Monday 9 June 2014

Practical 3

PRACTICAL 3: IDENTIFICATION OF PHASE DIAGRAM FOR PHENOL AND WATER

Objective:
  1. To understand the phase diagram of phenol and water.
  2. To determine the graphs phenol composition-temperature and the critical solution temperatures.

Introduction:

            Some of the liquids are completely miscible in all proportions, for example: ethanol and water. Other liquids which have certain miscible proportion in certain liquids, for example: ether and water, phenol and water. (Although phenol is not totally in liquid form, but it can be assumed as liquid because of the addition of water will decrease the melting point of phenol down below room temperature to produce a liquid-liquid system.) The factors that affect the miscibility are nature of solute/solvent, temperature and pressure. Temperature will depend whether the reaction results to an endothermic or exothermic process, if endothermic the solubility of the solution will increase with an increase in temperature. For an exothermic process, solubility will increase with a decrease in temperature. The nature of solvent and solute also take part in its solubility, such nature involves polarity and molecular size.
            Normally, the two liquids (phenol and water) will become more miscible in each other with the increasing of the temperature until where it reach a consulate temperature and beyond this temperature, the liquid-liquid system will completely miscible at any measurement. Most probably, any conjugated liquids can  form a close system, whereby there is the existence of two consulate temperature at the top and bottom, but it is not easily to determine the two temperature (before the substance become evaporated or solidified) except with nicotine and water.
            At any temperature below certain critical solution temperature, the compositions for two liquid phases in equilibrium are constant and are not affected by the relative amount of these two phases. The miscibility between two partial miscible liquids are normally affected by the existence of third component.


Procedures:

  1. 20ml mixture of phenol and water are prepared in 100cm³ conical flasks that contain 8% of phenol in volume.
  2. 20ml of each mixture is prepared by measuring with a burette.
  3. Water was filled into the conical flask before the addition of 8% phenol.
  4. Thermometer was inserted into the conical flask and the top of the conical flask was wrapped with film immediately to prevent the evaporation of carcinogenic phenol.
  5. The conical Flask that contains the mixture was put in a water bath and warmed to 70ºC.
  6. At the same time, conical flask was been shake slightly in the water bath to speed up the dispersion of two liquids.
  7. Conical flask was put in a beaker that contains cold water to cool down the temperature of the solution after taking out from hot water bath.
  8. The temperature where the mixture started becomes cloudy was recorded.
  9. Step 1-9 was repeated for 20ml mixture of phenol and water in such percentage of phenol: 11%, 37%, and 63% v/v.
  10. The graph temperature versus percentage of phenol was plotted to produce a phase diagram.


Results:

Percentage of phenol (%)
Volume of phenol (ml)
Volume of water (ml)
Temperature ( ºC)
Average temperature 
( ºC)
1
2

8.0
0.8
9.2
51.0
33.0
42.0
24.0
2.4
7.6
67.0
48.0
57.5
50.0
5.0
5.0
68.0
60.0
64.0
64.0
6.4
3.6
54.0
55.0
54.5
80.0
8.0
2.0
49.0
48.0
48.5

















Discussions:

