# An Investigation of Factors Affecting the Rate of Osmosis

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An Investigation of Factors Affecting the Rate of Osmosis

Introduction

Osmosis is the movement of water molecules across a semi permeable
membrane from a region of high water concentration to a region of low
water concentration.

[IMAGE]

A semi permeable membrane is a membrane with very small holes in it;
they are so small that only water molecules can pass through them.
Bigger molecules such as glucose cannot pass through it. In actual
fact water molecules pass both ways through the membrane, but because
there are more water molecules in the high concentration region than
the other there is a steady net flow into the lower concentration
region. The lower concentration is the stronger solution, such as a
glucose solution. This movement causes the glucose-rich region to fill
up with water. The water movement is diluting the solution so that the
concentration on both sides is equal.

Osmosis

Figure no.1

This diagram illustrates the net flow of water movement from a
hypotonic solution, low solute concentration, to an area of high
solute concentration. In other words this shows water movement across
the semi permeable membrane from a high concentration of water to a
low concentration of water.

Water moves from a high water potential to a low water potential. A
low water potential is a high solute concentration. Water potential
has the symbol Î¨, sigma.

When water potential on both sides is equal then it is known as
equilibrium. In this case there is not an absence of movement between
the equal water potentials but a movement of water in both directions,
maintaining the equilibrium. Osmosis is a continuous process which
does not require any energy to take place.

Factors Which Affect the Rate of Osmosis

1) Temperature: The higher the temperature is the faster the molecules
will move. This means that the movement of water molecules across the
semi permeable membrane will be faster.

2) Surface Area: When there is a larger surface area there will be

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### Popular Essays

more space for the molecules in and so they will be able to move
across more easily. In a small area the movement across the semi
permeable membrane will be restricted with less molecules being able
to get through at a time. So the rate will be faster with a bigger
surface area.

3) Difference in Water Potential: If the difference in water potential
is higher then the rate of osmosis will be faster. This is because if
there are not many water molecules in the low concentration region
then water molecules from the high concentration area can come in
quickly and easily. When there is less difference it will take longer
because there are already a lot of water molecules occupying the space
in the low concentration.

4) Pressure: When there is more pressure the molecules move faster.
This is because when there is not much pressure they are not being
forced to move quickly, but in high pressure they are being pushed
across to the low concentration.

Hypothesis

I believe that the bigger the difference in water potential the
quicker the molecules will move and the quicker the rate of osmosis
will be.

Why Use Potatoes

1) They are cheap

2) They have homogenous tissue. This means that they have the same
type of tissue throughout the potato, and so results won't be
indifferent or unfair.

3) The cell surface membrane acts as a semi permeable membrane,
allowing osmosis to occur.

Molarity of Sucrose

The molarity of sucrose is 0.27; we can tell this from the graph
above. The dotted line across the graph shows the equilibrium and if
we see where the curve joins this line and then trace it down from
that point to the x axis we find the solute concentration to be 0.27.
The Roberts textbook was our source for this information. Using this
we now know that the equilibrium is at 0.27M. This means when the
solution outside the potato is 0.27M it is matched by the
concentration of 0.27M by the potato. Therefore at this point no mass
change will occur. When the solution outside the potato is lower than
the equilibrium, then it will cause the potato to gain mass. This is
because the water moves into the potato from the concentration,
because the potato has less water than the solution. When the solution
is higher in concentration than 0.27M then the potato should lose
mass, because it has more water than the solution. To balance this out
it goes into the solution.

If the liquid that the potato is in has a solute concentration of 0.1;
the potato has a solute concentration of 0.27. This means that the
potato has higher sugar concentration than the outside solute, so
there are not many water molecules present there. Therefore water
molecules move into the potato, increasing its mass.

If for example the liquid outside the potato had a solute
concentration of 0.4 and the potato has its solute concentration of
0.27, then the mass will decrease. This is because there is not a high
concentration of water molecules in the liquid, so water molecules
will move out of the potato into the liquid.

At equilibrium the mass stays the same.

The rate at which the water moves into the potato is not always
constant. As it gets closer to equilibrium the difference of water
between the potato and the solute is less. Therefore the rate of
osmosis decreases. Also the rate of osmosis changes with time because
as osmosis happens the potato cells start to become turgor. This is
where the cells swell up and become rigid because so many water
molecules enter the cell. The cell has a rigid cell wall and so it
does not burst like animal cells, but the cell stops taking in water
molecules.

