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; 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) 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 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 Methanol-Water: Awm = 0.6174, Amw = 0.7279 Ethanol-Water: Awe = 0.7917, Aew = 1.6366 0.4 0.2 x-y diagram for methanol/water at 1 atm 0.2 0.6 0.4 Xm (molefraction of methanol in liquid) 0.8
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; 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) 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 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 Methanol-Water: Awm = 0.6174, Amw = 0.7279 Ethanol-Water: Awe = 0.7917, Aew = 1.6366 0.4 0.2 x-y diagram for methanol/water at 1 atm 0.2 0.6 0.4 Xm (molefraction of methanol in liquid) 0.8
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
ChapterA1: Evaluation Of Analytical Data
Section: Chapter Questions
Problem A1.22QAP
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Feedback - Q1
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).
Answer to Q1: a) (i) 3.48 atm; (ii) 7.21 atm, b) (i) 0.242 atm; (ii) 0.449 atm
PART 1a) (ii) PLEASE STEPS ON HOW TO REACH ANSWER
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