Molarity of Sucrose solution (mols +/- 0.1 mols) cant of Potato in the lead (g +/- 0.01 g)Weight of Potato After (g +/- 0.01 g) limiting in Weight of Potato (g +/- 0.01 g)Percentage Change in Weight (% +/- 0.1 %) 0.03.714.18+0.47+5.9 0.13.734.08+0.35+4.4 0.23.573.51-0.06-0.8 0.33.243.25+0.01+0.1 0.44.334.09-0.24-2.8 0.53.282.84-0.44-7.1 0.64.103.26-0.84-11.4 0.74.183.21-0.97-13.1 0.83.452.46-0.99-16.7 0.93.972.59-1.38-21.0 1.03.582.32-1.26-21.3 cultivation processing Calculation to settle the exchange in mound of the murphyes at 0.0 mols of saccharose (for example): 4.18 3.71=+0.47 Calculation to find the percentage change in kettle of fish of white spudes at 0.0 mols of saccharose (for example): ((4.18 3.71)/ (4.18+3.71) Ã100= +5.9% Conclusion The look into head word that I asked myself before the sample was What is the osmotic dominance of tater? The outcome I got after the try out was that the osmotic potential of potato is very steep because this experiment exactly showed what happened when in that compliments is Osmosis perplex and the following information will prove it. Our results greatly showed that there has been osmosis present.

Osmosis is the dissemination of wet through a partially permeable membrane from a region of take down solute doorway to a region of high solute concentration. And the results we got shows that osmosis occurred in this experiment. In 0.0 mols of saccharose the mass of potato has increase so there has been spreading of water from 0.0 mols of sucrose which is the lower concentration to 3.71 grams of potato which is the high concentration. Also in 1.0 mols of sucrose the mass of potato has fare and it should decrease when there is osmosis present because in this case 1.0 mols of sucrose is the higher concentration and the 3.58 grams of potato is the lower concentration. Basically the potato chips lose more mass in the higher lowering solution and gain mass in the lower strenuous solution....If you want to get a full essay, order it on our website:
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