface areas of the potato chips are different then osmosis will occur at different rates. If there is a larger surface area there will be more pores for the water to diffuse into or out so the rate of osmosis will be quicker. If the chip has a smaller surface area then osmosis will happen at a much slower rate due to there being less space for the water particles to diffuse into. To keep it fair the chips will be cut to approximately the same size.

Container size affects the rate of osmosis because if more water is contained then the rate will be quicker because there will be more water particles to diffuse into the chip. If there is less water there will be a slower rate of osmosis because there will less particles. The surface area must also be kept the same because otherwise the climate could make changes. On a warm day the solution will heat up quicker and the rate of osmosis will of course speed up. On a cooler day the solution would cool down faster with a smaller surface area. That would of course slow down the rate of osmosis. To keep it a fair test the container must be the same size throughout at 75ml filled up to 50ml.

The temperature of water will make a huge difference to the outcome of this investigation and that is why it must be kept constant. There is no easy way to keep it the same but any changes can be recorded to explain any anomalies. This can be achieved with a thermometer. Wrapping the containers in cling film stops any outside interference or air changes affecting the temperature of the water. This is important because if the water temperature increases the solutions particles will have more energy and will be moving faster and therefore will be diffusing in and out of the potato chip faster. If the water temperature drops the particles will have less energy and the rate of osmosis will slow down.

The amount of Time the chips are left in the solution for plays a big part in the outcome of this investigation. Increasing the amount of time they are left in the solution for causes a longer period for osmosis to occur and therefore more time for the water particles to diffuse in or out of the chip. This creates bigger differences in the amount of mass lost or gained. That is why it is necessary to keep the time they are left in the containers for the same otherwise if a chip is left in for longer it could lose or gain more mass compared to the chips take out earlier. The chips shall be taken out at the same time as each other. If a chip were left out in the air for a longer period of time then that would cause water loss. Weighing them at the same time would be difficult so to prevent the air affecting the mass of the chips, I will wrap them in cling film.

The actual accuracy of my method is of course one of the most important factors that affects the outcome of each experiment. I shall do my best to keep other factors the same apart from the concentration of course.

Results

1g potato chip experiment

2g potato chip 1st experiment

2g potato chip 2nd experiment

Average for the 2g experiment

Conclusion

The results indicate that the concentration of the potato is between 0.18 molar and 0.25 molar. I know this because the mass gained was or close to 0g when in a solution with that concentration.

The evidence suggests that as the lower the salt concentration of a solution becomes then the more mass will be gained by the potato chip. The lowest is obviously distilled water or 0 molar. This is because if there is a high water concentration in the solution the water will diffuse into the potato chip because water always diffuses to a lower water concentration/higher salt concentration.

The results show that the outcome is affected by the beginning mass of the chip. My first test was with 1-gram chips but the second and third tests were with 2-gram chips. I conducted the experiment with different beginning masses to show any differences between the mass gained or lost for different concentrations. The variation was not clear but there were differences. With a larger surface area the 2-gram chips I would of expected more mass to be gained. The 2-gram experiment was in between 0.1 molar and 0.2 molar so there weren’t any concentrations that matched the 1-gram experiment so I took the closest matching concentration measurements. That wasn’t 100% accurate but it gave me an idea. Because the 2-gram chip was in a concentration of 0.12 molar and the 1-gram chip was in a solution of 0.1 molar, the mass gained would be different. The 2-gram chip should have gained less but it didn’t because it had a larger volume and more pores for water to enter. Yet it still proves that my theory was correct.

I also predicted that as the concentration halved or doubled the mass gained or lost would be doubled. This is also true because if you look at the results for the 1-gram experiment you will see evidence to support this. Some of the results for the mass lost for 0.5 molar are half of the mass lost for double 0.5 molar – which is 1 molar. This justifies my prediction.

The results shown support my prediction that as the salt concentration increases the mass of the potato will decrease and as the water concentration increases (therefore the salt concentration decreasing) the potato mass will increase. My results prove my prediction therefore they must be correct.

Evaluation

The procedure used was, in my opinion, fairly accurate. The method was conducted with the factors in mind though it could have been more controlled. The anomalies were produced because of interference or inaccuracies from me. The anomalies were within each solution so it couldn’t have been anything that was the same for each container such as water temperature. I would put them down to slight differences in the starting mass, surface area or length of time they were left in the solution. The anomalies from my results were not included in the graphs but there were anomalies. I missed them out to give me a better idea of what the averages were. By including the anomalies in the equation I would cause the results to be misleading. There were two anomalous results in my 1-gram experiment, three in the first 2-gram experiment and none in the last 2-gram experiment.

The evidence supported my prediction so that proves that my results were correct. Therefore my method was accurate.

To improve my experiment I could make changes to control the environment each potato chip is in. I put the potato chips in the containers at the same time but there were slight differences between when each chip was taken out and put in. If this experiment were to be repeated I would place each chip in at exactly the same time. To do this more people would be needed for each container. Another improvement would be to conduct each experiment on the same day. The 1-gram experiment was carried out on a separate day to the 2-gram experiment and this could of affected the dehydration of the chips when they were out of the solutions. A warmer day would of dehydrated the chips more. It would of also warmed up the solution and therefore given it more energy to speed up the rate of osmosis. When the chip is as small as it is it is important to keep differences to a minimal because they could have a greater affect on the chip.

Further work could be to find out the moment when equilibrium occurs.