• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Biology - osmosis

Free essay example:

Biology coursework: Osmosis in potato cells

Planning:

Preliminary investigation:

We did a preliminary experiment to help us decide on the method and what concentrations should be used in the actual experiment. We are trying to discover what the best solution for a potato not to lose or gain any weight also the right sucrose concentration to keep it healthy.

So for the preliminary we decided to use four different concentrations they are:

• Distilled water
• 0.1 Molar sugar concentration
• 1 Molar sugar concentration
• 2 Molar sugar concentration

Method:

This is how we carried out the experiment:

• Using a corer I pulled out 4 pieces of potato
• Checked they were the same size, if not I cut them to size
• Placed each piece in a test tube
• Using a measuring cylinder, I measured the different types of solution required
• Filled the separate test tubes with the solution required up to the 2nd line
• Labeled the test tubes as distilled water, 0.1M, 1M and 2M
• Placed the 4 test tubes in the tub which had water heated at body temperature (37 degrees Celsius)
• Left it in the tub for 24 hours
• Took out the test tubes and placed them on a paper towel in the order I put them in so that I wouldn’t get confused
• Dried each potato chip with the towel to pat off excess moisture
• Put them one by one on the electronic scale and weighed them
• Recorded the results for each chip
• As I had time after doing the first set of results I redid the experiment under exactly the same conditions. This gave me a second set of results which gave me a more accurate view on and changes that occurred during osmosis.

Obtaining Results:

 Test Liquid in test tube Original mass before (g) Average mass before (g) Final mass after (g) Average mass after (g) Average change in mass (g) 1 Distilled Water 1. 1.432. 2.14 1.79 1. 1.792. 2.47 2.13 0.34 2 0.1M sugar solution 1. 1.432. 2.13 1.78 1. 1.652. 2.37 2.01 0.23 3 1M sugar solution 1. 1.442. 2.11 1.78 1. 1.032. 1.49 1.26 -0.52 4 2M sugar solution 1. 1.442. 2.11 1.78 1. 0.942. 1.26 1.1 -0.68 % Change in mass of chip (-/+cl) Gain/loss 19.00% Gain water 12.92% Gain water -29.21% Lose water -38.20% Lose water

Average mass before:

Distilled water = 1.43+2.14

= 3.57

= 3.57/2

= 1.79

0.1M = 1.43+2.13

= 3.56

= 3.56/2

= 1.78

1M = 1.44+2.11

= 3.55

= 3.55/2

= 1.78

2M = 1.44+2.11

= 3.55

= 3.55/2

= 1.78

Average mass after:

Distilled water = 1.79+2.47

= 4.26

= 4.26/2

= 2.13

0.1M = 1.65+2.37

= 4.02

= 4.02/2

= 2.01

1M = 1.03+1.49

= 2.52

= 2.52/2

= 1.26

2M = 0.94+1.26

= 2.20

= 2.20/2

= 1.10

Average change in mass:

Distilled water = 2.13-1.79

= 0.34

0.1M = 2.01-1.78

= 0.23

1M = 1.26-1.78

= -0.52

2M = 1.10-1.78

= -0.68

% change in mass:

Average change in mass/average mass before * 100

Distilled water = 0.34/1.79

= 0.19

= 0.19*100

= 19.00%

0.1M = 0.23/1.78

= 0.13

= 0.13*100

= 12.92%

1M = -0.52/1.78

= -0.29

= -0.29*100

= -29.21%

2M = -0.68/1.78

= -0.38

= -0.38*100

= -38.20% Using the results that I obtained, I drew this graph and as you can see there are no anomalous results, the solution with the highest average % change in mass is distilled water with about 20%. The solution with the lowest amount of % change is the 2M sugar solution with about    -40%. So we decided that for the actual experiment we wouldn’t use 2M as it is pointless because we can tell from 1M that there is a high concentration of sugar. Therefore for the actual experiment we are going to use 0M, 0.2M, 0.4M, 0.6M, 0.8M and 1M.

Actual Investigation

Aim:

My aim is to investigate how different concentrations of sucrose affect the mass of a potato chip.

Hypothesis:

Osmosis is the net movement of water molecules, through a selectively permeable membrane, from a region where there is a high concentration of water molecules to a region where there is a lower concentration of water molecules. Osmosis is a special sort of diffusion, where only water molecules can diffuse through a membrane. Larger molecules are unable to pass through small pores. The molecules will continue to diffuse until the area in which the molecules are found reaches a state of equilibrium, meaning that the molecules are randomly distributed throughout an object, with no area having a higher or lower concentration than any other.

