Partial Molal Volume

  1. A.     PURPOSE

Determine the partial molal volume of the components in the solution.

 

  1. B.     BASIC THEORY

Quantitative study of the solution has grown with the introduction of the concept of partial molal quantities. The nature of a solution, for example the volume of a mixture of alcohol and water, continues to change due to composition changes. G.N. Lewis developed and exact differential to obtain the partial molal volume quantities. If reviewed extensive property of a binary solution at constant temperature and pressure, G, the G is a function of 2 variables n1 and n2 which states the number of moles of component 1 and 2. Partial molal properties is defined through the relation:

=                                                                             (1)

=                                                                             (2)

At constant of temperatures (T) and pressure (P), the above concept can be mathematically expressed as follows:

The volume includes extensive property of a solution, so that a volume of a binary solution can be stated as follows:

=  +                                                             (3)

Partial molal volume of component 1 and 2 were determined by measuring the density of the solution.

Integration of the differential volume yields an expression which allows us to determine the solution’s volume from the partial molar volumes of its components (Castellan, 1983):

V =  +                                                                          (4)

Beyond being useful in its own right, this determination is useful because many other

thermodynamic quantities depend on system volume measurements.

Graphic method as described Lewis and Randall can be used as a method of processing data. This method is used of apparent molal volume Φ for treatment of a binary solution. Apparent molal volume is defined as:

                                                                                (5)

where V is the volume of a solution containing components of n1 and n2, while the  is the molar volume of pure solvent at T, P. from the apparent molal volume equation, the volume of the solution is:

                                                                     (6)

Consider a solution with the molality m using water as solvent. In this solution, for every 1000 grams of water (55.1 mol), there are m mol of solute. So n1 = 55.51 moles and n2 = m moles. Apparent partial molal volume becomes:…

                                                                          (7)

 is the molal volume of pure water which can be calculated from molecular weight (18.016 for water) divided by density at the situation observed. To the solution are met:

                                                                              (8)

and

                                                                               (9)

                                                                   (10)

With ,  is density of the solution and density of pure water consecutively, whereas  is the relative molecular mass or molecular weight solutes. So that the partial molal volume of all becoming,

                                                                 (11)

This equation is used if the specific gravity measurements are used pycnometer. In these equations W, ,  are weight of pycnometer filled with the solution, pycnometer filled with water and empty pycnometer consecutively. Partial molal volume of solvent (component 1) and solute (component 2) was calculated from the partial volume and the result obtained are as follows:

=  =  =                           (12)

 

 =           (13)

For a simple electrolyte solution, such as NaCL, it was found that plot  is linear against , to the concentration  that are not concentrated. Because :

 =                                             (14)

So that the partial molal volume second component becomes:

 =  +                                                                      (15)

If for the NaCl solution  is linear to  so :

                                                                  (16)

So

 =                                                                   (17)

For the partial molal volume of component 1 becomes:

 =                                                              (18)

Values  obtained from extrapolation of the graph  versus  at the concentration of m close to zero. By creating a linear graph of  versus , then the slope  can be searched and trhe partial molal volume of solvent can be calculated. Similarly from the slope  and  partial molal volume of solute can be calculated (Wahyuni, 2011).

Partial molal properties that are easy to describe is the partial molal volume contribution to the volume, from a single component of the sample to a total volume. Ideal behavior of the system with all the derived thermodynamic relations can be divided into two parts:

  1. Partial molal scale, such as the partial molal volume, enthalpy, and so on.
  2. Activity and activity coefficients, the application of Debye-Hückel limiting law.

Partial molal properties mathematically defined as:

                                       T,p,nj   =   Ji                                                          (19)

where Ji is the partial molal properties of the i-th component. Physically Ji means an increase in the amount of thermodynamic J observed when one mole of i is added to a system that large that its composition remains constant (Dogra and Dogra, 1990).

Partial molar volume of the components of a mixture are variable depending on the composition, because each type of molecule the change if the composition changes from A to B pure pure. Molecular environmental changes and changes in working styles between molecules that produces variations thermodynamic properties of the mixture if the composition changes.

Firstly, to get density of solution, equation that must be done is :

d =  x do                                                                       (20)

where :

We  = weight of empty pycnometer (g)

Wo   = weight of pycnometer + aquadest (g)

W     = weight of pycnometer + solution (g)                   

do    = density of aquadest =

d      = density of solution

then, we can found molality by equation :

m   =                                                                          (21)

so, the partial molal volume equation becomes :

=   ( BM –  )                                             (22)

                                                                                                     (Atkins, 1994).

