Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198727873
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 11, Problem 11A.4E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient and absorbance of the aqueous solution containing amino acid tyrosine has to be calculated where transmittance is 0.22 and cell length is 1.5 mm.

Concept Introduction:

  • According to Beer–Lambert law, intensity of radiation absorbed at a particular frequency depends on number of species absorbed (N), molar concentration [J] and the path length, L.

  I = I010ε[J]Lwhere,ε = Molar absorption coefficient I = Intensity of incident radiationI0 = Intensity of transmitted radiation

  • Equation for transmittance, T can be given as,

   T= II0T = 10ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

  • Equation for absorbance, A can be given as,

  A = log TT = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(a)

Expert Solution
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Explanation of Solution

Given information is shown below,

  Molar concentration, [J]= 0.080 mol.dm-3 Transmittance, T = 0.22Wavelength =  268 nm Path length = 1.5 mm

  • Molar absorption coefficient is calculated by the given formula.

  T = 10ε[J]Lε = 10(T)[J]L

Substitute the given values as shown,

  ε = 10(T)[J]L =  10(0.22)( 0.080 mol.dm-3)(1.5 mm)= 5.48 dm3.mol1.mm1

Molar absorption coefficient is 5.48 dm3.mol1.mm1.

  • Absorbance of the solution is determined as given,

  A = log T = log (0.22)= 0.66

Absorbance of the solution is 0.66.

(b)

Interpretation Introduction

Interpretation:

Transmittance through a cell with a length of 3.0 mm has to be determined.

Concept Introduction:

Equation for absorbance, A can be given as,

  A = log TT = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information is shown below,

  Molar concentration, [J]= 0.080 mol.dm-3 Transmittance, T = 0.22Wavelength =  268 nm Path length = 3.0 mm

Molar absorption coefficient is calculated by the given formula.

  T = 10ε[J]Lε = 10(T)[J]L

Substitute the given values as shown,

  ε = 10(T)[J]L =  10(0.22)( 0.080 mol.dm-3)(1.5 mm)= 5.48 dm3.mol1.mm1

Molar absorption coefficient is 5.48 dm3.mol1.mm1.

Absorbance of radiation through a 1.0 mm cell length is calculated as follows,

  A = ε[J]L(5.48 dm3.mol1.mm1)(0.080 mol.dm-3 )(3.0 mm)= 1.315

Transmittance of radiation through a 1.0 mm cell length is calculated as follows,

A = log TT = 10A = 10(1.315)= 0.048

Transmittance of radiation through a 3.0 mm cell length is 0.048.

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Chapter 11 Solutions

Elements Of Physical Chemistry

Ch. 11 - Prob. 11C.3STCh. 11 - Prob. 11C.4STCh. 11 - Prob. 11C.5STCh. 11 - Prob. 11C.6STCh. 11 - Prob. 11C.7STCh. 11 - Prob. 11D.1STCh. 11 - Prob. 11D.2STCh. 11 - Prob. 11D.3STCh. 11 - Prob. 11E.1STCh. 11 - Prob. 11E.2STCh. 11 - Prob. 11E.3STCh. 11 - Prob. 11E.4STCh. 11 - Prob. 11A.1ECh. 11 - Prob. 11A.2ECh. 11 - Prob. 11A.3ECh. 11 - Prob. 11A.4ECh. 11 - Prob. 11A.5ECh. 11 - Prob. 11A.6ECh. 11 - Prob. 11A.7ECh. 11 - Prob. 11A.8ECh. 11 - Prob. 11B.1ECh. 11 - Prob. 11B.2ECh. 11 - Prob. 11B.3ECh. 11 - Prob. 11B.4ECh. 11 - Prob. 11B.5ECh. 11 - Prob. 11B.6ECh. 11 - Prob. 11B.7ECh. 11 - Prob. 11B.8ECh. 11 - Prob. 11B.9ECh. 11 - Prob. 11B.10ECh. 11 - Prob. 11B.11ECh. 11 - Prob. 11B.12ECh. 11 - Prob. 11B.13ECh. 11 - Prob. 11B.14ECh. 11 - Prob. 11B.15ECh. 11 - Prob. 11B.16ECh. 11 - Prob. 11C.1ECh. 11 - Prob. 11C.2ECh. 11 - Prob. 11C.3ECh. 11 - Prob. 11C.4ECh. 11 - Prob. 11C.5ECh. 11 - Prob. 11C.6ECh. 11 - Prob. 11C.7ECh. 11 - Prob. 11C.8ECh. 11 - Prob. 11C.9ECh. 11 - Prob. 11D.1ECh. 11 - Prob. 11D.2ECh. 11 - Prob. 11D.3ECh. 11 - Prob. 11D.4ECh. 11 - Prob. 11D.5ECh. 11 - Prob. 11D.6ECh. 11 - Prob. 11E.1ECh. 11 - Prob. 11E.2ECh. 11 - Prob. 11E.3ECh. 11 - Prob. 11.1DQCh. 11 - Prob. 11.2DQCh. 11 - Prob. 11.3DQCh. 11 - Prob. 11.4DQCh. 11 - Prob. 11.5DQCh. 11 - Prob. 11.6DQCh. 11 - Prob. 11.7DQCh. 11 - Prob. 11.8DQCh. 11 - Prob. 11.9DQCh. 11 - Prob. 11.10DQCh. 11 - Prob. 11.11DQCh. 11 - Prob. 11.12DQCh. 11 - Prob. 11.13DQCh. 11 - Prob. 11.1PCh. 11 - Prob. 11.2PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Prob. 11.6PCh. 11 - Prob. 11.7PCh. 11 - Prob. 11.8PCh. 11 - Prob. 11.9PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Prob. 11.13PCh. 11 - Prob. 11.14PCh. 11 - Prob. 11.15PCh. 11 - Prob. 11.1PRCh. 11 - Prob. 11.2PRCh. 11 - Prob. 11.3PRCh. 11 - Prob. 11.5PR
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