            Two component systems containing liquid phases are discussed in this experiment. Phenol is partial miscible with water. Miscibility means how completely two or more liquids dissolve in each other. The curve plotted in the graph temperature versus percentage of phenol in water in volume per volume shows the limits of concentration and temperature within which two liquid phases exists in equilibrium. The region outside this curve contains systems having but one liquid phase.
For phenol and water system, it is a two component with two liquid phase system. Thus, the degrees of freedom are two as F=2-2+2=2 , which represents temperature and percentage of phenol in water in volume. A line drawn across the region containing two phases is called a tie line. It is always parallel to the base line in two-component systems. At equilibrium, all systems prepared on the tie line will separate into phases of constant temperature which termed conjugated phases. Tie line in a phase diagram use to calculate the composition of each phase in addition to the weight of the phases.
            Phenol and water have miscibility in limited proportion and form two phase liquid-liquid system below critical solution temperature. At 8% of phenol in water at 25ºC, single liquid phase is produced. This is due to the less percentage of phenol in water and it is miscible with water completely. From the graph plotted, at 53ºC, a minute amount of a second phase appears. The concentration of phenol and water at which this occurs is 11% by weight of phenol in water. At 68ºC, this is the upper consulate temperature which is the maximum temperature at which the two-phase region exists. Phenol-rich phase will lie below water-rich phase since phenol has a higher density than water.
            There are several precautions that should be taken to obtain an accurate result. First and foremost, after the addition of phenol into the conical flask, film should be wrapped on the top of conical flask with thermometer in the middle to avoid evaporation of phenol so that it will not reduce the concentration of phenol in tubes. Besides that, due to phenol is acidic and carcinogenic compound, thus extra care should be taken to avoid harm to the human. Moreover, pipette instead of measuring cylinder is used to obtain more accurate volume required. This can ensure the concentration of phenol obtained is accurate.
            In conclusion, phase diagram is used in practice to formulate systems containing more than one component where it may be advantageous to achieve a single liquid phase product.


Conclusion:
            The consulate temperature for phenol/water system is 68ºC. Phenol is partial miscible with water and produce one liquid phase system at certain temperature and concentration when pressure is fixed.


Reference:
1. E.A.Moelwyn-Hughes. (1961). Physical Chemistry, 2nd Ed.Pergamon.New York.
2. A.T.Florence and D.Attwood. (1998). Physicochemical Principals of Pharmacy, 3rd Edition. Macmillan Press Ltd.
           


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Practical 4

PRACTICAL 4 : DETERMINATION OF DIFFUSION COEFFICIENT

Date of experiment: 
22 May 2014

Objectives: 
 To study the diffusion of molecules in an agar medium.
 To determine the diffusion coefficient of crystal violet and bromothymol blue at different temperature.

Introduction:
Theory

Diffusion, which is the spontaneous movement of solutes from an area of high concentration to an area of low concentration can be explained by Fick's law which states that the flux of material (amount dm in time dt) across a given plane (area A) is proportional to the concentration gradient dc/dx.

                                          dc
dm = -DA ---- dt ----------------------------------      (i)
                  dx
D is the diffusion coefficient or diffusivity for the solute, in unit m2s-1
.

If a solution containing neutral particles with the concentration M0, is placed within a cylindrical tube next to a water column, diffusion can be stated as
M = M0 exp (-x2/4Dt) ------------------------------------------------------ (ii)
where M is the concentration at distance x from the intersection between water and solution that is measured at time t.

By changing equation (ii) to its logarithmic form, we get

ln M =1n M0 - x2 /4Dt

or            2.303 x 4D (log10 M0 – log10 M) t= x2 ---------------------------------------------------- (iii)

Thus a plot of x2 against t can produce a straight line that passes through the origin with the slope 2.303 x 4D (log10 M0 – log10 M). From here D can be calculated.

If the particles in the solution are assumed to be spherical, their size and molecular weight can be calculated by the Stokes-Einstein equation.

D = kT/6πηa

where k is the Boltzmann constant 1.38 x 1023 Jk-1, T temperature in Kelvin, π the viscosity of the solvent in Nm-2s and a the radius of particle in M. The volume of a spherical particle is 4/3 πa3, thus its weight M is equivalent to 4/3 πa3Nρ (ρ = density).
It is known that molecular weight M=mN (N is Avogadro’s number 6.023 x 1023 mol-1).
∴ M = 4/3 πa3Nρ ---------------------------- (v)

Diffusion for charged particles, equation (iii) needs to be modified to include potential gradient effect that exists between the solution and solvent. However, this can be overcome by adding a little sodium chloride into the solvent to prevent the formation of this potential gradient.

