1. Osmosis and Diffusion Lab Part 1
Unit 4 Day 3
Osmosis and Diffusion Lab Part 1

2. Opener
Get out the lab pre questions and the reading

3. Agenda
10 min: Explanation of lab 15 min: students set up lab 30 min: wait for results, discuss pre-lab questions 10 min: collect and share data, clean up

4. I can explain how the cell membrane controls the internal environment of the cell in order to maintain dynamic homeostasis
Learning Target #1

5. Osmosis v Diffusion
Diffusion is the movement of molecules from high to low concentration
Osmosis is the movement of water through a semipermeable membrane

6. Lab: part 1 Purpose: to determine unknown molarities of solutions
Materials: 6 cups labeled correctly, 6 pieces of 20 cm dialysis tubing, mL of different solutions, scale, syringe
Procedure: soak dialysis tubing in water for 30 seconds, tie a knot on one end, use syringe to add the correct solution ½ of the tubing, tie other end, take the mass, record, place in correct cup, add distilled water, wait 30 minutes, dry off, reweigh, share data with class

7. Pre-Lab questions

8. #2

9. #3

10. -i= ionization constant#4
Calculate the water potential of a 0.1 M NaCl solution at 25°C. Assume that the pressure potential is zero, and the solute potential=-iCRT.
Ψw = Ψs + Ψp = -(2)(.1 mol/L)(.0831 L-bar/mol-K)(298 K) + 0
Ψw = bar
Ψs = -iCRT
-i= ionization constant
C= concentration (M)
R= pressure constant
T= temp (K)

11. -i= ionization constant#5
If the concentration of NaCl inside of a plant cell is 0.15 M, which way will the water diffuse if the cell is placed into the 0.1 M NaCl solution? First step: calculate water potential for solution INSIDE cell
Ψw = Ψs + Ψp = -(2)(.15 mol/L)(.0831 L-bar/mol-K)(298 K) + 0
Ψw = bar
Ψs = -iCRT
-i= ionization constant
C= concentration (M)
R= pressure constant
T= temp (K)

12. #5 Ψw = -4.95 bar .1 M NaCl Cell Solution Ψw = -7.43 bar .15 M NaCl
Water will flow (by osmosis) INTO the cell, from HIGHER water potential, to LOWER water potential

13. #6
Turgid: the pressure from the vacuole swells and pushes against the cell wall

14. #6 Goal to make Ψ of cell = Ψ of solution (-4.95)
What must Turgor Pressure (ΨP ) equal if there is no net diffusion between the solution and the cell?
Goal to make Ψ of cell = Ψ of solution (-4.95)
Ψ = ΨP + ΨS (of cell)
Ψ = (-7.43)
Ψ = -4.95
ΨP in cell must equal 2.49