Procedure Activity 1: Fill 6 large beaker halfway with distilled water, making sure all beakers have equal amounts of water. Cut 6 30 cm dialysis bags and label each bag with a letter, A through F. Fill each dialysis bag with 15 mL of solution A through F that corresponds with the lab on each bag. For example, bag A is filled with solution A. Measure the mass of each dialysis bag and record masses of each bag in BILL. Cover the beakers with paper towel and leave the bags in the beakers overnight. Remove the dialysis bags from the beakers and let dry. Measure the mass of each dialysis bag and record it in BILL Order dialysis bags A through F from highest to lowest mass, which becomes 1.0, 0.8, 0.6, 0.4, 0.2, 0.0 molar solutions respectively. …show more content…
The molarity of sucrose is the amount of sucrose, ergo, each solution had different amounts of sucrose, the solute to this solution. As the amount of sucrose increased, the amount of water decreased. To have more of a solute, there must be less of a solvent, as the milliliters of the solutions tested remained the same. An increased molarity of sucrose led to an increased net diffusion of water into the cell, a greater mass change, because water potential of the cell decreased as sucrose increased. From these masses, a percent of mass change was calculated: final minus the initial, all divided by the initial. When the percents were ordered from least to greatest, the molarity of each solution can be deduced, as those with the least change had a molarity closest to zero and those with the most change had a molarity that was the highest. This is due to the fact that if they had 0.0M of solute, they are isotonic and should have no net diffusion of water in a single direction. If they have higher concentrations of sucrose, the water potentials are no longer equal and to decrease entropy of the system, water must diffuse to achieve equilibrium. In this way, it was deduced that Solution B is 0.0M, Solution D is 0.2M, Solution F is 0.4M, Solution A is 0.6M, Solution E is 0.8M, and Solution C is 1.0M. The graph seconded this data by portraying an almost constant …show more content…
As the molarity of the solutions as been solved, they serve as indicators of water potential. This is because when the potato slices are placed in each of the six solutions, some experience net diffusion of water into the potato cells, out of the cell, or no det diffusion at all. This is due to water potential moving down its concentration gradient:free water flows from high concentration to low concentration, to reach equilibrium. In the lab, the potato slices with sucrose solutions less than 0.28M increased in mass because their water potential was lower than the solution around them: water diffused into the potato samples to reach equilibrium. These were hypotonic sucrose solutions. Potato slices that were in solutions greater than 0.28M decreased in mass because the water potential of the potato was high than that of the solution around them: water diffused out of the potato cells and into the sucrose solution around them. These sucrose solutions were hypertonic to the slices. As the graph indicates, the potato is isotonic to the sucrose molarity at 0.28M. That indicates that the water potential reaches equilibrium at 0.28M of sucrose. That isotonic value provides the molar concentration of the solute in the potato, which can then be plugged into the equation for solute potential: multiply the negative ionization constant, the pressure constant, the temperature,
The tonicity of solution relative to eggs in water and the 0.5M glucose were hypotonic, because each egg increased in mass over 60 minutes. The tonicity of the eggs in 1.5M glucose and 2.0M glucose were hypertonic because each egg decreased in mass over 60 minutes. The water moved towards the hypertonic solution through the membrane of the egg in the water and 0.5M glucose solutions, causing the egg to increase in mass. The water moved out of the hypotonic membrane in a hypertonic 1.5M and 2.0M glucose solutions, which caused the egg to decrease in mass. Relevance
Using the data from the first two columns, an x-y scatterplot graph was created. Analyzing the graph, a set of points that formed a linear curve were identified, and the plot of the graph was reduced to these points. This is the initial cooling curve. A second series was then added to the graph, with points that correspond to the interval when t-butyl alcohol was freezing. A trendline was then created for each of the series to obtain the equation of the line and r values.
Additionally, it was difficult obtaining a piece of rhubarb that was thin and particularly red, therefore the effect could not be best observed in the cells. Part B: Design your own experiment Parts of this practical were taken and slightly altered from the following link http://www.markedbyteachers.com/gcse/science/investigate-the-effect-of-surface-area-on-osmosis-in-potato-tissue.html Aim: To observe the effect different surface area: volume ratios have on osmosis in potato tissue. Hypothesis: If the potato has a larger surface area: volume ratio, the quicker osmosis will take place and the larger the mass will be at the end of the experiment, therefore the difference in mass of the potatoes from the start of the experiment to the end of the experiment will be larger. Additionally, the potato pieces left in a saltwater solution will decrease in mass, whereas the pieces left in water will increase in mass.
When temperature is increased, the amount of obtainable energy increases; meaning that particles will move at faster pace at a higher temperature. Thus rate at which molecules diffuse will progressively speed up as the temperature increases. However if temperature of solution is decreased the rate of osmosis will decrease and rate at which molecules diffuse will be significantly less than that of higher
Pat McGurrin October 24, 2015 Period #1 Honors Biology Mr. Dinunzio Murder and Meal Lab Analysis Procedure: 1.) Gather all materials: Safety goggles, 250ml beaker, water, hot-plate, test-tubes, paper bag tear, stomach contents, pipette, Biruet solution, Benedict’s solution, and Iodine solution. 2.) Put on safety glasses.