Agar gels contain a partially strong network of molecules that is penetrated by water. The water molecules form a continuous phase around the gel. Thus, the molecules of solutes can diffuse freely in the water if chemical interactions and adsorption effects do not exist entirely. Therefore, the gel forms an appropriate support system to be used in diffusion studies for molecules in a medium of water.


Apparatus:
10ml, 50ml, 100ml and 250ml measuring cylinders, 1000ml beaker, 14 test tubes with stoppers, weighing boat, spatula, glass rod, electronic balance, hot plate and stirrer, test tube rack, dropper, filter funnel,marker and water bath.

Materials: 
Jelly Powder, Ringlet’s solution, 1 : 200 crystal violet indicator, 1 : 400 crystal violet indicator, 1 : 600 crystal violet indicator, 1 : 500 000 crystal violet indicator, 1 : 200 bromothymol blue solution, 1 : 400 bromothymol blue solution, 1 : 600 bromothymol blue solution, 1 : 500 000 bromothymol blue solution, and distilled water.

Procedures: 
      1.      Agar in Ringer’s solution was prepared as four tablets were crushed by using pestle and mortar.

      2.      Then, the crushed tablets were transfered into the 1000ml beaker and 500ml of distilled water was added into the beaker to dissolve the tablets.

      3.      7g of agar powders were weighed on electronic balance and 425ml Ringer’s solution was measured in a suitable measuring cylinders.

      4.       Agar powders and Ringer’s solution were mixed in the beaker. The mixed solution was heated until the solution completely dissolved and boiled.

      5.       The solution was continuously stirred during the heating process to prevent the jelly powder from embed in the mixture.

     6.       During the heating process, 14 test tubes were labelled according to the concentrations and types of the solutes, which are bromotyhmol blue and crystal violet. The heating process was stopped when the mixture become clear dissolved solution.

      7.       20ml of agar solution was prepared in 14 labelled test tubes.

      8.      5ml of 1:500 000 crystal violet solution was added into a test tube containing hot agar solution labelled with 1:500 000 crystal violet at 28°C whereas 5ml of 1:500 000 bromothymol blue solution was added into test tube containing the hot gel solution labelled with 1:500 00 bromothymol blue at 37°C. These two test tubes will be used as the standard to measure the colour distance resulting from the solute diffusion.

      9.      The agar solution in the rest of test tubes were allowed to cool at room temperature until they become solid agar.

     10.  5ml of each crystal violet solution was added into the gels that were prepared in 6 test tubes according to their concentration.

     11.  This step was repeated by using bromothymol blue as an indicator. All test tubes were closed to prevent evaporation and stored at temperature 28°C and 37°C.
  
     12.  The distance of crystal violet and bromothymol blue solution diffused in the agar were measured and recorded for seven days.




     Results


     a)  Crystal Violet

System
Time (seconds)
x, cm
x2, cm2
Slope of graph
D, cm2s-1
Temp., o C
Average Diffusion Coefficient, D cm2s-1
1:200
0
0
0
3.057 X 10-5
 9.7662 x 10-7
28
 2.8652 x 10-7
86400
1.3
1.69
172800
2.1
4.41
259200
2.8
7.84
345600
3.3
10.89
432000
3.7
13.69
518400
4.1
16.81
604800
4.3
18.49
1:400
0
0
0
1.911 X 10-5
6.6986 x
10-7
28
86400
1.2
1.44
172800
1.8
3.24
259200
2.4
5.76
345600
2.7
7.29
432000
3.0
9.00
518400
3.3
10.89
604800
3.4
11.56
1:600
0
0
0
1.488X 10-5
5.5281 x
10-7
28
86400
0.75
0.56
172800
1.6
2.56
259200
2
4.0
345600
2.3
5.29
432000
2.5
6.25
518400
2.8
7.84
604800
3.0
9.0
1:200
0
0
0
 3.652 X 10-5
 1.16669 x10-6
37
9.2097 x 10-7
86400
1.6
2.56
172800
2.4
5.76
259200
3.0
9.00
345600
3.6
12.96
432000
4.0
16.00
518400
4.4
19.36
604800
4.7
22.09
1:400
0
0
0
2.646 X 10-5
 9.27497 x 10-7
37
86400
1.4
1.96
172800
2.0
4.00
259200
2.6
6.76
345600
3.0
9.00
432000
3.4
11.56
518400
3.8
14.44
604800
4.0
16.00
1:600
0
0
0
1.800 X 10-5
6.6871 x 10-7
37
86400
1.0
1.00
172800
1.7
2.89
259200
2.2
4.84
345600
2.5
6.25
432000
2.8
7.84
518400
3.1
9.61
604800
3.3
10.89






