RP3

Question 1

Describe how to calculate dilutions

Answer 1

Use the formula: 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

Question 2

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. Volume of stock solution required, V1 = (C2/C1) x V2
   (0.15 ÷ 0.5) x 30 = 9 cm3
2. Volume of distilled water to top up with = V2 - V1
   30 - 9 = 21 cm3 distilled water

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)
   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
   ● 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 30
   oC
   control
   ● 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
   ● Blot dry to remove excess water before weighing
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

Why calculate %
change in mass?

Answer 4

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

Question 5

Why blot dry
before weighing?

Answer 5

● 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 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