Topic 2 : Required practical 3

Question 1

Describe how a dilution can be calculated

Answer 1

1. Calculate dilution factor = desired concentration (C2) / stock concentration (C1)
2. Calculate volume of stock solution (V1) = dilution factor x final desired volume (V2)
3. Calculate volume of distilled water = final desired volume (V2) - volume of stock solution (V1)

Question 2

worked example of how it can be calculated :

Answer 2

1. Calculate dilution factor = desired concentration (C2) / stock concentration (C1)
2. Calculate volume of stock solution (V1) = dilution factor x final desired volume (V2)
3. Calculate volume of distilled water = final desired volume (V2) - volume of stock solution (V1)

Question 3

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

Answer 3

Step:
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 variable : ● Volume of solution, eg. 20 cm3

2. Use scalpel / cork borer to cut potato into
   identical cylinders

Control variable : ● Size, shape and surface area of plant tissue
● Source of plant tissue ie variety or age

3. Blot dry with a paper towel and measure /
   record initial mass of each piece

Control variable :
● 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 30
   oC

Control variable :
● Length of time in solution
● Temperature
● Regularly stir / shake to ensure all surfaces exposed

5. Blot dry with a paper towel and measure /
   record final mass of each piece

Control variable : ● Blot dry to remove excess water before weighing

Part 2: processing data
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

Question 4

explain why % change in mass is calculated ( 2marks )

Answer 4

• enables comparison
• as plant tissue sample had different initial masses

Question 5

explain why potatoes are blotted dry before weighing (2marks)

Answer 5

• solution on surface will add to mass
• amount of solution on cube varies

Question 6

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

Answer 6

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