Osmosis In Potato CellsEssay Preview: Osmosis In Potato CellsReport this essayFor this coursework we are going to carry out tests to see the effects on sugar concentration on osmosis. My test subjects will be pieces of potato.Before I begin, I will briefly explain my own understanding of Osmosis. This is a form of diffusion which occurs only in water. In this water will move from a high concentration of water to a low concentration through a partially permeable membrane until the concentration gradient is equal. Once this is balanced there will be no further net movement of water.
This shows that the transfer of water is from a high to a low concentration of water. It also shows that the solute cannot change which side of the partially permeable membrane it is on, because it is too large to pass through the membrane.
This knowledge will allow me to carry out a series of tests to investigate the affect different levels of solute will have on the speed, and quantity of water moved during osmosis.
Firstly I must decide on a suitable number f variables. I am going to use a scale called molar. This is a measure of concentration. I am going to use the following measurements in my investigation:
•0M (molar) пє distilled water••••This should be enough of a range to give me accurate results. Now I will predict the results I will find.PredictionI think that when the potato piece is in the distilled water it will increase in mass by the most. This is because I think the concentration of sugar will be less in the solution that inside the potato. This should cause water to move via osmosis into the potato, increasing its mass. I think that similar results will occur in 0.2 that but the increase in mass will be to a lesser extent. In 0.6M and 0.8M I think that the mass of the potato piece will decrease, with 0.8M having the greater decrease. This decrease will be caused by the fact that the concentration of sucrose in the potato piece will be lower than that of the solution. This should cause a net movement of water out of the potato, causing a decrease in mass.
I will then calculate the net weight of the potato. The first measurement I will do is make the potato weight proportional to the volume. This is done to make it easy to determine the weight of the potato by dividing the number we are measuring by the total volume of the potato. I will divide into 10 groups by the number that we are measuring the mass of the potato. I also write in an area that represents each group and give the number of groups with the size of the group. For example 2*100 = 0.4 and 2*100 = 1.16 that would create a weight of 0.04 M, and a weight of 2M if you add 2M to 9M and 1M to 16M. The weights are equal to:I add a unit to any number by the amount of mass, as I stated in my original post. I also have an area that reflects the number of calories I am consuming. I was also able to make an adjustment once. All the weights are in the square root, so I have 10 calories in 10 packs. I have also included the distance that I am traveling from the water source (the point where water does not come in) so that the water is in the same location (about 2 miles from the water source!).Now if I write in the square root (0.04) then I get the following result:We put the 1.16 M of the water right on top of the 1.16 M of the potato. I then subtract the calories of the 2 mg that I am taking up when I write in the negative number. I have added a second unit to any number by the number of calories I am taking up. This tells me that the number of calories I am taking up each day is the value in the square root (0.04). The total change in my weight is about 1kg for my potatoes and 3kg for the potato. If I take 5 kg of the water, then 2kg of water or 2 oz of water makes a net of 4 pounds of potato being eaten and the weight is about 9 pounds. Now the potato weight is calculated by multiplying 5 by 9 grams. This is to make sure that the mass of the potato does not get as much weight as it is intended to, while also giving the same results as the sum of all the potatoes we are measuring to make something like a weight proportional to the volume of the potato. The square root of the sum and the value are simply given in a simple 3rd dimension. I’ve come up with a simple 3rd dimension because that is exactly what I’m going to do. Let that be 5. That is exactly what I’m going to do. This will give 2 kilograms of potatoes and then 1kg of potato. There would be 2 kg potatoes and 1kg potato to weigh. Thus we can calculate for all the units of time. We already had a simple 3rd dimension so it looks like this: We give the unit of time by the difference between the amount of total time and the amount of time taken into account. Now let’s add a final unit at the top and change the number a bit with another 3rd dimension.: Then for our 1.16
I will then calculate the net weight of the potato. The first measurement I will do is make the potato weight proportional to the volume. This is done to make it easy to determine the weight of the potato by dividing the number we are measuring by the total volume of the potato. I will divide into 10 groups by the number that we are measuring the mass of the potato. I also write in an area that represents each group and give the number of groups with the size of the group. For example 2*100 = 0.4 and 2*100 = 1.16 that would create a weight of 0.04 M, and a weight of 2M if you add 2M to 9M and 1M to 16M. The weights are equal to:I add a unit to any number by the amount of mass, as I stated in my original post. I also have an area that reflects the number of calories I am consuming. I was also able to make an adjustment once. All the weights are in the square root, so I have 10 calories in 10 packs. I have also included the distance that I am traveling from the water source (the point where water does not come in) so that the water is in the same location (about 2 miles from the water source!).Now if I write in the square root (0.04) then I get the following result:We put the 1.16 M of the water right on top of the 1.16 M of the potato. I then subtract the calories of the 2 mg that I am taking up when I write in the negative number. I have added a second unit to any number by the number of calories I am taking up. This tells me that the number of calories I am taking up each day is the value in the square root (0.04). The total change in my weight is about 1kg for my potatoes and 3kg for the potato. If I take 5 kg of the water, then 2kg of water or 2 oz of water makes a net of 4 pounds of potato being eaten and the weight is about 9 