An example of osmosis would be when a substance such as sugar dissolves in water, the sugar molecules attract some of the water molecules and stop them moving freely. This in effect, reduces the concentration of water molecules. If the water molecules were on the left and the sugar solution on the right, then the sugar molecules on the right would have “captured” half the water molecules. There are more free water molecules on the left of the membrane than on the right, so water will diffuse more rapidly from left to right across the membrane than from left to right. This is also because the cell membrane behaves like a partially permeable membrane.

This is shown in the diagram below: Prediction:

I predict that if I increase the concentration of sucrose the potato chip will decrease in weight and become turgid (they become hard and swollen). This is because osmosis causes water particles to travel from a high region to a low region through a partially permeable membrane, it only allows water molecules as the sucrose particles are too big.

The water moves to give an equal concentration of sucrose, the water from inside going outside until the concentration is equal.

Therefore a high concentration of sucrose will have less water particles causing more water to leave the potato to make the amount of water inside and outside the potato even.

So, 0m will have a higher mass as more water particles are outside the potato causing them to move inside whereas in 1m, there is more sugar solution causing the little water inside the potato to travel outside.

Fair Test:

To ensure that the experiment was carried out fairly we did the following things:

• We kept the same type of potato as we have to make sure that there is the same concentration of sugar and water.
• Volume of sugar needs to be the same to guarantee the rate of osmosis is the same.
• The lids have to be tightly closed to prevent evaporation.
• Test tubes have to be the same size
• Storing the tubs at the same temperature to make sure the rate of osmosis is the same.
• Make sure the scales are set at 0.
• The potato has to be pat dry to shake of any excess moisture.
• Washing out the measuring cylinder after each solution is used.
• Using the same corer so that the potatos are the same size.
• Repeat the experiment to gain more results.

Safety:

During the experiment we made sure we were in safe working environment by doing the following:

• Wore goggles
• Tied our hair back
• Stood up
• Used a chopping board when cutting the potatos to size
• We were careful when using the scalpel
• Every time I came in touch with the potato chips I would wash my hands and dry them because the chip could have become contaminated and extra water could have fallen on to the chip

Apparatus:

For the preliminary we used the following equipment:

• Distilled water (Tap water could have been contaminated)
• 0.1M sugar solution
• 1M sugar solution
• 2M sugar solution
• Potato
• Corer (To make in potato)
• Labels
• Scalpel
• Chopping board
• 4 test tubes
• A balance (Scales)
• Ruler (make sure the potatos are the same size)
• Measuring cylinder

Variables:

There are 3 variables that we are using, they are:

• Independent variable – we are changing the concentration, because a variety of concentrations will give me varied results which will help me come to a decent conclusion.
• Dependant variable – we are measuring the mass of the potato before and after we test it for osmosis.
• Controlled variable – we are keeping the potato, potato chips length, volume of solution, temperature and the length of time the same. For example if one of the potato chips were a cm longer than the others, it would have a larger surface area therefore a much bigger space for osmosis to occur.

Method:

This is how we carried out the experiment:

• Using a corer we pulled out 6 pieces of potato
• Checked they were the same size, if not I cut them to size
• Weighed each potato chip on the scale
• Recorded the results
• I carefully placed each piece in a test tube
• Labeled the test tubes as 0M, 0.2M, 0.4M, 0.6M, 0.8M and 1M
• Using a measuring cylinder, I measured the different types of solution required
• Filled the separate test tubes with their required solution up until  the 2nd line
• Placed the 6 test tubes in the tub which had water heated at body temperature (37 degrees Celsius)
• Left it in the tub for 24 hours
• Took out the test tubes and placed them on a paper towel in the order I put them in so that I wouldn’t get confused
• Dried each potato chip with the towel to pat off excess moisture
• Put them one by one on the electronic scale and weighed them
• Recorded the results for each chip
• As I had time after doing the first set of results I redid the experiment under exactly the same conditions. This gave me a second set of results which gave me a more accurate view on and changes that occurred during osmosis.

