The purpose of this experiment is to calculate the mass/change in mass of plant tissue, in our case potato tubers. The change in mass of the potato tuber disk determines the osmotic concentration. After soaking the potato tubers in different concentrations of sucrose, then calculating the percent change in mass, we can determine the osmotic concentration. The…
We took three piece of potato and placed them in three beakers with the same quantity of water but with different concentrations, results are:…
It was difficult to ensure that the same amount of potato was added to each test tube. Hard to add solution of H2O2 at the same time to see reliable results.…
Since the change in mass of the potato tubers was negative in the sucrose solutions from 0.2 mM to 0.5 mM, these solutions were hypertonic relative to the potato dicks. In the solution with only 0.1 mM of sucrose, the mass of the potato tubers increased which means that this…
Part B. Solubility of Different Alcohols. Use your observations to complete the following table, rating each system as soluble, insoluble, or partially soluble. Alcohol Methyl alcohol Water Hexane…
2. According to the graphed data (attached) gathered from this lab, the approximate concentration at which the solution and potato cytoplasm is isotonic is 0.4 mol/L. I can therefore conclude that the approximate solute concentration of potato cytoplasm is also 0.4 mol/L. This is because the concentration of potato cytoplasm never changed throughout the experiment, but the sucrose solution did. Seeing as isotonic means both concentrations are equal, I can conclude that the potato cytoplasm was always around 0.4 mol/L.…
Purpose: To determine the concentration of solute in the potato’s cytoplasm by measuring the change…
The hypothesis will be proven because when temperature is increased, energy is released and this energy breaks the forces that held up the atoms together of a substance. This being said, then when the temperature in a mixture increases, then the solute will be able to dissolve completely and each time temperature is increased, there should be more amount of solute that can be added. When the solute is saturated, then a precipitate will form and crystals should start appearing.…
For part A of the procedure we worked with the solubility of solid compounds in various solvents. The three solid compounds that were worked with during this procedure were benzophenone, malonic acid, and biphenyl. These three solids were then mixed with water (highly polar), methyl alcohol (intermediately polar), and hexanes (nonpolar). When benzophenone is mixed with water the results turned out to be insoluble because benzophenone is a pure hydrocarbon, which are very insoluble in water. When benzophenone was mixed with methyl alcohol, it was soluble because Methanol can hydrogen bond to the carbonyl oxygen of benzophenone. When benzophenone was added to hexanes it was partially soluble because “likes dissolve likes” benzophenone is polar and hexane in non polar. When the next compound, malonic acid was combined with water and with methyl alcohol it turned out to be soluble in both cases. However when malonic acid was combined with hexanes it was in soluble because Malonic acid is polar where as hexane is nonpolar. When the final compound, biphenyl was added to water it was insoluble because biphenyl is a non-polar molecule. Biphenyl was partially soluble when it was added to methyl alcohol but when hexane is added it is soluble because both are non-polar and "likes dissolves likes". For part B of the procedure we worked with the solubility of various alcohols in water and hexane. The three alcohols we used were 1-octanol, 1-butanol, and methyl alcohol. When 1-octanol was added to water it is insoluble because the longer the carbon chain, the more water is repelled. However when 1-octanol was added to hexane it was soluble because it is a polar molecule being that it is an alcohol. When 1-butanol is added to water it is partially soluble because 1-butanol slightly polar. Which means that it was soluble in hexane. When methyl alcohol was added to water it was…
The aim of this experiment was to test the effect of surface area on osmosis and the effect of solute concentration on osmosis. To test this aim to hypotheses were devised. 1) If potato pieces are immersed in various salt solutions, then the pieces with the greatest surface area to volume ratio will experience the greatest weight change, because more water can move by osmosis across the potato cell’s semi permeable membrane. 2) If potato pieces are immersed in various salt solutions, then the pieces immersed in the most concentrated solution will experience he greatest weight change, because more water must pass across the potato cells semi permeable membrane by osmosis to achieve an equilibrium. The results from testing these hypotheses did not support either and were shown to be flawed because they did not reflect the theories of osmosis relating to surface area to volume ration and salt concentration.…
As the concentration of a solute in water will, through osmosis, transfer until the concentration is nearly the same both inside the cell, or in this case potato, and outside of the cell, the potato strips both gained and lost mass due to the different concentrations of sugar in water. As expected, the potato strips placed in the concentration of sugar water closest to the concentration of sugar water in a potato had the least amount of change in mass. As the potato strips placed in substance Z and substance A both changed 7%, the concentration of sugar water in a potato must be somewhere in between a concentration of 0.4 and 0.6. One source of error for this lab is that the amount of substance placed into each Dixie cup was not measured and therefore were not consistent. If this lab were repeated, it is suggested that the amount of substance be measured in order to produce more consistent results.…
Through osmosis, I hypothesise that as the solution of the starch in each tube increases, the water content being reduced will cause the potato to release more water through the membrane to balance the starch solution, resulting in a decrease in the mass of the potatoes as the respective starch concentration increases, though maybe not for the 0% solution. Risk Assessment The potential danger of this experiment is relatively low. Nevertheless, lab coats and eye protection were worn when handling the starch concentration and extreme care was taken when cutting and weighing the potato using a scalpel on a slip-proof…
Conclusion: My hypothesis was correct; the potato cubes placed in distilled water and 0.2 M sucrose solution gained weight, starting with 5.97 grams and 7.35 grams respectively and ending with _____g and _____g. The ones in 1 M and 0.8 M…
Results- The solution had a Higher concentration the potato lost more mass then when the solution had medium concentration or no sugar concentration. The solution with the most sugar concentration lost 1.94 g of mass while the solution with medium concentration lost 1.13 g and the solution will with no sugar added tends gained 1.25 g of…
We started the lab by cutting the potato core length (44 mm). Each potato core are placed in test tubes. The test tubes have different salt concentration, test tube 1 has 0% salt concentration, test tube 2 has 0.625% salt concentration, test tube 3 has 1.25% salt concentration, test tube 4 has 2.5% salt concentration and test tube 5 has 5% salt concentration. The next day, we remove the potato core from the test tubes and measure their size. In test tube 1 with a 0% salt concentration, the potato core length after is 44 mm, the change in potato core length is 4 mm so the potato is hypertonic to that solution. In test tube 2 with a 0.625% salt concentration, the potato core length after is 42 mm, the change in potato length is 2 mm so the potato is hypertonic to that solution. In test tube 3 with a 1.25% salt concentration, the potato core length after is 39 mm, the change in potato length is -1 mm so the potato is…