b)  Bromothymol Blue


System
Time (seconds)
x, cm
x2, cm2
Slope of graph
D, cm2S-1
Temp, o C
Average Diffusion Coefficient, D cm2S-1
A (1:200)
0
0
0

2.917 X 10-5








9.31886 x 10-7
28

















 1.25302 x10-6
86400
1.5
2.25
172800
2.3
5.29
259200
2.7
7.29
345600
3.2
10.24
432000
3.6
12.96
518400
3.9
15.21
604800
4.2
17.64
B (1:400)
0
0
0
2.388 X 10-5



8.3703 x 10-7
28
86400
1.3
1.69
172800
1.9
3.61
259200
2.5
6.25
345600
2.9
8.41
432000
3.2
10.24
518400
3.6
12.96
604800
3.8
14.44
C (1:600)
0
0
0
 5.357 X 10-6
1.99017 x 10-6
28
86400
0.4
0.16
172800
0.8
0.64
259200
1.1
1.21
345600
1.3
1.69
432000
1.4
1.96
518400
1.6
2.56
604800
1.8
3.24
D (1:200)
0
0
0
3.652 X 10-5
1.16669 x 10-6 
37














1.3313 x 10-6
86400
1.7
2.89
172800
2.5
6.25
259200
3.1
9.61
345600
3.6
12.96
432000
4
16
518400
4.4
19.36
604800
4.7
22.09
E (1:400)
0
0
0
2.388 X 10-5
8.3703 x
10-7
37
86400
1.3
1.69
172800
2.0
4.00
259200
2.5
6.25
345600
2.9
8.41
432000
3.2
10.24
518400
3.5
12.25
604800
3.8
14.44
F (1:600)
0
0
0
8.003 X 10-6







1.9901 x 10-6
37
86400
0.7
0.49
172800
1.1
1.21
259200
1.4
1.96
345600
1.6
2.56
432000
1.9
3.61
518400
2.1
4.41
604800
2.2
4.84
______


 Graph
















Calculations:


From equation:
2.303 x 4D (log 10 Mo – log 10 M) t = X²
Hence the slope of the graph = 2.303 x 4D (log 10 Mo - log 10 M)

      1.      Crystal violet system with dilution 1:200 (28 0C)
    Slope= 3.057 x 10-5cm2/sec
    M = 1:500000                       M0 = 1:200
        = 1 / 500000                           = 1 / 200
        = 2 x 10-6                                          = 5 x 10-3
           
2.303x4D [log 10 (5x10-3)-log 10 (2x10-6)] = 3.057×10-5 cm2/sec
D = 9.7662×10-7 cm2/sec

     2.      Crystal violet system with dilution 1:400 (28ºC)
Slope = 1.911×10-5 cm2/sec
     M = 1:500000                      Mo = 1:400
         = 1 / 500000                          = 1 / 400
         = 2 x 10-6                               = 2.5 x 10-3

2.303x4D [log 10 (2.5x10-3)-log 10 (2x10-6)] = 1.911×10-5 cm2/sec
D = 6.6986×10-7 cm2/sec

     3.      Crystal violet system with dilution 1:600 (28 0C)
     Slope=1.488×10-5 cm2/sec
         M = 1:500000                      Mo = 1:600
             = 1 / 500000                          = 1 / 600
             = 2 x 10-6                                        = 1.67 x 10-3
    

