how much does a potato chip grow in moler solutions?

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Introduction.


Osmosis is the net movement of water between water molecules from a weaker solution to a stronger solution through a semi-permeable membrane.





Structure of a plant cell.


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What happens when a plat cell is placed in a solution weaker then the solution in a cell?


Osmosis takes place. Water diffuses into the cytoplasm and vacuole through the selectively permeable membrane. When this happens the solution in the middle is more concentrated and the vacuole expands while the cytoplasm moves to the side.


What happens when a plant cell is placed in a solution that is stronger then that inside the plant cell?


Again osmosis takes place. Water diffuses out of the cytoplasm and vacuole through the selectively permeable cell membrane. First the cell shrinks slightly and becomes flaccid; then the cell membrane pulls away from the cell wall. This causes spaces between the cell membrane and the cell wall. The solution is less concentrated.


First experiment


In this experiment I am going to fill three beakers with water and salt. The first beaker will just be water (week solution), the second with a small amount of salt (medium) and the third with even more salt then the second (strong). In to each of these I will place a potato chip that will be pre-measured in millimetres. I will then apply labels to the beakers to remind me which is what solution. After leaving the piece of potato chip in the beaker for twenty-four hours I will re-measure the chip (in mm) and record the results.





Weak Medium Strong


Prediction


I predict the potato in the weak solution will expand, the potato in the medium solution will shrink a little bit and the potato in the strong solution will shrink more then in the medium.





Results.


Solution length before the length after change in


Strength experiment. Experiment length


Weak 44 mm 4 mm +5mm


Medium 47mm 45mm - mm


Strong 47mm 44mm -mm


Conclusion


My results have proven that all of my predictions where correct.


. The length of the potato did increase in the weak solution.


. The length of the potato chips in the medium and strong solutions did decrease.


. The length of the potato in the strong solution decreased more then the potato in the medium solution.








Graph to show the increase and


decrease of potato chips after being


soaked in a certain strength solution.





Second experiment.


The first experiment showed me if a piece of potato is placed in a weaker solution it will grow but if placed in a stronger solution it will shrink, all because of osmosis.


In this experiment I will take a potato and cut it in to two equal pieces measured in mass (to two decimal places). One will be placed in distilled water and the other in a one molar (m) sugar solution. (4.g per litre. Much “stronger” then the “strong” solution in the first experiment.) We are using sugar for this one as salt may be damaging the cell membrane. I will then cover each solution in cling film so no evaporation takes place. I will then re-weigh it after twenty-four hours.


Find out


I will find out the change in mass, whether it’s more (+) or less (-).


I will then have to calculate the percentage increase.


Formula =


Change in mass (+ or -) x100


----------------------------------


Original mass.


Results table


Solution strength (m)


Original mass (g)


Final mass (g)


Change in mass (g)


Calculate percentage changing in mass.


Beaker 1.


0.0 m


4.55g


5.6g


1.14g


Beaker one percent change in mass


1.14 x100 = 144 / 4.55 = 5.054


Beaker


1m


4.g


.58g


1.74g


Beaker two percent change in mass


1.74x100 = 174 / 4. = 41.


Graph evaluation


This graph shows the isotonic point. This is the point on the graph that shows the predicted strength of the potato cell sap, this figure is 0.8 m. This means if my prediction is right, if I succeeded in making a 0.8 m solution then the potato would neither gain nor lose mass. This means it is identical to the potato cell sap.





Plan third experiment.


In this experiment I will find out accurately the strength of the solution inside a potato cell. To do this I will make six different strength solutions. These will be 0.0m, 0.m, 0.4m, 0.6m, 0.8m and 1m. In to each of these I will place three potato chips (cut with a potato cutter to make it fair) before we put them in I will measure the mass of them. When the three chips are each measured and in the beaker I will put cling film over each beaker to stop evaporation making it a fairer test. I will leave them for twenty-four hours and measure the mass again. I will then add together the mass of the three chips and divide by three to find the average. Through this I will be able to plot a better graph and to see if the solution is hypotonic (weaker then the cell sap) iceotonic (the same strength as the solution) or hypertonic (stronger then the solution).


