Concentration Of The Sugar Solution And Water example essay topic

OSMOSIS INA POTATO CHIP Aim: To investigate the effect of varying concentration of a certain sugar solution on the amount of osmotic activity between the solution and a potato chip of a given size. Aim: To investigate the effect of varying concentration of a certain sugar solution on the amount of osmotic activity between the solution and a potato chip of a given size. Osmosis is defined as the net movement of water or any other solutions molecules from a region in which they are highly concentrated to a region in which they are less concentrated. This movement must take place across a partially permeable membrane such as a cell wall, which lets smaller molecules such as water through but does not allow bigger molecules to pass through. The molecules will continue to diffuse until the area in which the molecules are found reaches a state of equilibrium, meaning that the molecules are randomly distributed throughout an object, with all areas having an equal concentration.

For this particular investigation I think that the lower the concentration of the sugar solution in the test tube, the larger the increase in mass of the potato chip will be. This is because the water molecules pass from a high concentration. Therefore, I believe that the chips that are in the tubes containing a higher concentration of water than sugar will have a larger mass than chips in tubes with higher sugar concentrations. Further information on potato plant cells: Plant cells always have a strong cell wall surrounding them. When they take up water via osmosis they start to swell, but the cell wall prevents them from bursting. Plant cells become 'turgid' when they are put in dilute solutions.

The definition of Turgid is when a cell becomes swollen and hard. The pressure inside the cell rises and eventually the internal pressure of the cell is so high that no more water can enter the cell. This liquid or hydrostatic pressure works against osmosis. Turgidity is very important to plants because this is what makes the green parts of the plant 'stand up' into the sunlight. When plant cells are placed in concentrated sugar solutions they lose water due to the process of osmosis, and they become 'flaccid. ' This is the exact opposite of 'turgid'.

The content of the potato cells shrinks, and consequently pulls away from the cell wall. These cells are said to be. When plant cells are placed in a solution, which has exactly the same osmotic strength as the cells they are in a state between turgidity and flaccidity. We call this incipient plasmolysis. 'Incipient 'means ' about to be'. In order to create a fair test certain aspects of the experiment will have to be kept the same whilst some key variables are changed.

In this test I will be changing the concentration of the sugar solution and water whilst keeping the time limit and the mass of the potato chip the same. This will give me a very varied set of results from which I hope to make a decent conclusion. If any of the non-variables below were not kept constant it would mean it would not be a fair test. For instance if one of the potato chips was 1 mg larger in mass than the others, the surface area of the chip would be larger and there would therefore be more space for osmosis (transportation of materials between two regions) to occur.

At the beginning of the experiment all the potato chips will be 2 mg in mass. There will be 6 test tubes each containing 3 2 mg potato chips. Each tube will contain 10 ml of solution, some with more water than others. The table below shows the details at the start of the experiment. Planned method: There will be 6 test tubes each containing a different amount of cold water and a sugar solution i.e. sucrose.

Sections of potato will be cut using a scalpel and will be measured using a ruler. This part of the preparation must be done very accurately as a change in the surface area may allow more or less osmosis to occur. The mass of each chip will be measured as well so that more results can be obtained. Three chips will be placed in each test tube each time so that I can take an average for each tube. Then the potato pieces will be removed, the surface solution removed using paper and then they will be reweighed.

If I then have time afterwards I will repeat this experiment again as to obtain a second set of results. This will hopefully produce more accurate results from which I will be able to draw a more accurate conclusion. Obtaining evidence Method: 1. Using a scalpel and weight machine I cut the potato chips into 18 small pieces, which were all 2.0 mg in weight. 2.

Taking a test tube rack I placed 6 test tubes and then labelled them H 2 o to 1.0 M. 3. Using a measuring cylinder I measured out different amounts of sucrose solution and water, which I then poured into the test tubes giving 10 M equal volumes. 4. I then weighed every potato chip on an electronic balance and recorded the weights. 5.

