Investigating the Factors that Affect Osmosis in Living Tissue

Investigating the Factors that Affect Osmosis in Living Tissue
Aim: To investigate the factors that effect osmosis in living tissue. Planning: Introduction Essentially, osmosis is the diffusion of water across a selectively permeably membrane. Osmosis is one of the ways by which substances enter and exit cells. Other ways include diffusion, the Donnan effect1, solvent drag, filtration, endocytosis, exocytosis and active transport. All of these methods are necessary to provide cells with the conditions necessary for their survival. Osmosis helps cells absorb the water that they need and also pass it on from one cell to another. Osmosis occurs in the uptake of water in root hair cells, it also occurs in the return of water from tissues to blood capillaries and is constantly occurring during the opening and closing of the stomata in plant leaves. Factors that shall be tested: Bearing in mind that we have limited time and shall be conducting our experiments in a laboratory, measuring certain things may therefore be impractical. We shall therefore limit our investigation to the effect of three things: Cross-sectional area of plant cylinder, concentration of solution and temperature of solution. Variables There are three main types of variable; Independent (input), control and dependent variables.
An independent variable, otherwise known as an input variable, is the variable that is to be tested by experiment and therefore deliberately changed. The control variable is the variable that is kept constant in order to test the independent variable fairly. The dependent variable is the one that depends on the control and independent variables. Since the control is to be kept constant, any effect on the control variable will therefore be due to the independent variable. Throughout our investigation the variables and there complexities shall be as follows: The constants or control variables for this experiment are going to be: ? The vegetable used-We shall be using the common white potato. Potatoes are produced by plants of the genus Solanum, of the family Solanaceae. The common white potato is classified as Solanum tuberosum. ? The volume of solution used-15cm3 ? The length of the cylinder of vegetable used-5cm ? The amount of time each vegetable is left in the solution-2 days ? The same vegetable used to obtain cylinders since two different vegetables(of the same type) may have been exposed to different surroundings and hence provide us with inaccurate results. These constants will make sure that all the tests performed will be fair and equal. A control experiment will also be set up to ensure fair testing and will be discussed later. The independent or input variables will be as follows: ? The concentration of the sucrose solution, a continuous variable ranging from pure water to a 1mol/litre solution in units of 0.1. ? The temperature of the solution, using three continuous values-room temperature- 22oC, 37 oC and 55 oC ? The cross-sectional area of the vegetable -using three different sized cork-borers- this too shall be of continuous complexity. The derived variables that shall be measured/taken note of are: ? Volume ? Mass ? Turgidity Equipment and Supplies ? Potato ? Cork borer ? Test tubes ? Test-tube racks ? Beakers ? Sucrose and distilled water -concentrations must be made up ? Ruler ? Top-pan balance ? Scalpel ? Water tank Predictions In order to attain a better understanding of osmosis and to make quantitative predictions regarding the factors that affect it, it may be of some use to consider the process of diffusion (since osmosis is a type of diffusion) and the structure of the cell membrane in some detail. The process of diffusion is defined as the random movement of particles2 from an area of high concentration to an area at lower concentration. This difference in concentration between two regions is known as the concentration gradient, the greater the difference between the two concentrations the steeper the concentration gradient becomes, resulting in an increased rate of diffusion. Diffusion always occurs when a concentration gradient exists and it proceeds until the particles are in a homogenous distribution, at which time they are said to be in equilibrium. Diffusion and hence osmosis follows various laws. It has been found that the rate of diffusion is proportional to the cross-sectional area and to the gradient of concentration. The amount of particles that diffuse or the distance that they diffuse is also proportional to the square root of time. Since diffusion depends on the random movement of particles, it follows that an increased average velocity of the particles will result in an increased rate of diffusion. Increasing the temperature of a particle increases it?s kinetic energy and so the rate of diffusion is proportional to the increase in temperature. Hence it follows that osmosis is the net movement of solvent molecules across a selectively permeable membrane into an area in which there is a higher concentration of a solute (i.e across a concentration gradient) to which the membrane is impermeable or less permeable. As we are discussing osmosis with regards to plants, I believe it is necessary that we