Diffusion and Osmosis

Diffusion and Osmosis of a Potato Introduction Diffusion labels the spread of molecules throughout unsystematic movement from expanses of higher concentration to areas of lesser concentration. The theory of diffusion is entwined to that of mass transfer propelled by a concentration gradient (Wikipedia). For example, when someone in a room passes gas, the odor will eventually spread out among the room until it has reached equilibrium with the air in the room. Instead of substances moving to balance out, osmosis is when water spreads to maintain balance.

If a watermelon is placed into a bowl with water with a higher concentration of salt then the watermelon, then the water will soon be depleted from the fruit into the bowl to grasp equilibrium. From these theories and definitions, a very significant question may be asked; How does the concentration of a solute affect the change in a mass? In this lab, potatoes are used to acquire the answer to this research question. 5 different concentrations of salt solutes are employed to see how a concentration of a solute really does affect the mass in a potato.It is predicted that the potato will gradually lose more mass as the salt in the solute increases. The independent variable of the lab is the concentration of the salt solutes.

They are purposefully being altered to try to reach a conclusion for the research question. The dependent variable of this experiment is the percent change of mass that is resulted in the potato. The lab includes controls such as the shape that the potato is cut into, the cup the solute is placed in, the period of time each potato is left in the cup and the amount of water in each solute.Materials and Procedure: 1. Set up 15 plastic cups in 3 rows and 5 columns. Label each cup with the percentage of salt solute using a black sharpie.

2. Figure 1: Cups of solute and potato lined up in rows depending on their concentration of salt. Figure 1: Cups of solute and potato lined up in rows depending on their concentration of salt. Obtain a potato and use the circular cork borer to make 15 cylinders of potato.

Then cut the cylinders against a cutting board so they are about an even size using a knife. 3. Mass each potato with a scale and put them next to the correct cup. 4.Create five salt solutions of water containing different percentages of salt (0, 2.

5, 5, 10 and 15%). For example, for the 10 % salt solute, measure 20 grams of salt and insert it into a large graduated cylinder of 200 ml (3 trials of 50 ml and one extra 50 ml in case of emergencies). 5. Pour 50 ml of salt solution into the correctly labeled plastic cup. Insert all potato cuts into their assigned cup at the same time and start the stopwatch.

6. The timer is set for 10 minutes and after 10 minutes, remove all potato cuts from the water. Then pat them dry using paper towels but DO NOT squeeze them.Make sure to keep each potato in the correct order with the correctly labeled cup. 7. Mass each potato once more then record all data into a data table.

Data Collection and Processing Table 1: Each of the five different solutes and trials (initial and final mass). Concentration of Salt Solute (%)| 0| 2. 5| 5| 10| 15| Trial 1| Initial Mass (g±0. 01)| 0. 76| 0. 79| 0.

77| 0. 75| 0. 78| | Final Mass (g±0. 01)| 0.

78| 0. 75| 0. 66| 0. 62| 0. 63| Trial 2| Initial Mass (g±0.

01)| 0. 75| 0. 77| 0. 78| 0. 78| 0. 73| | Final Mass (g±0.

01)| 0. 77| 0. 72| 0. 66| 0.

65| 0. 56| Trial 3| InitialMass (g±0. 1)| 0. 79| 0.

79| 0. 79| 0. 78| 0. 76| | Final Mass (g±0. 01)| 0.

81| 0. 73| 0. 69| 0. 65| 0. 63| Table 2: Data is processed into average mass lost or gained and the percent change. Concentration of Water (%)| Average Change in Mass(g±0.

03)| Average Mass Initial ( g±0. 03)| Average Mass Final( g±0. 03)| Percent Change| | | | | Trial 1(%)| Trial 2(%)| Trial 3(%)| Average(%)| 0| 0. 02| 0. 77±3. 90| 0.

79±3. 80| 2. 63| 2. 67| 2. 53| 2. 61±0.

06| 2. 5| 0. 05| 0. 78±3.

85| 0. 73±4. 11| -5. 06| -6. 49| -7.

59| -6. 38±0. 15| 5| 0. 11| 0. 78±3.

85| 0. 67±4. 48| -14. 3| -15.

4| -12. 7| -14. 1±0. 1| 10| 0. 13| 0. 77±3.

90| 0. 64±4. 69| -17. 3| -16. 7| -16.

7| -16. 9±0. 37| 15| 0. 15| 0.

76±3. 95| 0. 61±4. 92| -19. 2| -23.

3| -17. 1| -19. 8±0. 42| Sample Calculations Average Change in Mass: (Trial 1: Mfinal - Minitial) + (Trial 2: Mfinal - Minitial) + (Trial 3: Mfinal - Minitial) / 3 = Average 5% Ex: (0.

11) + (0. 12) + (0. 10) / 3 = 0. 11 Trial Percent Change: (Mass Final) – (Mass Initial) X 100 = Trial Percent Change (Mass Initial) 5% Trial 1 Ex: (0. 66) – (0. 77) X 100 = -14.

3% (0. 77) Average Mass Initial/Final Uncertainty: (0. 03)/(Mass Final) x 100 = Mass Final Uncertainty 0. 03)/ (Mass Initial) x 100 = Mass Initial Uncertainty Average Percent Change Uncertainty: 2(Mass Initial) + (Mass Final) x (Average Percent Change) (100) 5% Ex: 2(9.

78) + (0. 67) x (14. 1) = ± 0. 31 (100) Figure 2: This graph shows the relationship between the different salt solutes and the average percent change. It also includes error bars.

Figure 2: This graph shows the relationship between the different salt solutes and the average percent change. It also includes error bars. Conclusion and Analysis From the results of this experiment, one can observe that there is a definite orrelation between the percent of salt in the solute to the average percent change of mass in the potato. The solute with 15% lost 19. 8% of its mass due to osmosis as it tried to reach equilibrium.

The solute with 10% decreased its mass by 16. 9%, the solute of 5% lost 14. 1% and the solute with 2. 5% lost 6. 38%.

However, the trial with no salt in the solute had different results. Instead of experiencing osmosis, the potato went through diffusion. Since the salt percentage in the potato was greater than the salt percentage in the water, the potato actually gained mass as it tried to reach equilibrium.The hypothesis from above was correct; the difference in percentage of salt in solutes does affect the mass of a potato.

If the potato has less salt percentage than the solute, then it will experience osmosis as it drains its liquids in an attempt to equalize while if the potato has more salt then the water then it will absorb more water to struggle to grasp balance. From this experiment it can be predicted that the state in which the potato and the solute have the exact same amount of salt percentage is somewhere between the ranges of 0% and 2. 5% of salt.There are many weaknesses and limitations that were present throughout the lab.

Weaknesses included the procedure in which the potato chunks were removed from the plastic cups. There were fifteen cups in this experiment and only 3 pairs of hands to remove each potato. The time difference between each potato’s removal could have caused an inaccuracy. Also, when the salt was added, the amount of stirring of each cup wasn’t constant which could have left certain cups with amounts of salt at the bottom.

Without the salt fully dissolved, the accuracy of the lab is questioned. There were also limitations of the equipment.The plastic cups used were recycled and could have contained other substances beforehand that might slightly change the solute. If one was provided with new clean cups then the lab might have been more accurate.

The salt used was also common table salt which could have additionally affected the accuracy of the results. If one was to redo this experiment with better apparatus, uncontaminated cups and special controlled salt would be good to have in mind. Despite the limitations and weaknesses of the lab, the results still show that the higher salt percentage in a solute decreases the mass of a potato due to osmosis.