Experimental data for the excess molar volume of 1-propanol (1) + 1-hexene (2) system at 298.15K is provided in Table P9-29. The data range is from 0.0559 to 0.9490 for x 1 . If the pure componentdensities at 298.15 K are 0.79965 g/cm 3 for 1-propanol and 0.66828 g/cm 3 for 1-hexene, performthe following operations:A) Plot the solution molar volume as a function of composition of 1-propanol.B) Fit the solution molar volume to an appropriate functional form.C) Determine the partial molar volume for both substances when the system is equimolar.Table P9-29: Excess molar volume of 1-propanol (1) + 1-hexene (2) system at 298.15K(Treszczanowicz, 2010)x 1 V E (cm 3 /mol) x 1 V E (cm 3 /mol)0.0559 0.0747 0.5009 –0.0220.1058 0.0994 0.5518 –0.0520.1992 0.1002 0.6016 –0.06970.1507 0.1068 0.7038 –0.09510.2544 0.0863 0.8088 –0.10260.3009 0.0722 0.8497 –0.0950.352 0.0528 0.8990 –0.07090.3964 0.0317 0.9490 –0.0383Hints:a) The pure component molar volumes are not part of the data set directly, but the pure componentdensities are provided. Thus, it is necessary to convert the density to a molar volume. Use thefollowing data:MW of 1-propanol = 60.1 g/molMW of 1-hexene = 84.16 g/molb) The molar volume of the mixture is then calculated from the excess property data using thefollowing equation:푉 = 푉 퐸 + ( 푥 1 푉 1 + 푥 2 푉 2 )Using this, regenerate the data table in terms of V and carry on the steps discussed in class.
Experimental data for the excess molar volume of 1-propanol (1) + 1-hexene (2) system at 298.15
K is provided in Table P9-29. The data range is from 0.0559 to 0.9490 for x 1 . If the pure component
densities at 298.15 K are 0.79965 g/cm 3 for 1-propanol and 0.66828 g/cm 3 for 1-hexene, perform
the following operations:
A) Plot the solution molar volume as a function of composition of 1-propanol.
B) Fit the solution molar volume to an appropriate functional form.
C) Determine the partial molar volume for both substances when the system is equimolar.
Table P9-29: Excess molar volume of 1-propanol (1) + 1-hexene (2) system at 298.15K
(Treszczanowicz, 2010)
x 1 V E (cm 3 /mol) x 1 V E (cm 3 /mol)
0.0559 0.0747 0.5009 –0.022
0.1058 0.0994 0.5518 –0.052
0.1992 0.1002 0.6016 –0.0697
0.1507 0.1068 0.7038 –0.0951
0.2544 0.0863 0.8088 –0.1026
0.3009 0.0722 0.8497 –0.095
0.352 0.0528 0.8990 –0.0709
0.3964 0.0317 0.9490 –0.0383
Hints:
a) The pure component molar volumes are not part of the data set directly, but the pure component
densities are provided. Thus, it is necessary to convert the density to a molar volume. Use the
following data:
MW of 1-propanol = 60.1 g/mol
MW of 1-hexene = 84.16 g/mol
b) The molar volume of the mixture is then calculated from the excess property data using the
following equation:
푉 = 푉 퐸 + ( 푥 1 푉 1 + 푥 2 푉 2 )
Using this, regenerate the data table in terms of V and carry on the steps discussed in class.
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