consider the structure of plant cells. A simplified drawing of a leaf palisade cell3 is illustrated below: [image] Within the cell there are two selectively permeable membranes, "one lining the outer surface of the cytoplasm (in contact with the cellulose wall), and the other bordering the vacuole. The latter is called the tonoplast and the former is the plasma membrane."4 However, for simplicity we shall limit our discussion to the plasma membrane. As the plasma membrane is very small, it has been necessary to adapt an indirect method of working out it?s structure based on it?s physicochemical properties. There have been two major theories regarding the structure of the plasma membrane. The first, sometimes labelled the ?sandwich? theory, was put forward by J.F. Danielli and H. Davson in the late 1930?s. Danielli and Davson proposed that the cell membrane was made up of three main layers; "a bimolecular layer of lipid sandwiched between two layers of protein, the lipid molecules being set at right angles to the surface."5 This can be illustrated as follows: Protein molecules Lipid molecules Interior of cell However, in the early 1970?s a more sophisticated model of the plasma membrane was put forward by S.J. Singer and G.L. Nicholson. Their research showed that the membrane was actually far less rigid than earlier believed. It was also shown that the protein molecules are arranged in a type of mosaic arrangement. From these two properties the Singer-Nicholson model is often known as the fluid-mosaic model. Below is a diagram of the cell-membrane as proposed by Singer and Nicholson: The proteins found in the above model are not to give the membrane strength but are thought to act as pumps moving things across it. And ?so it seems that ?simple passive? diffusion, even of water molecules (osmosis) is not so simple or passive after all."6 It Is believed that the pores perforate the membrane at regular intervals. Their function is yet to be confirmed, however it is thought that they allow certain molecules that are insoluble in lipid to penetrate the membrane. The surface carbohydrates, a recent discovery, found only on the outside of the cell membrane are believed to play a role in allowing cells to recognise each other. Thus membrane permeability, the property of a membrane that determines its penetrability, and hence rate of osmosis, depends on the size of the pores of the membrane, the size of the particles of the substance attempting to pass that membrane, the solubility of the substance in the membrane and the presence of enzymes in the membrane to provide carrier-mediated transport. Having discussed diffusion and the structure of the cell membrane, it may be of use to use the principles of thermodynamics7 to explain osmosis. In terms of thermodynamics, the reason for the net movement of water molecules from a dilute solution to a strong solution, is that in the dilute solution there are more free water molecules, hence they have a greater potential energy than the strong solution. In the concentrated solution the water molecules cluster round the solute forming aquo-ions and impeding their movement. The diagram below shows this (the filled in circle represents water, unfilled represents sucrose and the partition is a selectively permeable membrane): Weak sucrose solution Strong sucrose solution [image][image] (hypotonic) (hypertonic) [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] [image] net movement of water molecules High water potential/low osmotic Low water potential/high pressure osmotic pressure The potential energy of water molecules is known as the water potential and is represented by the Greek letter psi-Y. Pure water has the value of Y 0, the addition of solute particles makes the value negative. The ?push? that osmosis gets its name from8 refers to the force of moving molecules in a solution against a membrane. This ?push? is known as the osmotic pressure. Whether a solution is hypotonic, hypertonic or isotonic with regards to the cell determines whether or not the osmotic flow of water will be into the cell (endosmosis), out of the cell (exosmosis) or balanced. If a cell is surrounded by a hypotonic solution that has a lower solute concentration and thus a lower osmotic pressure, it follows that endosmosis will occur, thus causing the cell to gain in mass and volume as well as turgidity9. If a cell is surrounded by a hypertonic solution in which the solute concentration and hence osmotic pressure of the external solution is greater than that of the cell, exosmosis will occur resulting in a decreased mass, volume and turgidity. If there is a homogenous distribution of solute concentration and hence osmotic pressure between the cell and its surroundings, i.e. isotonic solution, no change in mass or volume should occur as there shall be no net movement of water. The following shows the difference in plant cells due to osmotic changes: Cell in hypotonic Cell in isotonic Cell in hypertonic Cell in very hypertonic solution: turgid solution solution: flacid solution: plasmolysed10 Osmosis will not occur in cells whose membranes are not intact. For instance if the cell is denatured by heat, or plasmolysed, osmosis will not occur. Having given a somewhat comprehensive