I swiftly put 3 potato chips into each beaker and then started my stopwatch. 3 chips were used to create an average, which gave me a better set of results and more accurate graphs. 6. After 1 hour 15 minutes I drained out the solutions in the sink and placed all the chips on the paper in the order I had put them in the test tubes as to not confuse myself as to which chip came from which solution. 7. I dried each chip with the paper towel and then placed each one on the scales so that I could weigh them.

8. Each potato was measured accurately on the electronic scales and then the weights were recorded. This graph shown above gives the line of best fit for the percentage change in mass of the potato chips over the course of the 1 hour 15 minute experiment. The graph is a curve that slopes downwards and does not go through the origin.

Because the line is not straight and does not pass through the origin, it means that the percentage gain and loss in mass and concentration are not directly proportional. However, there is a pattern on my graph, and this is, as the concentration of the solution increases, the percentage change in mass decreases. The graph shows that the percentages gain and loss in inversely proportional to the concentration. The gradient does change in my graph. It gets less steep as the X axis gets bigger. This is because the potato chip is becoming as flaccid as it possibly can, and so the change in mass of each molar concentration is becoming closer and closer together.

From the line of best fit that has been added in, it can be seen that all of my points were very close to creating a perfectly smooth curve. This shows that my results are fairly reliable. My graph fits in with my prediction of the experiment graph. It shows that the potato cells increase in mass in solutions with a high water concentration and decrease in mass in solutions with a low water concentration.

When the concentration reaches above 0.8 M, there appears to be no further water loss, suggesting that the cell is fully. From the graph an estimate to the concentration of the potato cell can be made as 0.2 m, as this is the point where the potato is not increasing or decreasing in mass, this is known as the isotonic point. This is where no osmosis is taking place; both the potato and the solution have an identical molar concentration. The next point, 0.2 M looses approximately 4.0%. This shows that the water potential of the salt solution in the beaker is weaker than that of the potato chip. The next, 0.4 M, looses approximately 8.0% in mass.

This shows that the salt solution has an even weaker water potential than 0.2 M and that osmosis took place. This is why the potato lost even more mass, and it shows that the water potential in the beaker is less than that of the potato chip. This pattern carries on through the graph, and even more mass is lost, as more water moves out of the potato into the solution. My results also match with my initial predictions. This graph of the change in mass helps prove the point of complete plasmolysis, whereby the potato cannot expand and take in any more water.

As you can see as the molar concentration increases the change in mass decreases. From right to left the first two points on the graph are very spread out indicating that there was a large change in the mass. This decreases throughout the increasing molar concentration until the change is minuscule very small. This graph above indicates that there was an overall decrease in mass during the experiment. At the point 0.00 M the line for after the experiment is above the line for before the experiment unlike any of the others. This is because the water potential of the sugar solution is higher than that of the potato chip.

Evaluation The experiment was very successful in my opinion. I obtained a large quantity of very accurate results from which I was able to create informative graphs. I think I took easily enough results for the amount of concentrations that I was using, and the time that I used for the experiment to last was enough to allow sufficient osmosis to occur. However if I was to repeat the experiment I might think I would decrease the time of the result to allow more osmosis to happen and possibly find out the saturation point of the chips. The range of concentrations was adequate but I would possibly create more concentrations if I repeated the experiment so that I would have more varied results, i.e. 0.10 m, 1.15 m, 1.20 m, and so on. This way would have allowed me to also find out the isotonic point far more accurately as the one that I estimated is very approximate.

The cutting of the potatoes was the most difficult part of the experiment as although I was recording my results by mass, it could well have affected the surface area and so the overall rate of osmosis. If I were to repeat the experiment ideally if possible I would use a machine to cut the potatoes allowing perfect sizes and dimensions enabling a perfectly fair experiment. There were not any out of the ordinary results, but some were not as close to the line as others. Human may have caused this. When the potato chips were removed from the test tubes and dried I may well have dried some potatoes more thoroughly than others and so some would have more excess water, which would add to the mass.

If the experiment was repeated I could find another way to dry the potatoes that would ensure that all were dried in the same way for the same time. However with all this said I think that the experiment was truly successful and I was very pleased with the complete comparison of my results with my initial prediction. Andrew Steven Howell.