Cylinders or discs of fresh potato are often used to investigate osmosis in living cells. To carry out this type of experiment, you need to:
- cut equal-sized pieces of potato
- blot with tissue paper and weigh
- put pieces into different concentrations of sucrose solution for a few hours
- remove, blot with tissue paper and reweigh
The percentage change in mass can be calculated for each piece of potato:
A piece of potato has a mass of 2.5 g at the start and 3.0 g at the end.
percentage change in mass = (3.0 – 2.5) ÷ 2.5 × 100 = 0.5 ÷ 2.5 × 100 = +20%
The plus sign shows that it has gained mass - it will have gained water by osmosis.
A piece of potato has a mass of 2.5 g at the start and 2.0 g at the end.
percentage change in mass = (2.0 – 2.5) ÷ 2.5 × 100 = –0.5 ÷ 2.5 × 100 = –20%
The minus sign shows that it has lost mass - it will have lost water by osmosis.
A graph of change in mass (vertical axis) against concentration of sucrose (horizontal axis) can be plotted.
Where the line crosses the horizontal axis at 0% change in mass, the sucrose concentration is equal to the concentration of the contents of the potato cells. The sucrose concentration is isotonic with the cells' cytoplasm, so there is no net movement of water by osmosis.
Investigate the factors affecting the rate of Osmosis
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Investigate the factors affecting the rate of Osmosis
Osmosis is the diffusion of water from a lower concentration of solute
to a higher concentration of solute, through a partially permeable
membrane. In a high concentration of water the amount of solute (e.g.
sucrose) is low. This could be called a weak or dilute solution. In a
low concentration of water the amount of solute is high. This could be
called a strong or concentrated solution. When a partially permeable
membrane divides two such solutions, the water will move from the area
of high concentration to the area of low concentration until both
sides are equal. In plant cells water flows through the cell wall and
cell surface membrane into the vacuole from the outside. As a result,
the cell swells up. It doesn't burst as the cell wall stops the cell
expanding too much; at this point we say a cell is fully turgid.
The purpose of my investigation is to investigate the factors
affecting the rate of Osmosis. The key factors I have found affect
Length - of membrane
Temperature - of water
Concentration - of solution
The factor I am going to investigate is the concentration of the
solution in which the potatoes are placed. By doing this I will have
to control all other factors to ensure a fair test, these are; doing
all the tests at room temperature to control the temperature as a
higher temperature would mean diffusion would occur quicker. The mass
and surface area of the potatoes at the beginning of the experiment
are other controls I will need to monitor to ensure a fair test. The
mass of the potato is a dependent variable; this means that it will be
measured throughout the experiment. I will measure the mass in grams
correct to 2d.p. The potato chip will be measured before it is put in
the solution, and after. This will allow us to see whether osmosis has
taken place, and to what extent.
The volume of the solution that the potato chips are kept in must be
fair. The must be totally covered in the solution, and the amount of
solution will be kept the same because all the potato chips are the
same size. The amount of solution I cover each chip in will be 20cm.
I predict in my experiment that the higher the concentration of sugar
in the solution, the smaller the potato chip will become. I predict
this as the greater the concentration of water in the external
solution, the greater the amount of water that enters the cell by
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Rate Of Osmosis Key Factors Solute Potato Chip Plant Cells Stops Membrane Variable Potato Chips
osmosis. The smaller the concentration of water in the external
solution the greater the amount of water that leaves the cell. The
potato will become smaller showing the effects of osmosis. Likewise if
there were a lower concentration of sucrose I would expect this to
reverse and the potato chip would become larger. Sometimes there also
may be no change in mass as the solution inside the cells is equal to
that concentration of the solution outside; therefore there would be
no need for osmosis to occur. From my results prediction I can see my
results graph should look something like the sketch on the enclosed
q 6 Test tubes, to put the potato chips and sucrose solution in. I
have chosen to use test tubes, as they are the correct size to
completely submerse the potato in solution without using an excessive
q 6 bungs, to cover the test tubes.
q Test tube rack, to hold the test tubes upright preventing spillage
and maintaining the position of the potato cylinders ensuring coverage
by the solution.
q Pins, to support chips on underside of bung.
q Cork borer no.6, to cut uniform diameter chips.
q Sucrose solution, to make a range of concentrations (shown below).
q Scalpel, to adjust any potato to correct length.
q Tile, to cut potato chips on safely.
q Distilled Water, to make up different concentrations of sucrose
q Electrical digital balance, for weighing potato chips to 2.d.p for
obtaining accurate results.
q Potatoes, to be used as testable plant material.
q Potato peeler, to peel potatoes before investigation.
