feedback to help with question - Most of you managed this well once you realised that for Henry’s Law, the temperature should be approximated as the boiling temperature of the bulk component. This allowed the pure component vapour pressure of the dilute component to be evaluated. For the comparison with the x-y diagram, the determination of the initial gradient was approximate, but the value you found should certainly have corresponded more closely to the non-ideal Henry’s Law constant. This comment applies both to part (a) and part (b). ans = a) (i) 3.48 atm; (ii) 7.21 atm, b) (i) 0.242 atm; (ii) 0.449 atm Data:Antoine coefficients (P in mmHg, Tin K, log to base e):Margules parameters:1. a) Calculate the Henry's Law constant for dilute methanol in water at 1 atmosphere:i) assuming ideal solution behaviour;ii) using the Margules Coefficients provided to account for the non-ideality of the solution.Compare your answers with the value roughly obtained from the gradient in the x-y diagram below.Ym (molefraction of methanol in vapour)b) Now repeat part a) but this time to determine the Henry's Law constant for dilute water inmethanol at p=1 atm. Again, compare your results to the constant roughly obtained from the x-ydiagram.0.80.60.4Methanol: A = 18.588, B = 3626.6, C = -34.29Water: A = 18.304, B = 3816.4, C = -46.13Ethanol: A = 18.9119, B = 3803.98, C = -41.680.20Methanol-Water: Awm = 0.6174, Amw = 0.7279Ethanol-Water: Awe = 0.7917, Aew = 1.63660x-y diagram for methanol/water at 1 atm0.20.4Xm (molefraction of methanol in liquid)0.60.81

To calculate the Henry's Law constant for dilute methanol in water at 1 atmosphere, we can use the…

Answered: Data: Antoine coefficients (P in mmHg,…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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feedback to help with question - Most of you managed this well once you realised that for Henry’s Law, the temperature should be approximated as the boiling temperature of the bulk component. This allowed the pure component vapour pressure of the dilute component to be evaluated. For the comparison with the x-y diagram, the determination of the initial gradient was approximate, but the value you found should certainly have corresponded more closely to the non-ideal Henry’s Law constant. This comment applies both to part (a) and part (b).

ans = a) (i) 3.48 atm; (ii) 7.21 atm, b) (i) 0.242 atm; (ii) 0.449 atm

Data:
Antoine coefficients (P in mmHg, Tin K, log to base e):
Margules parameters:
1. a) Calculate the Henry's Law constant for dilute methanol in water at 1 atmosphere:
i) assuming ideal solution behaviour;
ii) using the Margules Coefficients provided to account for the non-ideality of the solution.
Compare your answers with the value roughly obtained from the gradient in the x-y diagram below.
Ym (molefraction of methanol in vapour)
b) Now repeat part a) but this time to determine the Henry's Law constant for dilute water in
methanol at p=1 atm. Again, compare your results to the constant roughly obtained from the x-y
diagram.
0.8
0.6
0.4
Methanol: A = 18.588, B = 3626.6, C = -34.29
Water: A = 18.304, B = 3816.4, C = -46.13
Ethanol: A = 18.9119, B = 3803.98, C = -41.68
0.2
0
Methanol-Water: Awm = 0.6174, Amw = 0.7279
Ethanol-Water: Awe = 0.7917, Aew = 1.6366
0
x-y diagram for methanol/water at 1 atm
0.2
0.4
Xm (molefraction of methanol in liquid)
0.6
0.8
1

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