Purpose: The purpose of this experiment is to investigate the movement of water into and out of a polymer. Hypotheses: • If Gummi Bears are submerged in tap water and distilled water, they will expand due to osmosis in order to create equilibrium in hypotonic solution • If Gummi Bears are submerged in salt water, they will decrease inside to create equilibrium in a hypertonic solution Materials: • 2 Gummi Bears • 2 plastic 8oz cups • 2 forks • Distilled water • Centimeter rule • Saturated salt solution • Paper towels • Electric scale • Permanent marker Procedures: 1. Obtain two plastic cups and two different color bears. 2.
We zeroed out the scale and weighed all four potato cores at once and recorded the mass. We then put those potato cores into the beaker of 75 mL of solution. With the potato cores in the beaker we then put a watch glass over the top of the beaker to minimize the amount of solution that evaporates. We let the potato cores sit in the solution overnight. The next day we then emptied the beaker of the solution by carefully draining the solution, while not letting the potato cores fall out.
Osmosis Lab Report Research Question: How does the change in the concentration of a sucrose solution affect the process of osmosis in a potato cell by measuring its mass? Background information: 1 Osmosis is the process by which a liquid passes through a semi-permeable membrane, moving from an area with a high concentration of water to a low concentration of water. There are various factors that affect osmosis such as: concentration, surface area and temperature. The concentration of solutions can affect the rate of osmosis, as there is more difference in the concentration of the solutions, which means osmosis, will take place quicker. Surface area could affect osmosis based on the ease by which molecules can get through the semi-permeable
Sodium Chloride, is soluble in water, because it is polar and has an alternating negative anion and positive cation. Water is also polar, so like will dissolve in like. Therefore, the negative part of the chloride ions are attracted to the positive side of the water molecule and the negative side of the water molecules is attracted to the positive sodium atom. The negative oxygen charges in water, attract to the positive sodium ions in sodium chloride. Sucrose is soluble in water, because it is a polar molecule and the polar molecules in water attract the positive and negative regions around the sucrose molecules, which allows sucrose to be soluble in water.
The hypothesis we came up with for this project was that in the distilled water there wouldn't be no change in the potato, we wouldn’t see a gain or loss with the water sitting in the beaker. The beaker with the 30% Sucrose and Distilled water we predicted that there was going to be weight gain to the potato. And for the distilled water with the 30% sucrose and we predicted weight loss. But the results came out to be the first bag content being isotonic the second bag came out to be hypotonic and lastly the result came out to be
In this experiment, the amount of water lost in the 0.99 gram sample of hydrated salt was 0.35 grams, meaning that 35.4% of the salt’s mass was water. The unknown salt’s percent water is closest to that of Copper (II) Sulfate Pentahydrate, or CuSO4 ⋅ 5H2O. The percent error from the accepted percent water in CuSO4 ⋅ 5H2O is 1.67%, since the calculated value came out to be 0.6 less than the accepted value of 36.0%.This lab may have had some issues or sources of error, including the possibility of insufficient heating, meaning that some water may not have evaporated, that the scale was uncalibrated, or that the evaporating dish was still hot while being measured. This would have resulted in convection currents pushing up on the plate and making it seem lighter by lifting it up
Practical I: Acid-base equilibrium & pH of solutions Aims/Objectives: 1. To determine the pH range where the indicator changes colour. 2. To identify the suitable indicators for different titrations. 3.
The experiment shall use several concentrations of sucrose solution and a substance known as Methylene blue. A piece of potato/ carrot shall be placed in a boiling tube and the solution shall be poured into it. This tube shall have Methylene blue added into it. After incubation some of this solution shall be taken out with a pipette and inserted into a separate boiling tube containing the same sucrose solution however this solution shall be known as the pre-incubated solution. The drop shall be watched so as to see if the density of the water and concentration of sucrose has increased or not, displaying the water
Biology Design Practical Joshua Edwards What are effects of the volume of a potato and the amount of weight it loses when placed in salt solution? Introduction This design practical uses a potato’s surface area to volume ratio to see what affects it has on osmosis in different concentrations. Osmosis is the movement of water molecules through a cell membrane into an area of a higher solute concentration. The movement goes the way of the solvent with more solute because the lower solute concentration is drifting through balancing the ratio of solute per solvent (En.wikipedia.org, 2018).
The membrane only allows certain molecules to pass through, water molecules are easily able to go through (Freeman 2011). Since sodium chloride is a normal saline it contains salt. The salt ions are not easily able to pass through the cell membrane like water. Compared to water, the sodium chloride should make the potatoes weigh less than in water, which should support our