· Collect all equipment.
· Peel potatoes to remove the skin, which is a different tissue. Cut
up 12 pieces of potato using no.6 cork borer. 2 for each test tube as
it will be more accurate than 1 and give a more accurate result, as I
will be able to then work out an average.
· Find the mass of the potato pieces in grams correct to 2.d.p using
electronic balance. If necessary cut a small piece off potato to make
all pieces of an even mass.
· Measure solutions into test tubes as follows;
1. 20cm distilled water, 0cm sucrose solution
2. 18cm distilled water, 2cm sucrose solution
3. 16cm distilled water, 4cm sucrose solution
4. 14cm distilled water, 6cm sucrose solution
5. 12cm distilled water, 8cm sucrose solution
6. 10cm distilled water 10cm sucrose solution
Put solutions into test tubes and place in test tube rack.
· Place 1 pin with 2 pieces of potato into the test tubes and place
· Leave at room temperature out of direct sunlight for 24 hours.
· After 24 hours I will take my results by re-measuring each potato
mass and recording it.
Results and Analysis
I can see here the first mass readings don't seem to change much. Or
have any regular pattern. There could be a number of reasons why this
has occurred, including mass values not being recorded correctly or
these potato samples may have not been completely submerged in the
The second readings seem to have a clearer pattern. The mass seems to
decrease as the sucrose solution increases though there seems one
anomalous result for 0.6 moles, which could have been caused by a
mistake in measuring potato or sucrose solution.
I have now worked out the percentage change for each reading using;
Average % change = change in mass x 100
And the average by calculating;
Reading1 + reading 2 = Average
In osmosis there will be a point where the concentrations of water
inside and outside the potato cells are equal (isotonic). At this
point there will be no change in the length, volume and mass of the
potato, as the net movement of water will be zero, no osmosis has
occurred. This means on my graph the line would meet the horizontal
axis showing no net movement and equal concentrations.
Using a graph I have now plotted for my results I could accurately
predict results for any sucrose solution between 0 and 1 moles using
my line of best fit.
I have managed to prove my original theory on how factors affect the
rate of osmosis. My results show the larger the concentration of
sucrose solution the smaller the mass of the potato chip became.
Really I should have used a series of more concentrated sugar
solutions as I can only prove so much of my theory as all, 5 readings
showed an increase in mass. For a more reliable investigation perhaps
I should have used some solutions, which would have produced a % loss
in mass. This would have furthered my experiment and improved the
quality of my evidence, as I would have been able to plot more points
on my graph.
I did find one anomalous result for 0.4 moles, which could have been
obtained for reasons described in my results. Looking at my results
either side of 0.04m, I would have expected the % change to be about
46% so it would fit on the graph with the line of best fit. I do
however, not believe this result to have had a detrimental effect on
my final results.
I also found my first set of readings unhelpful but as I had done 2
sets of readings I could take an average of the two to make my results
more reliable. This did mean my overall percentage increase might have
been slightly different, as these results would have affected them.
If I did this experiment again I would be more careful in my setting
up the experiment to ensure my results would be accurate. I had
problems cutting the potatoes so they may have ended up slightly
different sizes, which would account for any anomalous results.
Another reason the results could have occurred may have been because
the potato cores were not completely submersed in the sucrose
solution. The final weighing may have not been accurate as not all the
cores were blotted dry properly before weighing. Sugar solution
sticking to the core would have increased the final mass artificially.
To eliminate this bias I could have invented a universal blotting
regime or measured length increase instead of mass.
For a further experiment I could have investigated the effect other
factors had on the rate of osmosis. The temperature of the water in
the sugar solution is one. It would prove that the protein in the
potato membrane denatures after the optimum temperature (usually about
35 degrees c.) and osmosis would no longer occur as the membrane is
destroyed. Another factor that may have been investigated could be the
increase in length of potato chip.
Biology nelson science; by Michael Roberts.
Lonsdale/AQA study guide; edited by Mary James.
Encarta encyclopaedia; 2001 version.