treatment of osmosis and the factors that may affect it, we may now apply the key principles to make predictions regarding our three input variables: Concentration I believe that the vegetable will change mass, turgidity and volume depending on whether or not the solution is hypotonic, hypertonic or isotonic/isosmotic to the vegetable. I predict, having explained osmosis in terms of the concentration gradient and thermodynamics, that the pure water, which has the greatest water potential, will cause the vegetable to increase in mass, volume and turgidity the most. This is as the potato no doubt contains some solute particles, and thus causing endosmosis across the concentration gradient. The 1mol/litre concentration is likely to be hypertonic to the potato and hence result in the greatest decrease in mass, volume and turgidity of the vegetable due to exosmosis. Somewhere between these two concentrations will be the water potential of the potato, this isotonic solution will thus cause no change in mass volume or turgidity. As the rate of osmosis is proportional to the gradient of concentration, in a given time change in mass and volume will be inversely proportional to the change in concentration i.e. proportional to the change in water potential of the solution Hence, when plotting percentage increase in mass or volume, the graph should look similar to this: [image] [image] [image]% change in mass /volume [image][image][image][image][image][image][image] 0% [image] Concentration Cross-Sectional Area [image] I believe that increasing cross-sectional area will increase the rate of osmosis proportionally. This is as a larger cross-sectional area will mean that there is a greater surface area in contact with the solution thus increasing the rate of osmosis. Hence, if we are using pure water, doubling the cross sectional area should therefore double the rate of osmosis and hence double the increase in mass and volume that the vegetable undergoes. Similarly, when using the 1 molar solution, doubling the cross-sectional area should therefore double the decrease in mass and volume. Hence a graph of percentage change in mass or volume, for a given concentration, against cross-sectional area should look as follows: [image] % change in mass /volume [image] Surface area Temperature The three temperatures being considered are 220C , 370C and 550C. 220C is room temperature, 370C is the optimum temperature at which enzymes can act and at 550C many enzymes are denatured and thus inactivated. As mentioned earlier, enzymes are thought to play an important role in carrier-mediated transport even in osmosis, and so inactivating them will thus slow down the rate of osmosis. Many cell organelles are denatured at 550C, however the cell membrane is able to withstand a few more degrees. Yet with the inactivation of the nucleus and many enzymes, it is unlikely that osmosis will occur. Although increasing the temperature from 220C to 550C will have a negative effect on the rate of osmosis, increasing the temperature to 370C will increase the rate of osmosis. This is as osmosis depends on the movement of particles, increasing the temperature of a particle increases it?s kinetic energy and so the rate of osmosis is proportional to the increase in temperature. However, where living cells are concerned, the temperature rise to 370C will not only increase the average speed of the particles, but will allow the cell to function at it?s optimum temperature-also leading to an increased rate of osmosis. Due to the fact that there is more than one factor that will be altered when increasing the temperature from 220C to 370C, the increase in the rate of osmosis is not likely to be proportional to this increase in temperature. When illustrated graphically, the rate of osmosis against temperature should look something like this: [image] Rate of osmosis [image] 22 37 55 Temperature However, since the rate of osmosis determines the increase in length and volume, we may therefore assume that the graph of percentage change in mass/volume would be the same as above. And yet I believe that this is not so in the case of the experiment at 570C. I predict that there will be a change in mass and volume in this experiment, however, somewhat pedantically speaking, most of it will not be due to osmosis. This is as osmosis is very specifically diffusion across a selectively permeable membrane. However, since at this temperature the cell and thus the plasma membrane are denatured, there is no longer a selectively permeable barrier and so the likely change in mass and volume will be down to the porosity of the vegetable, i.e. it will thus be diffusion that occurs and not osmosis. Depending on the actual porosity of the vegetable, the change in mass or volume may even exceed the initial change at 370C as there will be no impeding barrier as well as the fact that the particles will be travelling at a faster average speed at this increased temperature. From our results we may thus asses the role of enzymes in osmosis concerning the potato. The effect of temperature on the osmotic pressure and hence the rate of osmosis is found by the equation: P = nRT [image] V Where n is the number of particles, R is the gas constant, T the absolute