  1. C.     THE EQUIPMENT AND CHEMICAL SUBSTANCE USED
  2. Equipments:
  • Pycnometer 25 mL
  • Measured Flask 50 mL
  • Erlenmeyer 250 mL
  • Beaker Glass 100 mL and 250 mL
  • Ball pipette
  • Volume Pipette 25 mL, 10 mL, 5 mL, 2 mL
  • Measurement pipette 2 mL
  • Stirrer
  • Thermometer
  1. Chemical substance:
  • NaCl 3 M dilute with concentrations 1.5 M, 0.75 M, 0.375M, 0.1875 M
  • Aquadest

 

  1. D.     PROCEDURE
                               
 

Ready for 6 erlenmeyer:

that contains NaCl solution with concentrations of 3.0 M, 1.5 M, 0.75 M, 0.375M, 0.1875 M

 

 

 

Weight the pycnometer filled with distilled water (W0)

 

 

   

Weight the empty pycnometer (We)

 

 

     
         
 
 
 
 
 
     
 
   

Weight the pycnometer filled with a solution of NaCl(W) for each concentrations

 

 

   

Record the temperature inside the pycnometer

 

 

       
 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. E.     OBSERVATION DATA

T      =   27 oC = 300 K

We   = weight of empty pycnometer (g)                          =   20.5558 gram

Wo   = weight of pycnometer + aquadest (g)                  =   45.8768 gram       

do  = density of aquadest =   =  1.01284 g/cm3

Table 1. Data of Pycnometer Weight

No

Concentration

(M)

We

Wo

W

W-Wo

W-We

Wo-We

1

3.0

20.5558

45.8768

47.542

1.6652

26.9862

 

 

25.321

 

2

1.5

46.9968

1.12

26.441

3

0.75

46.1702

0.2934

25.6144

4

0.375

45.9927

0.1159

25.4369

5

0.1875

45.8948

0.018

25.339

 

  1. F.     RESULT AND DISCUSSION

In this experiment we conducted experiments to determine the partial molar volume. Partial molal volume is a volume in which there is a comparison between the solvent with the solute, which is determined by the number of moles of solute substances contained in 1000 grams of solvent. The purpose of the experiments conducted is to determine the partial molar volume of the solution components. This experiment uses material NaCl and distilled water, NaCl serves as the solute and distilled water as a solvent.

Before trial Pycnometer weighed first. Pycnometer is a tool used to measure the density of the solution. Pycnometer empty weight of 20.5558 grams and heavy when filled with distilled water at 45.8768 grams. Pretreatment is a dilute solution of NaCl 3.0 become 1.5 M, 0.75 M, 0.375 M, 0.1875 M. Each solution is included in Pycnometer, then Pycnometer NaCl is fully weighed and measured temperature. Each solution in Pycnometer weighed directly proportional to the concentration of NaCl, if the smaller the concentration of NaCl it weighs less too. In addition, the density of NaCl is greater than the density of distilled water. However, by and large, the changes do not affect the concentration of NaCl solution temperature. NaCl density obtained from the product of the density of distilled water by weight solution of NaCl (weight Pycnometer containing NaCl Pycnometer minus blank) is then divided by the weight of distilled water (containing distilled water Pycnometer weight minus empty weight  Pycnometer.

The difference in the concentration of NaCl solution produce different  density anyway. The higher the concentration of the solution, its density will also increase. This is because the higher the concentration of a solution, showing the number of particles in the solution more and more. In other words, the concentration of a solution is directly proportional to the density of the solution. Partial molal volume is strongly influenced by the concentration of the solution. The higher the concentration the higher the partial molal volume anyway or in other words is proportionally (solute).

Molal volume of the solution components can be measured by dividing the total volume of the solution to the number of moles of the components of the solution.

The equation above shows that the relationship between the partial molal volume is inversely proportional to the molarity. The concentration of a substance greatly affect weight Pycnometer that will be weighed. The higher the concentration the more severe the Pycnometer. This can happen because the drafters of the NaCl concentration of NaCl contains more substance than water so that the weight becomes larger, we all know that NaCl is a solid that is made into a solution, NaCl has a higher molecular weight than water (solvent).

In weighing Pycnometer, carried a small concentration of the solution to a large concentration. This is done so that the weight of the weighed to a small concentration is not affected by the large concentration. Large concentrations can affect small concentration turned out to be somewhat greater, although not the same. But small does not affect the concentration of great concentration. This is done because Pycnometer used only 1 piece, so avoid large errors in the experiment.

After data analysis and graphing  vs √m proven that if value of m is high so the value of   high too. After we get price of  then be looking for a V1 or partial molal volume of the solvent and the price of V2 or partial molal volume of solute. From the graph between  vs V1 and  vs V2 obtained the price that the greater the price of m, the greater the price of V2 too, but otherwise the greater the price of m, the smaller the price of V1.

 

 

 

 
   

Figure 1. The graph of The Graph of F vs Öm

 

Figure 2. The Graph of V2 vs m

 

 

 

 

 

 

 

 

 

 

 
   

Figure 3. The Graph of V1 vs m

 

Dilution performed to observe how large the effect of adding the volume of water that occurs in a wide variety of solution concentration. Thus it would be known how much influence the concentration of the solution of the partial molal volume of the solution.