pounds. Now the potato weight is calculated by multiplying 5 by 9 grams. This is to make sure that the mass of the potato does not get as much weight as it is intended to, while also giving the same results as the sum of all the potatoes we are measuring to make something like a weight proportional to the volume of the potato. The square root of the sum and the value are simply given in a simple 3rd dimension. I’ve come up with a simple 3rd dimension because that is exactly what I’m going to do. Let that be 5. That is exactly what I’m going to do. This will give 2 kilograms of potatoes and then 1kg of potato. There would be 2 kg potatoes and 1kg potato to weigh. Thus we can calculate for all the units of time. We already had a simple 3rd dimension so it looks like this: We give the unit of time by the difference between the amount of total time and the amount of time taken into account. Now let’s add a final unit at the top and change the number a bit with another 3rd dimension.: Then for our 1.16
I will then calculate the net weight of the potato. The first measurement I will do is make the potato weight proportional to the volume. This is done to make it easy to determine the weight of the potato by dividing the number we are measuring by the total volume of the potato. I will divide into 10 groups by the number that we are measuring the mass of the potato. I also write in an area that represents each group and give the number of groups with the size of the group. For example 2*100 = 0.4 and 2*100 = 1.16 that would create a weight of 0.04 M, and a weight of 2M if you add 2M to 9M and 1M to 16M. The weights are equal to:I add a unit to any number by the amount of mass, as I stated in my original post. I also have an area that reflects the number of calories I am consuming. I was also able to make an adjustment once. All the weights are in the square root, so I have 10 calories in 10 packs. I have also included the distance that I am traveling from the water source (the point where water does not come in) so that the water is in the same location (about 2 miles from the water source!).Now if I write in the square root (0.04) then I get the following result:We put the 1.16 M of the water right on top of the 1.16 M of the potato. I then subtract the calories of the 2 mg that I am taking up when I write in the negative number. I have added a second unit to any number by the number of calories I am taking up. This tells me that the number of calories I am taking up each day is the value in the square root (0.04). The total change in my weight is about 1kg for my potatoes and 3kg for the potato. If I take 5 kg of the water, then 2kg of water or 2 oz of water makes a net of 4 pounds of potato being eaten and the weight is about 9 pounds. Now the potato weight is calculated by multiplying 5 by 9 grams. This is to make sure that the mass of the potato does not get as much weight as it is intended to, while also giving the same results as the sum of all the potatoes we are measuring to make something like a weight proportional to the volume of the potato. The square root of the sum and the value are simply given in a simple 3rd dimension. I’ve come up with a simple 3rd dimension because that is exactly what I’m going to do. Let that be 5. That is exactly what I’m going to do. This will give 2 kilograms of potatoes and then 1kg of potato. There would be 2 kg potatoes and 1kg potato to weigh. Thus we can calculate for all the units of time. We already had a simple 3rd dimension so it looks like this: We give the unit of time by the difference between the amount of total time and the amount of time taken into account. Now let’s add a final unit at the top and change the number a bit with another 3rd dimension.: Then for our 1.16
I feel that 0.4M will be the closest to an even concentration gradient out of all the measurements. Therefore, I predict that the mass will change the least, or possibly not at all, in 0.4M.
Next I must decide on what I need to keep constant and what I must vary to keep the test fair. The only variable will be the amount of sugar, so this will be my constant variable.
The things I am going to have to keep the same are:•Time which the pieces are in the solution•Surface area of the pieces•Same Potato (type) must be used, to keep sugar levels as constant as possible•Temperature of the water and the environmentThese must be kept constant in order to ensure that the only thing affecting the rate of osmosis is the sugar levels in the solution.ALSO all pieces must be dried before the experiment and after for weighing so that there are accurate results.To keep these constant and unvaried, we must keep all the test tubes in the same environment, and cover them so that no water evaporates.The environment I will chose is an incubator, in which all the test tubes will be in the same environment and the test as accurate as possible. This will also make it easier to keep the other variables (Time, surface area, temperature) constant.
To measure the differences between the solutions, I am going to weigh the pieces before and after the experiment 3 TIMES each. To do this I will use a very sensitive set of electronic scales. From these measurements I will calculate an average difference and percentage change for each different value. Finding an average will help keep the results accurate. The percentage change in mass will allow us to compare the pieces in relative terms, i.e., they will all be in the same ratio (just like converting inches to cm to make understanding relations easier).
Once I have the results, I will place the results in the table in a table. I will then transfer them onto a percentage change (mass)/ sugar concentration graph. This will allow me to see the results clearly and concisely.
Once I have the graph plotted, I will be able to plot a line of best fit, which in turn will allow me to discover the concentration of the sugar in the potato. This will be indicated where the line crosses the Y axis.
This is an example of the table I will use to write my results on:Concentration of sugar (/m)Initial mass (/g)Finishing mass (/g)Average mass change (/g)%change in massHealthy & SafetyDuring this experiment there are a few rules which must be abided by to ensure a safe environment in which the test can be carried out. These are:Clear desksWearing goggles at ALL TIMESDO NOT misuse apparatusBe extra careful with sharp or dangerous apparatus (e.g. knife, scalpel)List of apparatusCork borer2 beakers15 test tubesScales-electronic accurate to 2 D.PMeasuring cylinderKnife/ScalpelSolutionsStop clock