temperature i.e. in Kelvin11 and V is the volume. With an increased temperature we may thus expect an increased osmotic pressure and thus a greater rate of osmosis (although, as explained earlier a rise too high would kill the organism). Similarly, this equation applies to concentration as concentration is number of particles divided by volume. Plan Of Investigation Before we begin the investigation it is essential that we take various safety precautions. These shall include handling the equipment, in particular the scalpel, with care. It is also crucial that we keep our area used for experimentation tidy to prevent accidents. The investigation shall proceed in three main parts, each dealing with one different independent or input variable. We shall first consider that of concentration. Given that we have sufficient time in which to carry out a thorough investigation, we shall conduct the investigation with regards to the effects of concentration using eleven different concentrations on the potato. The concentrations shall range from pure water to a 1 molar solution of sucrose in units of 0.1. In addition to this we shall also set up a control experiment. A control experiment is one that is set up to ensure that only the condition being tested has affected the results.. In this investigation we shall place the three different vegetables into a test-tube without adding the solution, if our predictions are correct, there should be no change in the mass or volume of the vegetable (-perhaps a minor reduction in mass/volume due to evaporation). With these results, I believe that we will be able to adequately evaluate our predictions. However our data would be more accurate if we used the method of averaging, i.e. repeat each experiment several times which is useful in order to reduce experimental error. However, it is unlikely that we will have sufficient time to use this method. In order to obtain a fair set of results regarding the effect of concentration, the following variables shall be constant: ? Type of vegetable used: Solanum tuberosum ? The volume of solution used-15cm3 ? The length of the cylinder of vegetable used-5cm ? The amount of time each vegetable is left in the solution-2 days ? The same vegetable used to obtain cylinders since two different vegetables may have been exposed to different surroundings and hence provide us with inaccurate results. ? The temperature of the solution -room temperature 22oC. ? The cross-sectional area of the vegetable. The steps that shall be taken are as follows: Step1: I shall obtain one vegetable, and using a cork borer of radius 1cm, the volume of the cylinder, given that it shall be of length 5cm can thus be worked out using the formula: Volume of Cylinder = p x r2 x h \ Volume = p x 12 × 5 =15.7 cm3 I shall cut cylinders (eleven used for the different concentrations and one for the control experiment) from the vegetable, The cork borer is used to ensure that all the pieces are of the same diameter. Step 2: The cylinders shall be aligned against a straight edge where they shall be cut to a length of 5cm. The straight edge is to ensure that the pieces are of exactly the same length. This may be illustrated as follows: [image]The length shall be determined using a ruler. A ruler is accurate to the nearest millimetre and so the accuracy is 5 + 0.001. It is thus very accurate. The percentage error may be worked out using the following equation: Degree of accuracy x 100 [image]Total measurement \ 0.001 × 100 = 0.02% [image] 5 Step3: The cylinders shall be blotted dry using filter paper, before being weighed on a top-pan balance. They shall be blotted dry in order to receive an accurate reading of their weight. As all the cylinders should be of the same length, and as they are from the same vegetable, their densities should also be the same, we shall thus weihg them all and obtain the mean. A top-pan balance is accurate to 0.01 of a gram. Step4: We must then obtain 15ml of distilled water. This shall be done, as in all the experiments by pouring the distilled water from the container into a measuring cylinder as shown below: [image]It is important that we take our reading from the bottom of the liquid meniscus to reduce error. A measuring cylinder is accurate to the nearest 1cm3. Therefore our measurement shall be 15cm3 + 1. We may also find out the percentage error using the following equation: Degree of accuracy x 100 [image]Total measurement [image]\ 1 × 100 = 6.7% (to 2.s.f) 15 Thus we may expect an error of approximately up to 6.7%. Although this does seem rather high, the only other measuring cylinder that we have, which is accurate to 0.1cm3, only has a capacity of 10cm3, and therefore is of no use. Step4: We next have to make up a 15 ml solution of correct concentration and osmotic pressure. To make up a specific concentration of a solution of sucrose we use the formula: [image] Equation1) Concentration moles/litre =Amount of solute in moles12 Volume of solution in litres Equation2) Number of moles = mass [image] molecular mass As we know the required concentration and the volume required, we therefore alter the formula to find the number of moles of sucrose required: Amount of solute (mol) = Volume of solution (l) x Concentration (mol/l) For