Investigating the Equivalent Molarity of Potato Tissue Compared to Sucrose Solutions

Investigating the Equivalent Molarity of Potato Tissue Compared to Sucrose Solutions

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Investigating the Equivalent Molarity of Potato Tissue Compared to Sucrose Solutions

Plan

Osmosis occurs across a semi-permeable membrane where there is a
difference in concentrations on the two sides of the membrane, and
knowing that the cells will either become turgid (when water passes
into them)or plasmolysed (when water passes out of them) and will
therefore alter their volume. Our aim is to try and find out the
equivalent molarity by the change in length of a potato chip when
submerged in sucrose solution.

Prediction

We have studied turgidity and plasmolosis in a text book (biology for
life) and in a preliminary experiment, where we tested osmosis on
onion cells and saw them become turgid or plasmolysed. Although, we
used only 2 concentrations. Quite simply one solution to turn it
turgid and the other plasmolysed, and did not actually measure the
amount of concentration or expansion. Therefore my prediction is that
the lower the concentration of the sucrose solution the lager the mass
of the potato will be. This is because the water molecules pass from
the lower sucrose solution to the higher sucrose solution, the potato.
Therefore, the chips in higher sucrose concentration will have a
smaller mass equal to the equivalent concentration.

Further information on potato plant cells

Source: Nelson Advance Science

Plant cells always have a strong cell wall surrounding them. When they
take up water by osmosis they start to swell, but the cell wall
prevents them from bursting. Plant cells become "turgid" when they are
put in dilute solutions. Turgid means swollen and hard. The pressure
inside the cell rises and eventually the internal pressure of the cell
is so high that no more water can enter the cell. This liquid or
hydrostatic pressure works against osmosis. Turgidity is very
important to plants because this is what make the green parts of the
plant "stand up" into the sunlight.

When plant cells are placed in concentrated sugar solutions they lose
water by osmosis and they become "flaccid." This is the exact opposite

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of "turgid". The contents of the potato cells shrink and pulls away
from the cell wall. These cells are said to be plasmolysed.

When plant cells are placed in a solution which has exactly the same
osmotic strength as the cells they are in a state between turgidity
and flaccidity. We call this incipient plasmolysis. "Incipient" means
"about to be".

Variables

To construct a fair test I will have to keep certain aspects of the
experiment the same. For this investigation I intend to vary the
concentration of the sucrose solutions. This ought give me a good set
of results from which I will draw a decent conclusion. If any of the
non-variables below were not kept the same it would not be a fair test
or entirely alter my results. For instance if one potato chip was 1cm
wider, longer, shorter or thinner than the other the surface area for
osmosis to occur would be altered

· For the purpose of my experiment I am going to do all the
experiments at room temperature.

· To keep the water potential of the potato initially will be kept the
same by using the same type of potato

· throughout the experiment. I will measure the length in 'mm'. The
potato chip will be measured before it is put in the solution, and
after. This will allow us to work out the percentage change of the
potato after osmosis has taken place.

· The volume of the solution that the potato chips are kept in must be
fair. They 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.

Apparatus

For the experiment we will require:

Large potato

Sharp knife

Petri dishes

Ruler

Sucrose solutions (0.2mol, 0.4mol, 0.6mol 0.8mol)

Distilled water

Method

i) set up five Petri dishes containing the above sucrose solutions and
A zero fluid should be include using distilled water. Label each base
on its lid.

ii) Using a sharp knife cut the large potato (checking it was health
and hard) into 15 4cmlong by approx.0.5cm wide by approx.0.5cm depth.
Making sure length is the main measurement priority. Place 3 chips to
each Petri dish making sure they are completely bathed in the solution
(making sure this is kept as constant as possible for each chip, to
keep the fair test conditions.)

iii) At the start of the next biology class remove the chips from
their dishes and measure the new length (if changed at all) and record
them in a well constructed table to clearly show the new length of
each chip(3chips per concentration to get an average) for every
concentration and the average and the percentage increase.

Obtaining

Original

Length(mm)

Solution

(mol)

Size increase(m)

New size

% increase

40

0.0

+9.25

49.25

23.1%

40

0.2

+1.12

41.12

2.8%

40

0.4

-5.30

34.70

-13.2%

40

0.6

-11.90

28.10

-29.7%

40

0.8

-16.90

23.10

-42.2%

40

1.0

-19.11

20.89

-47.7%

% Increase = Actual Increase

Initial Value

% Decrease = Actual Decrease

Initial Value

Analysis

Results show in accordance with my prediction that 'the lower the
concentration of the sucrose solution the lager the mass of the potato
will be.' This is due water passing from weak solutions to strong
solutions across a semi-permeable membrane. In the graph % change
against solution strength show the results form a curve that slopes
downwards and does not go through the origin. Because the line is not
straight and does not pass through the origin, it means that the
percentage gain and loss in mass and concentration are not directly
proportional. However there is a pattern in the curve as the solution
increases the percentage change in length decrease thus making it
inversely proportional.

This curve crosses the y axis (where there is no percentage change) at
approximately 0.24mol sucrose solution. This concentration is the
equivalent molarity of the sap inside the cell.

In my graph the gradient changes it gets less steep as the sucrose
solution decreases. This is because the potato chip is become as
flaccid it can. Due to my line of best fit it shows that my results
were very nearly a completely smooth curve, therefore making my
results very reliable and fitted my prediction.

None of my results became completely plasmolysed or turgid other wise
one of my results in percentage change would have been the same as the
other. To conclude, my prediction was correct and we were able to
discover the equivalent concentration of potato tissue.

Evaluation

I think the experiment went very successfully I gathered 5 accurate
results that enabled me to draw up a reliable graph to make comments
on and evaluate. If I were to repeat the experiment there would be a
number of things that I would like to change to improve my results
further, i.e. keeping the Petri dishes in a controlled temperature and
pressure, leaving them for longer to allow the solution to fully
penetrate the membrane and taking many more results at different
concentrations so as that I can find out when its fully plasmolysed
and fully turgid and get more exact to the equivalent concentration. A
disadvantage to our experiment is that, because of having so little
different concentrations, one anomalous result affects the line of
best fit significantly. Luckily I did not have any!

For other corresponding experiments we could repeat it but use
solutions closer the equivalent concentration result that we found to
define its exact value. We could also use other plan tissues to test
if our prediction applied to them to and we could also compare the
osmotic activity between plants.
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