2.303x4D [log 10 (1.67x10-3)-log 10 (2x10-6 )] = 1.488×10-5 cm2/sec
D = 5.5281×10-7 cm2/sec

Average of Diffusion Coefficient, m²/sec for Crystal violet system at 28ºC
= (9.7662×10-7 cm2/sec+6.6986×10-7 cm2/sec+5.5281×10-7 cm2/sec) / 3
= 2.8652×10-7 cm2/sec

      4.      Crystal violet system with dilution 1:200 (37ºC)
           Slope =3.652×10-5 cm2/sec
               M = 1:500000                       M0 = 1:200
                   = 1 / 500000                           = 1 / 200
                   = 2 x 10-6                                         = 5 x 10-3
           
 2.303x4D [log 10 (5x10-3)-log 10 (2x10-6 )]  = 3.652×10-5 cm2/sec
 D=1.16669×10-6 cm2/sec

      5.      Crystal violet system with dilution 1:400 (37ºC)
Slope =2.646×10-5 cm2/sec
    M = 1:500000                      Mo = 1:400
        = 1 / 500000                           = 1 / 400
        = 2 x 10-6                               = 2.5 x 10-3
           
2.303x4D [log 10 (2.5x10-3)-log 10 (2x10-6)] = 2.646×10-5 cm2/sec
D = 9.267497×10-7 cm2/sec

6. Crystal violet system with dilution 1:600 (37ºC)
Slope = 1.8×10-5 cm2/sec
     M = 1:500000                      Mo = 1:600
         = 1 / 500000                          = 1 / 600
         = 2 x 10-6                                        = 1.67 x 10-3 

2.303x4D [log 10 (1.67x10-3)-log 10 (2x10-6 )] = 1.8×10-5 cm2/sec
D = 6.871×10-7 cm2/sec

     Average of Diffusion Coefficient, m²/sec for Crystal violet system at 37ºC
      = (1.16669×10-6 cm2/sec + 9.267497×10-7 cm2/sec +6.871×10-7 cm2/sec) / 3
      = 9.2097×10-7 cm2/sec

# Value of D37°C for crystal violet using the equation
D28°C/D37°C = T28°C/T37°C
2.8652×10-7/D37°C = (28°C + 273.15K)/(37°C + 273.15K)
D 37°C = 2.9508×10-7 cm2/sec

      7.      Bromotymol blue system with dilution 1:200 (28ºC)

Slope  = 2.917×10-5 cm2/sec
      M = 1:500000                      Mo = 1:200
      = 1 / 500000                          = 1 / 200
      = 2 x 10-6                                        = 5 x 10-3
           

2.303x4D [log 10 (5x10-3)-log 10 (2x10-6)] = 2.917×10-5 cm2/sec
D = 9.1886×10-7 cm2/sec

      8.      Bromotymol blue system with dilution 1:400 (28ºC)
Slope = 2.388×10-5 cm2/sec
     M = 1:500000                      Mo = 1:400
         = 1 / 500000                          = 1 / 400
         = 2 x 10-6                                        = 2.5 x 10-3

 2.303x4D [log 10 (2.5x10-3)-log 10 (2x10-6 )] = 2.388×10-5 cm2/sec
D= 8.3703×10-7 cm2/sec

9. Bromotymol blue system with dilution 1:600 (28ºC)

Slope =5.357×10-6 cm2/sec
     M = 1:500000                      Mo = 1:600
         = 1 / 500000                          = 1 / 600
         = 2 x 10-6                                        = 1.67 x 10-3

2.303x4D [log 10 (1.67x10-3)-log 10 (2x10-6 )] = 5.357×10-6 cm2/sec
D = 1.9907×10-7 cm2/sec

Average of Diffusion Coefficient, m²/sec for Bromothymol blue system at 28ºC
 = (9.1886×10-7 cm2/sec + 8.3703×10-7  cm2/sec +11.9907×10-7 cm2/sec) / 3
 = 1.253×10-6 cm2/sec

