rp3 -> Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissue.

how to calculate 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)

describe how you would use a 0.5 mol dm-3 solution of sucrose (stock solution) to produce 30cm3 of a 0.15 mol dm-3 sucrose solution.

method to produce of a calibration curve with which to identify the water potential of plant tissue (eg. potato)

Part 1: collecting data

1. Create a series of dilutions using a 1 mol

dm-3 sucrose solution (0.0, 0.2, 0.4, 0.6, 0.8,

1.0 mol dm-3 )

● control - Volume of solution, eg. 20 cm3

2. Use scalpel / cork borer to cut potato into

identical cylinders

● control - Size, shape and surface area of plant tissue

● control - Source of plant tissue ie variety or age

3. Blot dry with a paper towel and measure /

record initial mass of each piece

● control - Blot dry to remove excess water before weighing

4. Immerse one chip in each solution and

leave for a set time (20-30 mins) in a

water bath at 30oC

● control - Length of time in solution

● control - Temperature

● control - Regularly stir / shake to ensure all surfaces exposed

5. Blot dry with a paper towel and measure /

record final mass of each piece

● control - Blot dry to remove excess water before weighing

Repeat (3 or more times) at each concentration.

Part 2: data processing

6. Calculate % change in mass = (final - initial mass)/ initial mass

7. Plot a graph with concentration on x axis and percentage change in mass on y axis (calibration curve)

● Must show positive and negative regions

8. Identify concentration where line of best fit intercepts x axis (0% change)

● Water potential of sucrose solution = water potential of potato cells

9. Use a table in a textbook to find the water potential of that solution

Explain the changes in plant tissue mass when placed in different concentrations of solute

Increase in mass:

● Water moved into cells by osmosis

● As water potential of solution higher than inside cells

Decrease in mass:

● Water moved out of cells by osmosis

● As water potential of solution lower than inside cells

No change:

● No net gain/loss of water by osmosis

● As water potential of solution = water potential of cells

Why calculate %change in mass?

● Enables comparison / shows proportional change

● As plant tissue samples had different initial masses

Why blot dry before weighing?

● Solution on surface will add to mass (only want to measure water taken up or lost)

● Amount of solution on cube varies (so ensure same amount of solution on outside)