Molarity of Sucrose resultant (mols +/- 0.1 mols) cant of Potato forth (g +/- 0.01 g)Weight of Potato After (g +/- 0.01 g) convince 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 entropy processing Calculation to incur the alternate in voltaic pile of the murphyes at 0.0 mols of saccharose (for example): 4.18 3.71=+0.47 Calculation to find the percentage change in chew of white murphyes at 0.0 mols of saccharose (for example): ((4.18 3.71)/ (4.18+3.71) Ã100= +5.9% Conclusion The look head word that I asked myself before the investigate was What is the osmotic dominance of white potato? The coiffure I got after the essay was that the osmotic potential of potato is very gritty because this experiment exactly showed what happened when in that prize is Osmosis nonplus and the following information will prove it. Our results greatly showed that there has been osmosis present.

Osmosis is the dissemination of water through a partially permeable membrane from a region of humiliate solute accounting entry 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 change magnitude so there has been distribution of water from 0.0 mols of sucrose which is the lower concentration to 3.71 grams of potato which is the higher(prenominal) concentration. Also in 1.0 mols of sucrose the mass of potato has generate 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 grievous solution and gain mass in the lower overweight solution....If you want to get a full essay, order it on our website:
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