example, to make a ½ molar solution, as we are using 15 ml of water i.e 0.015 litres, using equation 1, we thus multiply ½ by 0.015 which equals 0.0075 moles. Now, to find the mass required we rearrange equation two placing mass as the subject: mass = Number of moles x molecular mass \ given that the molecular mass of sucrose is 324g mass = 0.075 × 324 = 2.43g. The osmotic pressure of the solution may be found using the formula: P = nRT [image] V It has thus been found that a 1molar solution of sucrose has an osmotic pressure of 245kPa. The osmotic pressure that leads to no increase or decrease in mass or volume of the vegetable, is thus equal to the osmotic pressure of the vegetable. All the masses of sucrose required to be added to the 15ml of water and osmotic pressures are worked out in a similar way and are shown on the following table: [image] Concentration (mol/l) Osmotic Pressure (kPa) Mass (g) [image] [image] 0 (pure water) 0 0 0.1 [image] 24.5 0.49 0.2 49 0.97 0.3 [image][image] 73.5 1.46 0.4 98 1.99 0.5 [image] 122.5 2.43 0.6 [image] 147 2.92 0.7 [image] 171.5 3.40 0.8 [image][image] 196 3.89 0.9 220.5 4.37 [image] 1 245 4.87 The concentrations will be made up in a beaker and not in a test tube, this is as the test-tube may break whilst we are using the glass-rod to stir the solution and to make sure that all the solute particles dissolve. Stage5: The vegetable cylinder will then be added to the required concentration inside a test-tube where a rubber bung will be placed to prevent the solution evaporating. This may be illustrated as follows: [image][image][image][image][image][image] Rubber bung Test-tube [image] [image] Vegetable [image] Required concentration [image] The test-tube is then placed in a rack and left for two days. It should not be placed in an area where it is likely that there will be a drastic change in temperature i.e. not directly exposed to sunlight. The above stages will be done for all the ten required concentrations. A control experiment will also be set up to make sure that the changes are due to the solution and nothing else. The control experiment will thus be a potato in a test-tube with no solution present: [Rubber bung Test-tube Potato [image] Air [image] Our results for this experiment are to be plotted on a table with values for the concentration, the change in mass, volume and turgidity. We shall also plot a graph of percentage change in mass against concentration as well as percentage change in volume against concentration. Having considered concentration we may consider the method to be used to find the effect of temperature. The temperatures we shall use are 220C , 370C and 550C. Two different concentrations will be used, i.e. 1mol/l and pure water in order to test the effect of temperature on endosmosis and exosmosis, thus allowing us to gain a broader insight in to the effect of temperature on osmosis. Therefore 6 test-tubes will be required. With these results, I believe that we will be able to adequately evaluate our predictions. However our data would be more accurate if we used the method of averaging, yet it is again unlikely that we will have sufficient time to use this method. In order to obtain a fair set of results regarding the effect of concentration, the following variables shall be constant: ? Type of vegetable used: Solanum tuberosum ? The volume of solution used-15cm3 ? The length of the cylinder of vegetable used-5cm ? The amount of time each vegetable is left in the solution-2 days ? The same vegetable used to obtain cylinders since two different vegetables (of the same type)may have been exposed to different surroundings and hence provide us with inaccurate results. ? The cross-sectional area of the vegetable. The steps that shall be taken are as follows: Step1: I shall obtain one vegetable, and using a cork borer of radius 1cm, cut a total of 6 cylinders . Step2: As described earlier, the cylinders shall be aligned against a straight edge where they shall be cut to a length of 5cm. Step3: The cylinders shall be blotted dry using filter paper, before they are weighed on the top-pan balance. Step4: We must then obtain 15ml of distilled water. This shall be done, as described earlier. Required concentrations will the be made up. The concentrations shall be pure water and a 1molar solution. Stage5: The 6 cylinders will be placed in the required solution of sucrose. Two for room temperature-at two different concentrations, another two will be placed in a water tank heated to 37 0C-again at two different concentrations. The water is there to make sure that all parts are heated to the same temperature. The remaining two will be placed in a water tank heated to 570C- again at the two different concentrations. Our results for this experiment are to be plotted on a table with values for the temperature, concentration, the change in mass, volume and turgidity. We shall also plot a graph of percentage change in mass against temperature as well as percentage change in volume against temperature. Having considered the two independent variables of concentration and temperature we may now consider the method to be used for the experiment dealing with cross-sectional area. Given that the radius? used shall be 0.5cm,0.7cm and 1cm the