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

Answer 2

1. Calculate dilution factor (desired concentration / stock concentration): 0.15 / 0.5 = 0.3
2. Calculate volume of stock solution (dilution factor x final volume): 0.3 x 30 cm3 = 9 cm3
3. Calculate volume of distilled water (final volume - stock solution volume): 30 cm3 - 9 cm3 = 21 cm3

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

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)
2. Use scalpel / cork borer to cut potato into
   identical cylinders
3. Blot dry with a paper towel and measure /
   record initial mass of each piece
4. Immerse one chip in each solution and
   leave for a set time (20-30 mins) in a
   water bath at 30oC
   exposed
5. Blot dry with a paper towel and measure /
   record final mass of each piece
   6.repeat 3 times or more at each concentration

Question 4

what are the control variables?

Answer 4

● Volume of solution
● Size, shape and surface area of plant tissue
● Source of plant tissue i.e. variety or age
● Blot dry to remove excess water before weighing
● Length of time in solution
● Temperature
● Regularly stir / shake to ensure all surfaces exposed
● Blot dry to remove excess water before weighing

Question 5

how to process data of the results?

Answer 5

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 6

Explain why % change in
mass is calculated.

Answer 6

● Enables comparison / shows proportional change
● As plant tissue samples had different initial masses

Question 7

Explain why the potatoes
are blotted dry before
weighing.

Answer 7

● 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)

Question 8

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

Answer 8

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