Osmosis: Definition, Types & Examples

Procedure I - Test Solution 1: Water 

Complete the tables and questions below using your data and information found under the Background tab 

Data Table I Note: Difference in Final Volumes = Final Volume of Test Sol - Final Volume of Water 

Trial	Starting Volume of Test Solution  (L)	Starting Volume of Water  (L)	Final Volume of Test Solution  (L)	Final Volume of Water  (L)	Difference  in Final Volumes  (L)
1	1.28	1.64	1.46	1.46	0.0
2	1.28	2.04	1.66	1.66	0.0

[1] 

Given that the final heights (and volumes) are the same for the water and test solution, what can you conclude about the amount of solutes in these two solutions? Explain your reasoning. 

Response: The amount of solute molecules in the two solutions is the same. 

Reasoning: When the two solutions are separated with a semipermeable membrane, solvent molecules migrate from a region where the solute concentration is low to a region where solute concentration is high until a balance in solute concentration or osmotic equilibrium is reached between the two solutions. Since the volume of the test solution was lower than that of water at the start of the experiment, the concentration of solute molecules was high in the test solution compared to the concentration of solute molecules in water. This resulted in the migration of water molecules across the semipermeable membrane to the test solution until when the concentration of the solute molecules in the two solutions was the same. Also, since the amount of solute molecules was equal in the two solutions, the concentration of the respective solute molecules equalized, balancing the osmotic pressure when the final volume of the two solutions became the same. 

[2] 

As discussed in the Background material, water is an important biological molecule. Do you expect water to continue to flow across the semipermeable; lipid bilayer after osmotic equilibrium is reached? Why or why not? 

Response : Water will continue flowing across the semipermeamble membrane at the same rate in either direction once the osmotic equilibrium is reached. 

Reasoning : Once the osmotic equilibrium is reached, there will be balanced osmotic pressure leading to further movement of solvent molecules at equal rates in either direction across the semipermeable membrane. 

Procedure II - Test Solution 2: Guanine solution 

Complete the tables and questions below using your data and information found under the Background tab 

Data Table II Note: Difference in Final Volumes = Final Volume of Test Sol - Final Volume of Water 

Trial	Starting Volume of Test Solution  (L)	Starting Volume of Water  (L)	Final Volume of Test Solution  (L)	Final Volume of Water  (L)	Difference  in Final Volumes  (L)
1	1.28	1.41	1.78	0.91	0.87
2	1.28	1.53	1.84	0.97	0.87

[3] 

Why did the height (and volume) change in the test solution? What is the basis for the increase in the volume in the test solution? 

The height of the test solution changed because osmotic pressure exerted in the water forced solvent molecules to move across the semipermeable membrane. The osmotic pressure was exerted because the concentration of solute molecules in the test solution was higher than the concentration of solute molecules in the water solution. This causes the movement of solvent molecules from a region of low concentration (water solution) to a region of high concentration (test solution) across the semipermeable membrane until when an osmotic equilibrium was reached. Therefore, the basis for the increase in the volume of the test solution is the difference in the concentration of solute molecules between the two solutions, which exerts osmotic pressure. 

[4] 

What do you think would happen if both starting volumes were the same? Test your hypothesis by doing a data run. Explain your observation. 

Null hypothesis: If both volumes were the same, the final volume of the test solution would increase. This is because, at similar volumes, the concentration of solute molecules will be higher in the test solution than in water. The consequence of this is the build-up of osmotic pressure which will make solvent molecules migrate across the semipermeable membrane to the test solution until an osmotic equilibrium is reached, thus raising the final volume of the test solution. 

[5] 

Why is the difference in final volumes the same (or very close to the same) for both trials? 

The difference in final volumes remained the same because the concentration of solute molecules in both solutions in both trials was close or the same. 

Procedure III - Test Solution 3: Cytochrome C solution 

Complete the table and questions below using your data and information found under the Background tab (see the Summary of Needed Formulas section) and Activity Form tab 

Data Table III Note: Difference in Final Volumes = Final Volume of Test Sol - Final Volume of Water 

Trial	Starting Volume of Test Solution  (L)	Starting Volume of Water  (L)	Final Volume of Test Solution  (L)	Final Volume of Water  (L)	Difference  in Final Volumes  (L)
1	1.28	1.69	1.82	1.15	0.67
2	1.28	1.42	1.68	1.02	0.66

[6] 

Based on your data and observations from procedure II and procedure III, which solution has the highest concentration? Explain your answer. 

From the data captured in procedures II and III above, guanine solution has a higher concentration of solute molecules compared to cytochrome C solutions. This is because a higher difference in final volumes (0.87) was recorded for guanine solution compared to the final volumes (0.67 or 0.66) for cytochrome C solutions. 

[7] 

Concentration Calculation : Using your Trial 1 data from procedure II and procedure III, and the known concentration of guanine ( Summary of Formulas and Concepts Needed for Calculations ) at bottom of TableTop Procedures 1 – 3 attachment. 

Setup the ratio are as follows 

Concentration of guanine solution /concentration of cytochrome C solution = difference in final volumes for guanine solution/difference in cytochrome C solution final volumes 

Inserting the known values 

0.225 mmol/L /concentration of cytochrome C solution = 0.87/0.67 

0.225 mmol/L /concentration of cytochrome C solution = 1.299 

concentration of cytochrome C solution = 1.299/0.225 

concentration of cytochrome C solution = 0.173 mmol/L