We know that when the solution is above equilibrium, 0.27M, then the
potato will not stop losing water until it cannot lose anymore. This
means that the rate of osmosis will continue to increase, until it
gets to a point where it cannot increase anymore and has to stop. For
example, a plant cell when the cell becomes turgid. A graph to
illustrate this is a graph of rate of osmosis against concentration.
This is ideally what our results should show.

[IMAGE]

The lower the concentration of the solution the more weight it should
gain and the higher the concentration of the solution the more weight
it should lose.

A graph of what the mass change should look like ideally will look
this:

% mass change

[IMAGE]

This graph shows that as the solution reaches equilibrium then the
percentage mass change begins to slow down, and does not change quite
so much.

* Apparatus List

* Ã¼ Potato: we need the potato because this is what will be
conducting the experiment on

* Ã¼ Solute 0M - 1M: we need the solution to put the potato in so we
can test the rate of osmosis

* Ã¼ Weighing scales: we need this because to calculate the rate of
osmosis we need to weigh the potato at intervals to see how
quickly it is gaining or losing water molecules

* Ã¼ Stopwatch: this is necessary to know when to measure the potato
at the right time interval

* Ã¼ Cork borer: we need this to obtain the potato sample

* Ã¼ Tile: we need this to cut the potato on, so we don't damage the
work benches

* Ã¼ Scalpel: we need this to cut the potato samples skin off

* Ã¼ Beakers: we need 6 different labeled beakers to put the 6
different solutions into

* Ã¼ Permanent pen: we need this to label the beakers

* Ã¼ Tweezers: we need this to pick up the potato sample out of the
beaker to weigh it

* Ã¼ Paper towel: we need this to dry the potato sample of any excess
liquid after it has been taken out of the solution. If it was not
dried then it would effect the results

* Ã¼ Recording table: this is to record the results quickly and
clearly

Planning the Preliminary Experiment

Firstly I will do a preliminary experiment, to test my method and then
see how I can improve it for the final experiment. Also a preliminary
experiment will show me what results to expect if it goes correctly.
For this preliminary experiment I will use a cork borer to produce the
sample potato cells. They should have approximately the same surface
area and mass. If possible the samples should be taken from the same
potato. The amount of solute but into each beaker will be 40cm3.

The range of sucrose values I will use will be from 0-1 molar at o.2
intervals. I will leave the cylinders in the beakers for at least half
an hour allowing enough time for osmosis to happen and to see a change
in mass.

The potato should be weighed every 5 minutes to give an idea of how
fast osmosis is occurring. We need repeats at each concentration
because one set of data is not 100% reliable, we need to take several
to make sure our readings are accurate.

* Method of my Preliminary Experiment

* Â· First get six beakers and label them all accordingly, from OM
(distilled water) to 1M in 0.2 intervals.

* Â· Now measure 40cm3 of each solute into each appropriate beaker

* Â· Get a potato and a cork borer of appropriate size, if the potato
small you cannot have a large size cork borer as you will not be
able to get enough potato cylinders from the potato

* Â· Cut the skin off the potato cylinders as the skin is not
homogenous

* Â· Weigh the potato and make sure all the samples are off similar
weights and shapes

* Â· Now put the cylinders into each beaker and begin to time

* Â· Every five minutes take the potato samples out and weigh them,
then put them back into the beaker, and record the weight

* Â· Continue to weigh the samples at 5 minute intervals for 30
minutes

* Problems Encountered and How to Solve Them

* The preliminary experiment helped me very much in conducting my
final experiment. My preliminary experiment had many errors which
I identified and corrected in my final experiment. Firstly I
didn't use the 0M solution; this was because this solution had run
out as many people were not economic in pouring it into the
beaker. This can be solved by using only the amount needed in your
experiment and pouring the solution cautiously as to avoid
spilling any.

* I did not in this experiment use more than one potato sample, to
check if my results were consistent or if they were a one off.
Therefore in the final experiment I will take more than one sample
at each solution.

* Practically doing the experiment was a huge rush when it was time
to weigh the potato sample. I did not put each potato sample in at
different times, meaning I had to weigh them all at the same time,
which was not possible. To overcome this in the final experiment I
will put first sample in, and then wait 1 minute before putting in
the next potato sample. By doing this I will have enough time to
weigh all the potato samples correctly. Also I have decided to
take the weight reading every 10 minutes. This is due to the fact
that measuring the weights every 5 minutes was too chaotic and
could easily lead to errors. Also by staggering the time of
weighing the samples taking the last sample of the first five
minutes would overlap with taking the first sample at the next
five minutes. Therefore it would not be possible to take the
sample every 5 minutes.