Obtaining results:

 Concentration (M) Mass before (g) Average mass before (g) Mass after (g) Average mass after (g) Average change in mass (g) % change in mass (%) 0 1. 1.932. 2.483. 2.62 2.34 1. 2.802. 3.033. 3.03 2.95 0.61 26.07% 0.2 1. 2.002. 2.523. 2.64 2.39 1. 1.992. 3.013. 2.82 2.61 0.22 9.21% 0.4 1. 2.042. 2.533. 2.62 2.40 1. 1.952. 2.703. 2.52 2.39 -0.01 -0.42% 0.6 1. 2.002. 2.503. 2.51 2.34 1. 1.522. 2.303. 2.13 1.98 -0.36 -15.38% 0.8 1. 2.042. 2.533. 2.55 2.37 1. 1.382. 2.143. 1.84 1.79 -0.58 -24.47% 1 1. 2.122. 2.493. 2.44 2.35 1. 1.302. 1.683. 1.63 1.54 -0.81 -34.47%

Calculations:

Average mass before:

0M = 1.93+2.48+2.62

= 7.04

= 7.04/3

= 2.34

0.2M = 2.00+2.52+2.64

= 7.16

= 7.16/3

= 2.39

0.4M = 2.04+2.53+2.62

= 7.19

= 7.19/3

= 2.40

0.6M = 2.00+2.50+2.51

= 7.01

= 7.01/3

= 2.34

0.8M = 2.04+2.53+2.55

= 7.12

= 7.12/3

= 2.37

1M = 2.12+2.49+2.44

= 7.05

= 7.05/3

= 2.35

Average mass after:

0M = 2.80+3.03+3.03

= 8.86

= 8.86/3

= 2.95

0.2M = 1.99+3.01+2.82

= 7.82

= 7.82/3

= 2.61

0.4M = 1.95+2.70+2.52

= 7.17

= 7.17/3

= 2.39

0.6M = 1.52+2.30+2.13

= 5.95

= 5.95/3

= 1.98

0.8M = 1.38+2.14+1.84

= 5.36

= 5.36/3

= 1.79

1M = 1.30+1.68+1.63

= 4.61

= 4.61/3

= 1.54

Average change in mass:

0M = 2.95-2.34

= 0.61

0.2M = 2.61-2.39

= 0.22

0.4M = 2.39-2.40

= -0.01

0.6M = 1.98-2.34

= -0.36

0.8M = 1.79-2.37

= -0.58

1M = 1.54-2.35

= -0.81

% change in mass:

Average change in mass/average mass before * 100

0M = 0.61/2.34

= 0.26

= 0.26*100

= 26.07%

0.2M = 0.22/2.39

= 0.09*100

= 9.21%

0.4M = -0.01/2.40

= 0.04

= 0.04*100

= -0.42%

0.6M = -0.36/2.34

= 0.15

= 0.15*100

= -15.38%

0.8M = -0.58/2.37

= -0.24

= -0.24*100

= -24.47%

1M = -0.81/2.35

= -0.34

= -0.34*100

= -34.47%

Analysis: I have drawn a graph using the results that I obtained from the experiment. As you can see there are no anomalous results, my graph proves that my prediction is right because there is a pattern which shows that as the concentration of sucrose increases the average percentage change in mass decreases.It also shows that the potato cells increase in mass in solutions with a high water concentration and decrease in mass in solutions with a low water concentration.
The graph shows that the amounts of percentage gain and loss in inversely proportional to the concentration. The gradient does not change in my graph. But as the percentage change in mass decreases the points come closer together. This is because the potato chip is becoming as flaccid as it possibly can. This results in the mass of the molar concentration becoming closer together. My graph proves that my results are fairly reliable as it fits in with my prediction of the experiment graph.

Evaluation:

The cutting of the potatoes was the most difficult part of the experiment as although I was recording my results by mass, it could well have affected the surface area and so the overall rate of osmosis. If I were to repeat the experiment I would have possibly found a machine to cut the potato as it would ensure that all potatoes would be the same weight and dimensions. As well as the potato I could have found a more accurate way to measure out the solutions and to determine the molar concentrations. Perhaps I could have used a burette. This would ensure that I have an accurate amount of fluid in each test tube. I could also weigh each chip on a more accurate scale, e.g. not to 0.00g but to 0.0000g. Because of this I had to change my plan several times at the beginning because of things like wanting the potato chips to be originally 6-7cm in size. When it came down to the actual experiment I realised that the test-tube won’t be able to hold the potato chips that big so I had to cut it down to 5cm. I also had to change the size because of the fact that the potato chip caused the solution to overflow.

The experiment was very successful in my opinion. I obtained a large quantity of very accurate results because I repeated the experiment 3 times from these valid results I was able to create an informative graph. The repeated results are fairly similar to the original this also helps to prove that my prediction was right. I think I took easily enough results for the amount of concentrations that I was using, and the time that I used for the experiment to last was enough to allow sufficient osmosis to occur.