cross sectional area may thus be found by the formula p x r2. The cross-sectional areas are thus 0.79, 1.54 and 3.14cm2. Two different concentrations will be used, i.e. 1mol/l and pure water in order to test the effect of surface area on endosmosis and exosmosis, thus allowing us to gain a broader insight in to the effect of surface area on osmosis. Therefore 6 test-tubes will be required. With these results, I believe that we will be able to adequately evaluate our predictions. In order to obtain a fair set of results regarding the effect of concentration, the following variables shall be kept constant: ? Type of vegetable used: Solanum tuberosum ? The volume of solution used-15cm3 ? The length of the cylinder of vegetable used-5cm ? The amount of time each vegetable is left in the solution-2 days ? The same vegetable used to obtain cylinders since two different vegetables may have been exposed to different surroundings and hence provide us with inaccurate results. ? The temperature of the solution. The steps taken shall be as follows: Step1: I shall obtain one vegetable, and using cork borers of radius 0.5cm,0.7cm and 1cm, cut a total of 6 cylinders-two of each cross-sectional area. Step2: As described earlier, the cylinders shall be aligned against a straight edge where they shall be cut to a length of 5cm. Step3: The cylinders shall be blotted dry using filter paper, before they are weighed on the top-pan balance. Step4: We must then obtain 15ml of distilled water. This shall be done, as described earlier. Required concentrations will the be made up. The concentrations shall be pure water and a 1molar solution. Stage5: The vegetable cylinder of required cross-sectional area will then be added to the required concentration inside a test-tube where a rubber bung will be placed to prevent the solution evaporating. Our results for this experiment are to be plotted on a table with values for the surface area, concentration, the change in mass, volume and turgidity. We shall also plot a graph of percentage change in mass against cross-sectional area as well as percentage change in volume against surface area Observations Having performed the experiments, we noticed that in all the experiments their was a smell of ethanol. Ethanol is formed by anaerobic13 respiration and is formed, in plants, in the cytoplasmic matrix: [image][image][image] Glucose enzymes in cytoplasmic matrix Ethanol + Carbon dioxide + Energy C6H12O6 C2H5OH CO2 210kJ/mol The strongest smell of ethanol was observed in the potato that was placed in the water-tank at 370C. There was a feint smell in the test-tubes at room temperature and surprisingly a feint smell in those at 550C. The fact that ethanol was produced in the water tank at 550C suggests that some enzymes were still activated. This leads me to suspect an inadequacy in my method. Regarding the variable of concentration, the total mass of potato was found to be 69.72g. Thus one potato weighs 69.72 / 12= 5.81g. In the second independent variable, that of temperature, the total mass was found to be 35.58. Thus one potato weighs approximately 35.58 / 6 = 5.93g For cross-sectional area the total mass of 3.14cm2 was 10.44. Thus one potato of cross-sectional area 3.14cm2 weighs 11.58 / 2 = 5.79. Using the cross-sectional area of 1.54 cm2 the mass of two potatoes was found to be 5.70. Thus the individual mass is 5.70 / 2 = 2.85g. Using the cross-sectional area of 0.79cm2, the mass of two was found to be 2.94g. Thus the individual mass is 2.94 / 2 = 1.47g. In order to asses the change in turgidity, I have devised a convenient scale based on approximate estimations. The scale is from between 1-5 in which 5 is fully turgid and 1 is fully flaccid. All cylinders have been found to be, before being placed in solution, approximately 4 on this scale. Results Our results may be tabulated and illustrated graphically as follows: Conclusion The results obtained show much of what was predicted to be correct. With regards to the effect of concentration, the results clearly show that the change in mass, volume and turgidity was largely determined by whether or not the external solution was hypotonic, hypertonic or isotonic to the potato. From looking at our results showing the change in mass against concentration, we see that the pure water solution produced the greatest increase in mass (16.35%) whilst the 1molar solution produced the greatest decrease in mass (-33.39) as predicted. The line of best-fit shows negative correlation- percentage change in mass seems to be inversely proportional to the concentration of the external solution, i.e.: % Change in Mass a 1 [image] Concentration Þ % Change in Mass = k [image] Concentration Most points correspond to the linear trend-line within an x and y-axis error margin of 5%. However, the point that seems least accurate is that obtained from the 0.1molar solution. This is most likely to be due to inadequacies in taking the initial mass and volume measurements. From the trend-line it appears as though the water potential of the potato is approximately equal to the water potential of the 0.36 molar solution. Thus: hypotonic solution <0.36

Investigating the Factors that Affect Osmosis in Living Tissue 8.6 of 10 on the basis of 1042 Review.