Required Practical 3

• • Producing a dilution series of a solute - Produce a calibration curve
  • Calibration curve -> Identify the water potential of plant tissue
• Dilutions = C1 x V1 = C2 x V2

  • C1 = concentration of stock solution
  • V1 = volume of stock solution used to make new concentration
  • C2= concentration of solution you are making
  • V2 = volume of new solution you are making
• V2 = V1 + volume of distilled water to dilute with
• Create a dilution series

  1. Use a 1 mol dm-3 sucrose solution

  2. Dilute to 0.0, 0.2, 0.4, 0.6, 0.8 1.0 mol dm-3
• Cut potato into identical cylinders

  1. Use a cork borer

  2. Control: size, shape and surface area of plant tissue

  3. Control: source of plant tissue ie variety or age
• Measure and record initial mass of each piece

  Blot dry with a paper towel
• Immerse 1 chip in each solution

  1. Leave for 20-30 mins

  2. In water bath at 30 degrees

  3. Control: length of time in solution

  4. Control: temperature

  5. Control: regularly stir and shake to ensure all surfaces are exposed
• Measure and record final mass of each piece

  Blot dry with a paper towel
• • Calculate the percentage change in mass - (final mass - inital mass)/ initial mass
  • Plot a graph with concentration on x axis and percentage change in mass on y axis (calibration curve) -> must show positive and negative regions
  • Identify conc where line of best fit intercepts the x axis (0% change) -> water potential of sucrose solution = water potential of potato cells
• Percentage change in mass is used to enable comparison and show the proportional change this is because plant tissue samples had different initial masses.
• Blot drying is used before weighing so the solution on the surface will add to mass, as we want to measure the water taken up or lost. The amount of solution on a cube varies so it is important to ensure the same amount of solution is on the outside
• The changes in plant tissue mass when placed in different concentrations of solute:

  • Increase in mass - water moves into cells by osmosis, as the water potential of the solution is higher than inside cells
  • Decrease in mass - water moves out of cells by osmosis, as the water potential of the solution is lower than inside cells
  • No change - no net gain or loss of water by osmosis, as the water potential of the solution is equal to the water potential of the cell