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Determination of Partial Molar Quantities Experiment (((((((( Solve the questions on page 2 in detail )))))))) we send it twice and you gave us random

Determination of Partial Molar Quantities Experiment

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Aim The experiment aims to determine the density and specific volumes of non-ideal solutions, and partial molar volumes of their components. Theory Thermodynamics deals with macroscopic properties of a body that can be classified in two groups as intensive and extensive properties. The intensive properties are those that do not depend on the system size and might vary from point to point. Such properties are not additive. Examples include temperature, pressure and density. Extensive propertics, like mass and volume, on the other hand, are additive for subsystems. "Depare, on mas.". Partial molar quantities are key elements in the thermodynamics of solutions. The most important of these is the partial molar free energy or chemical potential. However, partial molar volume is the easiest to visualize and measure. The partial molar volumes of the components of a mixture containing with composition because the environment and the interaction of each type of molecule changes as the composition changes. The partial molar volume, Vj, of a substance J at some general composition is defined formally as follows: Vj=(njV)P,T,n=j where the amounts of all other substances present are constant as well as the pressure and temperature. The final volume of a mixture with components A and B can be calculated by integration since the composition and partial molar volumes are constant as the amounts of A and B are increased. V=0nAVAdnA+0nAVBdnB=VAnA+VBnB For ideal solutions, For non-ideal solutions, nTV=nTVAnA+nTVBnB V=VAnA+VBnB It should be noted that quantities relating to pure substances are denoted by superscript * When the equations [3] and [4] are divided by number of moles of solutions, the above equations become Vm=VAxA+VBxB for non-ideal solutions Here, Vm is the molar volume of the solution and xA and xB are the mol fractions of A and B, respectively. In a binary mixture of A and B, if xB is very small, the mixture can be treated as a dilute solution of solvent A and solute B. In this case, as XB approaches zero, VB approaches a certain limiting value that is the volume increase per amount of B mixed with a large amount of pure A. In the resulting mixture, only solvent molecules surround each solute molecule. Materials and Che micals Balance, pipette, flask, distilled water, cthanol, acetic acid Procedure 1. Prepare 50ml acetic acid and ethanol solutions with 20%,40%,60% and 80%w/w. Do necessary calculations beforehand. 2. Weigh empty flask. =27.4522 ? 3. Fill the flask completely with distilled water. Make sure the water is free of any air bubbles and the outer surface of the flask is completely dry. Weigh the flask. 68.92g 4. Empty the flask. 5. Fill the flask completely with the prepared solutions successively and record the weights. Wash and dry the flask prior to each specific measurement. 30j2 whash and dry the Report Objectives 1. Calculate the specific volume (ml/g) of each solution and plot the specific volume vs. percentage weight of solutions. 2. Draw a smooth curve through points, and draw tangents to the curve at different concentrations. The intercepts of these lines on 0% and 100% ordinates give the partial specific volumes of the components respectively at various concentrations. 3. Calculate the partial molar volume (mV/gmol) of components in solution, from partial specific volume (ml/g) and molecular weight (g/gmol). 4. Calculate the molar volumes of the solutions at different concentrations. 5. Calculate the specific volume of the solutions using partial specific volumes and weight fractions of components within the solution. Compare the results with the experimental data. 6. Compare the results obtained for ethanol and acetic acid solutions. Comment on the differences. The recorded values: (Calculations will be done for Ethanol) - Volume for the empty flask =41.47ml. For 20\% ethanol: - Weight of the solution =40.165g - Volume of the solution =41.47ml - Density of the solution =0.968g/ml For 40% ethanol: - Weight of the solution =38.85g - Volume of the solution =41.47ml - Density of the solution =0.936g/ml For 60% ethanol: - Weight of the solution =37.05g - Volume of the solution =41.47ml - Density of the solution =0.893g/ml For 80% ethanol: - Weight of the solution =35.53g - Volume of the solution =41.47ml - Density of the solution =0.856g/ml For 100% ethanol: - Weight of the solution =33.58g - Volume of the solution =41.47ml - Density of the solution =0.809g/ml

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