Conclusion:

From this investigation I can conclude that the potato chip that was placed in the lowest concentration - 0M sucrose solution gained the most weight at 26.07%. The higher concentrations tend to lose weight as the 0.8M lost 24.47% and the 1M lost 34.47% as the potato could not expand further to take in more water, this is when plasmolysis occurred. The graph also shows a clear indication that was an overall decrease in mass during the experiment.

Further investigation:

If I was to repeat this experiment I would use a different root vegetable such as a carrot, sweet potato or a turnip. I would also use a different soluble solution like salt, this will prove that osmosis can occur in anything that has water.

I would use an alternate method preferably the visking tube method otherwise known as the dialysis tubing. This is a good method to use as it is a semi permeable membrane that allows small molecules like water to pass through but does not allow larger molecules like sugar. If solutions of different concentration are on either side of the visking membrane, water molecules will pass through and tend to dilute the more concentrated solution. Osmosis doesn’t allow large molecules to pass through either that is why this is an ideal method to use.

This is how I would carry out the experiment using the visking tube method:

• The visking tubing is softened under a running tap until it opens up. A knot should be tied at one end to make a bag
• A bung that is just small enough to fit delicately inside the visking tubing is inserted into the unknotted end
• Then a ring of suitable bore rubber tubing is cut to make a band.
• This band is then slipped over the visking tubing so that it grips the tubing to the bung
• The reason that food dye is added to the sucrose solution is to make the level in the capillary tube easy to see
• The bag is filled completely to the brim with the coloured sucrose solution by squirting a pipette full at a time through the hole in the rubber bung
• No air bubbles can remain
• At this stage the knot and the attachment to the bung is checked for leaks
• The capillary tubing is then carefully inserted into the bung
• The fit needs to be tight and the sucrose should come up about half way
• If the level is too high the visking bag can be gently squeezed to force a little solution out of the top of the capillary. If the level is too low, a dialysis clip is attached just above the knot to reduce the volume of the bag.
• A ring from some narrow bore rubber tubing is cut and slipped over the capillary tube to act as a marker for the level
• You then clamp the apparatus to the stand
• Place into a large beaker of 0.5M sucrose solution (Make sure the visking bag is hanging freely in the solution. I.e. not touching the sides or bottom. This stops it from being squashed and distorting the results)
• Slide the rubber band to the top of the capillary level in order to mark the starting position
• Put the apparatus into 2.0M sucrose solution
• The level in the capillary tube will fall as the water inside passes through the visking tubing in order to dilute the stronger solution outside
• Put the apparatus into distilled water
• The level in the capillary tube will rise as the water passes into the visking bag to dilute the stronger solution inside.

During the experiment the precautions to be taken are as follows:

• The apparatus cannot be stored for more than a few days with the sugar solution inside as it will go mouldy
• The capillary tube should be removed and thoroughly rinsed as any sugar left inside will harden and block the tube
• The visking bags can be left attached to the bungs but should be rinsed and stored in water or a mixture of water and ethanol to keep the bag soft
• If the visking dries out it may become brittle and crack

Expected results:

I expect osmosis to take place and to see the pattern that of the concentration of sucrose to increase as the average percentage mass to decrease.

Ayshah Tariq 10:8

This student written piece of work is one of many that can be found in our GCSE Life Processes & Cells section. Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Life Processes & Cells essays

1.  ## Investigating the effect of Sucrose Concentration on the Rate of Osmosis in Potato Chips.

5 star(s)

These differences might have been caused because each potato had a different weight before the experiment. It is unlikely that one could repeat this experiment using potatoes of exactly the same weight in order to retrieve exact results. This is because we assume that potatoes own unique genetic make - up dissimilar to other potatoes.

2. ## Diffusion in Agar Block

This could lead to errors of up to 0.5cm per side. This error will also affect the volume and surface area and calculations suggest that each may have had an error of 1.5cm3 and 1.5 cm2. * When the agar was being taken out, gases in the air like carbon

1. ## Osmosis is defined as 'the movement of water molecules from an area of high ...

The chip must be totally covered in the solution, and the amount of solution will be kept the same because all the chips are the same size. The amount of solution I cover each chip in will be 25ml. The type of potatoes I will use is going to be

2. ## The Effect of Different Concentrations of Sucrose On Visking Tubing

I will then measure them again and record. I will then work out the mass change and percentage change. I will record my results in a table like this. Solution (M) Number Original mass (g) Final mass (g) Mass change Percentage change Prediction: In my experiment, I am going to • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 