Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 4, Problem 4.21P

(a)

Interpretation Introduction

Interpretation:

The overall mole fraction of phenol in the given mixture has to be determined.

Concept introduction:

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.  Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(nA+nB+nC

The lever rule is mainly used to determine compositions of phases and the relative proportions of phases to each other in Binary diagrams of compounds and using the lever rule we can determine quantitatively the relative composition of a mixture in a two phase region in a phase diagram.

According to lever rule, we can write,

n=n'+n''

Where,

n': Total amount of molecule in one

n'': Total amount of molecule in other phase

The total amount of A in the sample is nxA where xA it is the overall mole fraction of A in the sample.  The overall amount of A is also the sum of its amounts in the two phases, where it has the mole fractions xA'andxA'' respectively.

Thus,

nxA=n'xA'+n''xA''n'(xA'xA)=n''(xAxA'')

(b)

Interpretation Introduction

Interpretation:

The relative amounts of two phases in the given mixture has to be determined using lever rule.

Concept introduction:

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.  Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  χA=numbersof moles ofmolecule A(nA)total number of moles(nA+nB+nC

The lever rule is mainly used to determine compositions of phases and the relative proportions of phases to each other in Binary diagrams of compounds and using the lever rule we can determine quantitatively the relative composition of a mixture in a two phase region in a phase diagram.

According to lever rule, we can write,

n=n'+n''

Where,

n': Total amount of molecule in one phase

n'': Total amount of molecule in other phase

The total amount of A in the sample is nxA where xA it is the overall mole fraction of A in the sample.  The overall amount of A is also the sum of its amounts in the two phases, where it has the mole fractions xA'andxA'' respectively.

Thus,

  nxA=n'xA'+n''xA''n'(xA'xA)=n''(xAxA'')

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Done 11:14 ⚫ worksheets.beyondlabz.com 5 (a). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 6 5 (b). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 7 6. Are there any differences between the spectra you obtained in Beyond Labz and the predicted spectra? If so, what were the differences? <
2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled** Peak Chemical Shift (d) 5.7 1 Multiplicity multiplate .......... 5.04 double of doublet 2 4.98 double of doublet 3 4.05 doublet of quartet 4 5 LO 3.80 quartet 1.3 doublet 6 Peak Chemical Shift (d) Multiplicity

Chapter 4 Solutions

Elements Of Physical Chemistry

Ch. 4 - Prob. 4D.1STCh. 4 - Prob. 4D.2STCh. 4 - Prob. 4D.3STCh. 4 - Prob. 4D.4STCh. 4 - Prob. 4E.1STCh. 4 - Prob. 4E.2STCh. 4 - Prob. 4F.1STCh. 4 - Prob. 4A.1ECh. 4 - Prob. 4A.2ECh. 4 - Prob. 4A.3ECh. 4 - Prob. 4A.4ECh. 4 - Prob. 4A.5ECh. 4 - Prob. 4A.6ECh. 4 - Prob. 4B.1ECh. 4 - Prob. 4B.2ECh. 4 - Prob. 4B.3ECh. 4 - Prob. 4B.4ECh. 4 - Prob. 4B.5ECh. 4 - Prob. 4B.6ECh. 4 - Prob. 4B.7ECh. 4 - Prob. 4B.8ECh. 4 - Prob. 4C.1ECh. 4 - Prob. 4C.2ECh. 4 - Prob. 4C.3ECh. 4 - Prob. 4C.4ECh. 4 - Prob. 4C.5ECh. 4 - Prob. 4C.6ECh. 4 - Prob. 4C.7ECh. 4 - Prob. 4D.1ECh. 4 - Prob. 4D.2ECh. 4 - Prob. 4D.3ECh. 4 - Prob. 4D.4ECh. 4 - Prob. 4D.5ECh. 4 - Prob. 4D.6ECh. 4 - Prob. 4D.7ECh. 4 - Prob. 4D.8ECh. 4 - Prob. 4D.9ECh. 4 - Prob. 4D.10ECh. 4 - Prob. 4D.11ECh. 4 - Prob. 4D.12ECh. 4 - Prob. 4D.13ECh. 4 - Prob. 4E.1ECh. 4 - Prob. 4E.2ECh. 4 - Prob. 4E.3ECh. 4 - Prob. 4E.4ECh. 4 - Prob. 4F.1ECh. 4 - Prob. 4F.2ECh. 4 - Prob. 4F.3ECh. 4 - Prob. 4.1DQCh. 4 - Prob. 4.2DQCh. 4 - Prob. 4.3DQCh. 4 - Prob. 4.4DQCh. 4 - Prob. 4.5DQCh. 4 - Prob. 4.6DQCh. 4 - Prob. 4.8DQCh. 4 - Prob. 4.9DQCh. 4 - Prob. 4.10DQCh. 4 - Prob. 4.11DQCh. 4 - Prob. 4.12DQCh. 4 - Prob. 4.13DQCh. 4 - Prob. 4.14DQCh. 4 - Prob. 4.15DQCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.1PRCh. 4 - Prob. 4.2PRCh. 4 - Prob. 4.3PRCh. 4 - Prob. 4.4PRCh. 4 - Prob. 4.5PRCh. 4 - Prob. 4.6PR
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY