The Net Movement Of Solvent Molecules

Published Date: 02 Nov 2017

From this definition it can be concluded that when sucrose solution has low concentration and its water potential is higher than the potato’s, water will diffuse into the potato, resulting in the increase in mass of the sample. On the opposite manner, when the sucrose concentration is high but the water potential is lower that of potato, there will be an increase in the volume of solution because in the process of osmosis, water will be moved from the potato sample into the solution. The water potential of potato tuber cells can be calculated as when the water potentials for sucrose solution and potato are the same, there will be no change in the mass of the sample.

Data Collection and Processing

Table 1.: The initial and final masses of the potato cylinders for different concentrations of sucrose solution.

Concentration of sucrose solution [mol dm-3]

trial 1

trial 2

trial 3

trial 4

trial 5

trial 6

initial weight [g ±0.01g]

final weight [g ±0.01g]

initial weight [g ±0.01g]

final weight [g ±0.01g]

initial weight [g ±0.01g]

final weight [g ±0.01g]

initial weight [g ±0.01g]

final weight [g ±0.01g]

initial weight [g ±0.01g]

final weight [g ±0.01g]

initial weight [g ±0.01g]

final weight [g ±0.01g]

0.0

2.85

3.21

2.16

2.53

1.94

2.24

1.94

2.33

1.70

1.88

2.00

2.10

0.05

2.06

2.18

2.12

2.29

1.93

2.08

1.93

2.07

1.90

2.07

1.75

1.98

0.10

2.33

2.42

2.54

2.65

1.91

2.06

1.96

2.05

1.99

2.06

2.20

2.34

0.15

2.16

2.24

1.73

1.79

1.93

2.08

1.91

1.97

2.15

2.46

1.81

2.03

0.20

2.00

2.00

2.12

2.18

1.91

1.97

1.92

1.96

2.04

2.16

2.13

2.27

0.25

2.02

1.97

2.03

2.04

1.90

1.90

1.92

1.94

1.75

1.81

2.12

2.18

0.30

2.15

2.09

2.12

2.04

2.18

2.13

1.99

1.96

2.17

2.14

-

-

The following formula was used to calculate the percentage change in mass of the potato cylinders:

Percentage change in mass = (∆mass / initial mass) ∙ 100%

An example is shown below for 0.05 molarity concentration for trial 1.:

[(2.18g - 2.06g)/2.06] ∙ 100% = [0.12/2.06] ∙ 100% = 0.0583 ∙ 100% = 5.83 %

Next step is calculating the percentage mass change of potato tuber cells for different sucrose solutions. The initial masses of samples vary greatly, so presenting the data in a manner of percentage change will make the analysis process easier.

The following formula was used to calculate the mean percentage change in mass:

Mean percentage in mass = ∑mass change in valid trials / number of valid trials

Standard deviation and the mean percentage change will be calculated using Microsoft Office Excel (=AVERAGE,=STDEV).

Table 2.: The percentage change in mass of the potato cylinders, the mean percentage change in mass and the standard deviation for different concentrations of sucrose solution.

Concentration of sucrose solution [mol dm-3]

Percentage change in mass [%]

Mean [%]

Standard deviation

trial 1

trial 2

trial 3

trial 4

trial 5

trial 6

0.00

12.63

17.13

15.46

20.10

10.59

5.00

13.49

5.33014

0.05

5.83

8.02

7.77

7.25

8.95

13.14

8.49

2.49746

0.10

3.86

4.33

7.85

4.59

3.52

6.36

5.09

1.67546

0.15

3.70

3.47

7.77

3.14

14.42

12.15

7.44

4.88368

0.20

0.00

2.83

3.14

2.08

5.88

6.57

3.42

2.4451

0.25

-2.48

0.49

0.00

1.04

3.43

3.83

1.05

2.33538

0.30

-2.79

-3.77

-2.29

-1.51

-1.38

-

-2.35

0.982354

Graph 1.: The relationship between the concentration of sucrose solution [mol dm-3] and the mean percentage change in mass of the potato samples [%]. Error indicate the standard deviations of observed values.

Graph 1.2.: The same relationship, but excluding values of mean percentage change in mass for 0.15 molarity of sucrose solution, as the standard deviation showed the dispersions too big to set an accuarate trend line. Determing the most accuarate trend line is relevant because its equation will be later used to calculate the final solute potential.

Table 3.: Solute potential of different sucrose solutions at 20°C.

Sucrose solution concentration [mol dm-3]

Sucrose solution potential [kPa]

0.05

-130

0.10

-260

0.15

-410

0.20

-540

0.25

-680

0.30

-860

0.35

-970

0.40

-1120

0.45

-1280

0.50

-1450

0.55

-1620

0.60

-1800

0.65

-1980

0.70

-2180

0.75

-2370

0.80

-2580

0.85

-2790

0.90

-3000

0.95

-3250

1.00

-3500

(Source:  the instruction sheet of the experiment)

Graph 2.: The relationship between the concentration of sucrose solution and its water potential.

The solute potential of the potato tuber cells equals the solute potential of sucrose solution in which samples didn’t change their masses. Such concentration of sucrose solution can be calculated using the graph 1 (equation for trend line):

y = 65.571x2 – 65.436 + 12.44

Solve for x:

65.571x2 - 52.996 = 0

(8.09759 x - 7.27984) (8.09759 x + 7.27984) = 0

8.09759 x - 7.27984 = 0 or 8.09759 x + 7.27984 = 0

8.09759 x = 7.27984 or 8.09759 x + 7.27984 = 0

x = 0.899012 or 8.09759 x + 7.27984 = 0

x = 0.899012 or 8.09759 x = -7.27984

x = 0.899012 [mol dm-3] or x = -0.899012 [mol dm-3]

First graph shows that the point where there is no mass change of the potato cylinder occurs is 0.899012 mol dm-3, as this is the value at which the trend line intersects the x-axis.

Last step is determining the solute potential of the potato tuber cells, by solving the equation of trend line for Graph 2. for x = 0.90 mol dm-3 (sucrose concentration.

y = -1225.6x2 – 2201.1x – 43.237

y = - (1225.6 ∙ 0.92) – (2201.1 ∙ 0.9) – 43.237

y = - 992.736 – 1980.99 – 43.237= - 3016.963

The solute potential of potato tuber cells was calculated at about -3017 [kPa]

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