     10.  Bromotymol blue system with dilution 1:200 (37ºC)
Slope = 3.625×10-5 cm2/sec
     M = 1:500000                       Mo = 1:200
         = 1 / 500000                           = 1 / 200
         = 2 x 10-6                                         = 5 x 10-3

2.303x4D [log 10 (5x10-3)-log 10 (2x10-6 )] = 3.625×10-5 cm2/sec
D = 1.1669×10-6 cm2/sec

11. Bromotymol blue system with dilution 1:400 (37ºC)
      Slope  =2.388×10-5 cm2/sec
           M = 1:500000                      Mo = 1:400
               = 1 / 500000                          = 1 / 400
               = 2 x 10-6                              = 2.5 x 10-3
          
2.303x4D [log 10 (2.5x10-3)-log 10 (2x10-6)] = 2.3888×10-5 cm2/sec
D = 8.3703×10-7 cm2/sec

12. Bromotymol blue system with dilution 1:600 (37ºC)
Slope  =8.003×10-6 cm2/sec
      M = 1:500000                      Ma = 1:600
          = 1 / 500000                          = 1 / 600
          = 2 x 10-6                                        = 1.67 x 10-3

2.303x4D [log 10 (1.67x10-3)-log 10 (2x10-6 )] = 8.003×10-6 cm2/sec
 D = 1.9901×10-6 cm2/sec
  
 Average of Diffusion Coefficient, m²/sec for Bromotymol blue system at 37ºC
      = (1.1669×10-6 cm2/sec +8.3703×10-7cm2/sec +1.9901×10-6 cm2/sec) / 3
      = 1.3313×10-6 cm2/sec


# Value of D37°C for bromothymol blue using the equation
D28°C/D37°C = T28°C/T37°C
1.253×10-6 /D37°C = (28°C + 273.15K)/(37°C + 273.15K)
D 37°C = 1.2904×10-6 cm2/sec



Discussion: :

                Diffusion is the movement of solute particles in a solid from a high concentration area to a low concentration area, resulting in the uniform distribution of the substance. Diffusion is a process which is not due to the force action but due to the random movements of atoms. In each diffusion reaction the flux of matter is equal to the conductivity multiplied by a driving force. Conductivity is the mobility of the diffusing species or known as diffusivity. The presence of concentration gradient acts as the driving force for the solute particles to move.

            Fick’s First Law states that the flux, dm, across a membrane of unit area, A, is proportional to the concentration gradient, dc/dx, and is expressed by

dm = -DA(dc/dx)dt

D is the diffusion coefficient or diffusivity for the solute. The value of diffusivity describes the rate of diffusion and therefore its unit is m2s-1. Diffusion coefficient depends on several factors such as temperature, pressure, composition, physical state, structure of the phase, and oxygen fugacity. The higher the value of diffusion coefficient, the easier for that solute to penetrate through the continuous phase.

The negative sign indicates that the direction of the diffusive flux is down the concentration gradient. Diffusive flux goes from high concentration region to low concentration region with the gradient going the opposite direction, from low to high concentration. Fick’s First Law only applies to steady state flux where the concentration gradient is uniform.

            In this practical, a solution containing neutral molecules with the concentration M0 is placed within a cylindrical tube. The diffusion can be expressed as

M = Mo exp (x²/4Dt) 

The equation is derived into logarithmic form and produce new equation,

ln M = ln Mo – (x²/4Dt)

                                          or    2.303 x 4D (log 10 Mo- log 10 M) t = x²

Therefore, a graph of x2 against time produces a straight line that passes through origin with the slope of 2.303 x 4D (log 10 Mo- log 10 M). D can be calculated from the gradient obtained.

            The results show that the diffusion coefficient, D, for both crystal violet and bromothymol blue is increasing from 1:600<1:400<1:200 which mean that 1:200 concentration is the easiest to diffuse as 1:200 concentration is the most concentrated solution. Therefore, the solute with high concentration is easier to diffuse through the solid medium compared to the low concentration. This is because the concentration gradient is high for concentrated solute and forces the solute particles to diffuse further through the medium.