* Furthermore in this experiment my potato was too small for the
size of my cork borer, and I ran out of potato. I then had to use
a different potato to obtain more sample cylinders. This is not at
all ideal as this would have affected my results as the potato
would be different to the original. Consequently in my final
experiment I will select a smaller cork borer size and a large
potato to ensure I don't run out of potato to take my sample from.

* Lastly I decided that I didn't get enough results from conducting
the experiment over half an hour only. So I have decided to do the
experiment over a longer period of an hour.

* Method

* Â· First get six beakers and label them all accordingly, from OM
(distilled water) to 1M in 0.2 intervals.

* Â· Now measure 40cm3 of each solute into each appropriate beaker

* Â· Get a potato and a cork borer of appropriate size, if the potato
small you cannot have a large size cork borer as you will not be
able to get enough potato cylinders from the potato

* Â· Cut the skin off the potato cylinders as the skin is not
homogenous

* Â· Weigh the potato and make sure all the samples are off similar
weights and shapes

* Â· Now put the first cylinder into the 0M beaker and begin to time,
after one minute put in the next potato sample into the 0.2M
beaker. Continue to put in the samples into the beakers in
ascending order, until there are none left

* Â· At ten minutes take the 0M potato sample out, dry it of excess
solution, weigh it, return it to the beaker and record the weight.
At 11 minutes take out the 0.2M and repeat the procedure. Do this
for the rest of the potato samples at the right time

* Â· Continue to weigh the samples at 10 minute intervals for 60
minutes

* Results

Time (minutes)

Concentration (M)

0

0.2

0.4

0

2.17

2.58

2.08

2.57

2.15

2.25

10

2.21

2.63

2.1

2.59

2.13

2.22

20

2.22

2.64

2.1

2.6

2.09

2.19

30

2.27

2.68

2.1

2.62

2.07

2.18

40

2.27

2.68

2.1

2.61

2.01

2.13

50

2.27

2.71

2.08

2.6

1.97

2.07

Time (minutes)

Concentration (M)

0.6

0.8

1

0

2.84

2.59

2.31

2.5

2.17

2.31

10

2.82

2.55

2.23

2.46

2.1

2.21

20

2.75

2.47

2.13

2.36

1.98

2.11

30

2.72

2.4

2.05

2.27

1.87

2

40

2.61

2.32

1.98

2.2

1.79

1.93

50

2.53

2.24

1.9

2.11

1.73

1.87

Because I did the experiment using two samples in each beaker to
ensure my results were correct, I can now take the average of both
sets of data for the most accurate results.

Time (minutes)

Concentration (M)

0

0.2

0.4

0.6

0.8

1

0

2.38

2.33

2.20

2.72

2.41

2.24

10

2.42

2.35

2.18

2.69

2.35

2.16

20

2.43

2.35

2.14

2.61

2.25

2.05

30

2.48

2.36

2.13

2.56

2.16

1.94

40

2.48

2.36

2.07

2.47

2.09

1.86

50

2.49

2.34

2.02

2.39

2.01

1.80

* Using these results I can now draw three graphs to analyze the
rate of osmosis. These will be: 1) Percentage mass change Vs.
solute concentration

* 2) Rate of osmosis Vs. Solute concentration

* 3) Percentage mass change Vs. time

* The equation to find the percentage mass change is:

* Change in mass (new mass - original mass) divided by the original
mass multiplied by 100.

* To calculate the rate of osmosis you will need to use the
equation:

* Final mass - starting mass divided by starting mass, multiplied by
100.

* Using these equations I produced the following tables from which I
produced my graphs:

* 1) Percentage mass change Vs. solute concentration

Concentration of solution, M

Percentage Mass change, %

0

4.62

0.2

0.43

0.4

-8.18

0.6

-12.13

0.8

-16.6

1

-19.64

2) Rate of osmosis Vs. Solute concentration

Solution Concentration (M)

0

0.2

0.4

0.6

0.8

1

rate of osmosis

4.842105

0.645161

-8.18182

-12.1547

-16.632

-19.6429

3) Percentage mass change Vs. time

Time (minutes)

Percentage mass change (%)