The temperature of all solution is the same 27 oC. We were used only one pycnometer in order to minimize errors and calculation. Partial molal volume is strongly influenced by the concentration of the solution. The higher the concentration the higher the partial molal volume as well or in other words proportional.

 

  1. G.    CONCLUSION AND SUGGESTION
  2. Conclusion
    1. The concentration of a solution is proportional to the density of the solution.
    2. Partial molal volume is the volume in which there is a comparison between the solvent and solute.
    3. Partial molal volume is strongly influenced by the concentration of the solution. The higher the concentration the higher the partial molal volume as well or in other words proportional.
    4. The partial volume of this experiment is 2.112 that can be shown at the graph.
    5. Suggestion
      1. Students should understand the working principles first before doing practical activities in the laboratory to minimize errors.
      2. Before and after practicum, these tools preferably be clean already.
      3. Accuracy and precision greatly affects the outcome of the observation.

 

  1. H.     REFERENCES

Atkins, P. W.. 1994. Kimia Fisika. Jakarta: Erlangga.

Castellan, Gilbert W. 1983. Physical Chemistry. 3rd Ed.  Addison-Wesley.

Dogra, S. K. dan Dogra, S. 1990. Kimia Fisik dan Soal-Soal. Jakarta: Universitas Indonesia.

Wahyuni, Sri and Team Lecturer of Physical Chemistry. 2011. Practicum Guide of Physical Chemistry. Semarang: UNNES.

 

 

                                                                                                Semarang, 6 November  2012

            knowing,

            Practicum Lecture                                                                  Practitioner,

 

 

 

           Ir. Sri Wahyuni, M.Si                                                               Falasifah Aulia

          NIP.                                                                                                 4301410044

 

 


 

APPENDIXS

 

  1. 1.    Calculation of Solution Density

Concentration of NaCl (M)

W-Wo

Density (g/mL)

W-We

Wo-We

do

3.0 1.6652 1.0794 26.9862

 

 

25.321

1.0128

 

1.5 1.12 1.0576 26.441
0.75 0.2934 1.0246 25.6144
0.375 0.1159 1.0175 25.4369
0.1875 0.018 1.0136 25.339

 

d          = x do

d1         =  x 1.0128    = 1.0794 g/cm3

d2         =  x 1.0128      = 1.0576 g/cm3

d3         =  x 1.0128    = 1.0246 g/cm3

d4         =  x 1.0128    = 1.0175 g/cm3

d5         =  x 1.0128      = 1.0136 g/cm3

 

  1. 2.    Calculation of Solution Molality

m   =         

m1       =          =  3.3188 mmol/g

m2        =            = 1.5466 mmol/g

m3        =          = 0.7648 mmol/g

m4        =          = 0.3767 mmol/g

m5        =           = 0.1870 mmol/g

 

  1. 3.    Calculation of Partial Molal Volume of each Concentration

W-Wo

Density g/mL

1/d

(mL/g)

Wo-We

Molality (mmol/g)

1.6652 1.0794 0.9264

25.321

3.3188

1.12 1.0576 0.9455

1.5466

0.2934 1.0246 0.9760

0.7648

0.1159 1.0175 0.9828

0.3767

0.018 1.0136 0.9866

0.1870

  =   ( BM –  )

1     =            58.5 M/mol –

= 33.8866 cm3/mol

2     =            58.5 M/mol –

= 27.4308 cm3/mol

3     =            58.5 M/mol –

= 42.0177 cm3/mol

4     =            58.5 M/mol –

= 45.4989 cm3/mol

5     =            58.5 M/mol –

= 53.9768 cm3/mol

 

  1. 4.    Calculation of V2

m

F

Öm

Fregretion

F0

V2

V01

V1

3.3188

33.8866

1.8218

2.1685

-1.68

4.0922

17.78639

17.6714

1.5466

27.4308

1.2436

2.1506

-1.68

2.2606

17.78639

17.7498

0.7648

42.0177

0.8745

2.1391

-1.68

1.0913

17.78639

17.7737

0.3767

45.4989

0.6137

2.1310

-1.68

0.2652

17.78639

17.7820

0.1870

53.9768

0.4325

2.1254

-1.68

-0.3091

17.78639

17.7849

 

Slope               = 2.112

Intercept          = 0.031

 

 = 2.16685 – (1.8218 x 2.112) = -1.68

 

      = 4.0922 mL

            = 2.2606 mL

           = 1.0913 mL

           = 0.2652 mL

           = -0.3091 mL

 

  1. 5.    Calculation of V1

 

 

  =

= 17.78639 mL

 

          = 17.6714 mL

          = 17.7498 mL

          = 17.7737 mL

          = 17.7820 mL

          = 17.7849 mL

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