            If we compare the diffusion coefficient in the term of temperature between 28°C and 37°C, the D for 37°C has the larger value than that of 28°C.  The same phenomena occurred for both crystal violet and bromothymol blue solution. Temperature is one of the factors that affect the rate of diffusion. As the temperature increases, the amount of energy available for diffusion is increased. There would be increase in molecules' mobility (kinetic energy). The molecules move faster and there will be more spontaneous spreading of the material which means that diffusion occurs quicker. Thus, the diffusivity increases as the temperature increases.

The value of D37°C can be determined if we already know the value for D28°C by using the equation below

D28°C/D37°C = T28°C/T37°C

where T is the temperature in Kelvin.
However, there are the differences between the value of D37°C that we get from the experiment and from the theoretical value. For crystal violet, 9.2099×10-7 cm2s-1 is the experimental value of D37°C whereas 7.5502×10-7 cm2s-1 is the theoretical value that obtained from the equation above. Meanwhile, for bromothymol blue the D37°C from theoretical and experimental is 6.7559×10-7 cm2s-1 and 7.6703×10-7 cm2s-1 respectively. Generally, the experimental value for D37°C is higher than theoretical value. This is maybe due to the some errors that occurred during the experiment is carried out. The temperature may be set higher than 37°C that cause the rate of diffusion to increase. Besides, the errors also due to the room temperature at 28°C that is not constant throughout the experiment and the viscosity of agar in the test tube that is not uniform. Misreading or parallax error might occur.    

According to the equation:
M   = 4/3πa3 N Þ
a3   = 3M / 4πN Þ
a     = 3√ 3M / 4πN Þ    --------------------------- (1)
Where M= molecular weight, a = radius of particle, N=Avogadro's number (6.02x1023),
Þ =density.
Another equation:                    
D = kT / 6 πηa --------------------------- (2)
Substituting (1) into the equation (2)
D = kT / 6 πηa
    = kT / 6 πη 3√ 3M/4πNÞ
    = kT 3√4πNÞ / 6 πη 3√ 3M ------------------------- (3)
Where D=diffusion coefficient, η= viscosity, T=temperature, k=Boltzmann constant                (1.38 x1023 Jk-1)

According to equation (3), D is inversely proportional to M. So, D increases when M decreases. Crystal violet with molecular formula C25N3H30Cl has molecular weight of 407.979 g mol-1 while bromothymol blue solution with molecular formula C27H28Br2O5S has molecular weight of 624.38 g mol−1. Molecular weight is how much mass each particle has or how heavy it is. The heavier the particle, the slower it diffuses into solidified agar solution, assuming energy of the system remains constant. Crystal violet solution diffuses faster than bromothymol blue since the value of diffusion coefficient for crystal violet is higher than bromothymol blue at both temperatures. Crystal violet solution diffuses easily because its molecular weight is much smaller compared to that of bromothymol blue and thus easy for them to penetrate through gel medium. 

Conclusion

From the experiment, the diffusion coefficient of crystal violet at 28°C is 7.3311×10-7 cm2s-1 and at 37°C is 9.2099×10-7 cm2s-1. Meanwhile, for bromothymol blue solution the diffusion coefficient at 28°C is 6.5599×10-7 cm2s-1 and at 37°C is 7.6703×10-7 cm2s-1. The diffusion coefficient of crystal violet is generally higher than bromothymol blue because of the difference inmolecular weight. In term of temperature, diffusivity at 37°C is much higher than diffusivity at 28°C. This is due to the rapid movement of the solute particles as the temperature increases.

References
  1. http://urila.tripod.com/mole.htm
  2. http://en.wikipedia.org/wiki/Molecular_mass
  3. http://www.pojman.com/mg_materials/Diffusion/Diffusion.html

Appendix

Continuous stirring until a clear solution is formed which is shown below




The test tubes before being separated into water bath and left to room temperature



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