0

0.2

0.4

0.6

0.8

1

0

0.00

0.00

0.00

0.00

0.00

0.00

10

1.86

0.85

-1.15

-1.12

-2.56

-3.94

20

2.32

1.08

-2.73

-3.87

-6.65

-8.71

30

4.04

1.48

-3.53

-6.05

-11.34

-15.76

40

4.04

1.27

-6.28

-10.14

-15.07

-20.43

50

4.62

0.64

-8.91

-13.84

-19.95

-24.44

* Analysis

* [IMAGE]

* 1) This graph shows us that the weight of the potato decreases as
the concentration of the solution increases. We know this is
correct as water moves across a semi-permeable membrane from a
high concentration to a low concentration, trying to achieve
equilibrium. In a high concentration inside the potato there is
more water molecules than in the solution, so they move into the
solution, and consequently the potato loses mass. From this graph
we can see that the equilibrium is around 0.3M. This is very
accurate as the correct value is 0.27m.

* [IMAGE]

* 2) This graph shows how the rate of osmosis is affected by the
different solute concentrations. The graph shows that as the
concentration of solution increases the rate of osmosis increases.
This is as there is a bigger difference between the solution and
the potato. This means that the water can easily leave the potato
to go into the sucrose solution, because there is a lot of free
space in the concentration.

* The graph also shows that the equilibrium was around 0.23M, again
close to the actual figure. This is the point where the line meets
the axis and where we no that no water movement is occurring
because both sides outside and inside the potato are balanced.

* [IMAGE]

3)

This graph shows how the percentage mass changes over time, comparing
them with all the other solute concentrations. From this we can
analyze the rate of osmosis. We can see that both the first and second
lines, of 0M and 0.2 M are relatively flat. This suggests that the
rate of osmosis slows down towards the end of the experiment. This is
expected as it becomes harder for the molecules to pass through the
membrane as it is getting closer to equilibrium. There is less space
inside the potato for the water molecules to move into, so it is
taking them longer to push through.

The other lines which go into the negative half of the graph do not
show that their rate of osmosis is slowing down. Therefore this is
probably because they have not yet reached equilibrium, and longer
time would be needed to measure this in more detail.

Conclusion

In conclusion we have found that the rate of osmosis increases as the
difference in water potential increases. This is evident from the
graphs produced. We can that as the solutes get more and more
concentrated the rate increases more and more. The reason being for
this is that as the solution gets more concentrated, there are fewer
water molecules in the solution. This means that the water entering
the solution can do so readily, without being slowed down by many

Also, although we did not set about intending to prove the molarity of
sucrose as being 0.27M, we found that it accurate as we produced
readings of 0.23 and o.28 for the molarity of sucrose.

We can also conclude that in solute concentrations lower than 0.27 the
potato will gain mass and in solute concentrations above 0.27 it will
lose mass.

Evaluation

Overall I would say that the experiment was a successful one, proving
my hypothesis to be correct. However the results may not be entirely
accurate due to some problems which came up in the experiment. The
main problem was weighing the potato samples, as there were not many
scales. Therefore you could have easily missed your time to put the
sample back into the beaker. Also another problem was drying the
potato samples. This is because there was no way whether we could be
sure if we had dried all the samples equally. Some samples may not
have been dried enough and other may have been dried too much. Doing
this would give rise to anomalous results.

To carry out the experiment safely it was imperative we were extremely
cautious when using the scalpel knives. This could have easily cut
someone.

To further improve the experiment we could conduct it with a longer
time limit, so that we can specifically see how long it takes for the
samples in different solutions to reach equilibrium.

Hypothesis

I believe that the higher the concentration of the solute the longer
it will take for it to reach equilibrium.

Method

* Â· First get six beakers and label them all accordingly, from OM
(distilled water) to 1M in 0.2 intervals.

* Â· Now measure 40cm3 of each solute into each appropriate beaker

* Â· Get a potato and a cork borer of appropriate size, if the potato
small you cannot have a large size cork borer as you will not be
able to get enough potato cylinders from the potato

* Â· Cut the skin off the potato cylinders as the skin is not
homogenous

* Â· Weigh the potato and make sure all the samples are off similar
weights and shapes

* Â· Now put the first cylinder into the 0M beaker and begin to time,
after ten minutes put in the next potato sample into the 0.2M
beaker. Continue to put in the samples into the beakers in
ascending order, until there are none left

* Â· Every hour measure the sample, by taking it out and then drying
it, weighing it and putting it back into the beaker. This should
be done for until all the solutions have reached equilibrium

* I made the time interval an hour because it is clear this
experiment would take much longer than the previous one.