To make this fairer I will


Use a chipper so the potatoes are all the same size.


Measure the solution in to the beakers using a measuring cylinder (75 cm)


Cover the beaker with cling film to stop evaporation.


Use three pieces of potato in each beaker to get the average.


Prediction


I predict that the concentration of the cell sap to be about 0.4m


Results


Before


Molar mass (g) mass (g) mass (g)


Bathing solution chip 1 chip chip


0.0 5.41 5.16 5.6


0. 4.74 4.81 4.4


0.4 4.50 4.8 4.0





0.6 .0 .50 .68


0.8 .8 .70 .60


1.0 5.60 5.70 5.67


Molar average


Bathing solution


0.0 5.7


0. 4.65


0.4 4.6


0.6 .6


0.8 .74


1.0 5.65


After


Molar mass (g) mass (g) mass (g) average


Bathing solution chip 1 chip chip


0.0 5.15 5.80 5.4 5.


0. 4.45 4.80 4.78 4.67


0.4 4. 4.07 4.8 4.4





0.6 .08 .58 .46 .70


0.8 .4 .08 .11 .8


1.0 4.00 4.1 . 4.01


Molar average average change in mass


Bathing solution before after


0.0 5.7 5. +.1


0. 4.65 4.67 +.0


0.4 4.6 4.4 -.1


0.6 .6 .70 -.60


0.8 .74 .8 -.75


1.0 5.65 4.01 -1.64


Percentage change.


Percentage change in mass=


Average change in mass(+ or -) x 100


-----------------------------------------------


average initial mass


0.0m = 0.1 x 100


------------- =+.46%


5.7


0.m = 0.0 x 100


------------- =+.40%


4.65


0.4m = 0.1 x 100


------------- =-.75%


4.6


0.6m =- 0.60 x 100


------------- =-17.85%


.6


0.8 = -0.5 x 100


---------- =-8 %


.74


1.0m = 1.64 x 100


------------- =-.0%


5.65


Analysis and Conclusion


Although my prediction was a lot of the actual isotonic point graph gives me an excellent set of results. I suppose I am pleased with my unusual results as it gives me the chance to try to find what was wrong. In my experiment the potato in the water had a .46% gain in length. The potato in the 0. molar sucrose solution had an increase in length of 0.40%, the potato in the 0.4 molar solution had a decrease in length of .75 %. The potato in the 0.6 molar solutions had a decrease of 17.85%, the potato in the 0.8 molar solution decreased by 0.8% and the solution in the 1 molar solution decreased by .0%. These prove that the higher the concentration of water in a cell the more it grows. Also the stronger the solution the more it loses.


Evaluation


I think that the experiment went very well, even though there were odd results they didn’t produced a good graph. Still there were a few areas where there could be improvement. Firstly, when I dried off the water on the potato chips after the experiment and before I weighed them, I used a paper towel. This might have either taken some water out of the potato or it might of left some excess water on the potato. This part of the experiment is difficult to come up with an accurate and fair method, as other ways would also lead to some slight mistakes. I.e. a hair dryer will heat the chip and my evaporate some of the water and I would have to dry each potato for an equal amount of time with the hair dryer.


Also the potato itself was not exactly the same size, although I did try to cut them to 5cm each.


Another way of improving the results would have been to leave the experiment running longer, this would have enabled me to find the saturation point (when the potato can no longer take in any more water) and dehydration point (when the potato cannot lose any more water) and therefore get a more accurate result.


I could have kept all the chips the same constant temperature. I could of done this in an incubator but I did not have the fertility’s to do so.


Also I could of chosen potato’s that where different ages (old, young etc) and repeated the experiment several times with the potato’s of different ages to see if that would off effected the posses.


Apart from different aged potato’s I could of used different breeds of potato’s


Finally, I could extend the experiment to a more exact level by looking at the potato cylinders under a microscope, and then I would be able to see the cells in greater detail and draw some more observational results.











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