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 = I 0 10 − ε [ J ] L w h e r e , ε = M o l a r a b s o r p t i o n c o e f f i c i e n t I = I n t e n s i t y o f i n c i d e n t r a d i a t i o n I 0 = I n t e n s i t y o f t r a n s m i t t e d r a d i a t i o n Equation for transmittance, T can be given as, T= I I 0 T = 10 − ε [ J ] L w h e r e , I = I n t e n s i t y o f i n c i d e n t r a d i a t i o n I 0 = I n t e n s i t y o f t r a n s m i t t e d r a d i a t i o n N = N u m b e r o f s p e c i e s a b s o r b e d [ J ] = M o l a r c o n c e n t r a t i o n L = P a t h l e n g t h Equation for absorbance, A can be given as, A = − log T T = ε [ J ] L w h e r e , I = I n t e n s i t y o f i n c i d e n t r a d i a t i o n I 0 = I n t e n s i t y o f t r a n s m i t t e d r a d i a t i o n N = N u m b e r o f s p e c i e s a b s o r b e d [ J ] = M o l a r c o n c e n t r a t i o n L = P a t h l e n g t h

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Answer 1

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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 = I010−ε[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 = 10−ε[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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

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Answer 2

Question

100%

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 = I010−ε[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 = 10−ε[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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

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Answer 3

Question

100%

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 = I010−ε[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 = 10−ε[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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

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Answer 4

Question

100%

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 = I010−ε[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 = 10−ε[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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

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Answer 5

Question

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Answer 6

Question

100%

Answer 7

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 = I010−ε[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 = 10−ε[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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

Answer 8

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 = I010−ε[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 = 10−ε[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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

Absorbance of the solution is determined as given,

A = −log T = −log (0.22)= 0.66

Absorbance of the solution is 0.66.

Answer 9

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 = I010−ε[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 = 10−ε[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

Answer 10

(a)

Expert Solution

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

Absorbance of the solution is determined as given,

A = −log T = −log (0.22)= 0.66

Absorbance of the solution is 0.66.

Answer 11

(a)

Expert Solution

Answer 12

(a)

Expert Solution

Answer 13

Expert Solution

Answer 14

(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

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

Answer 15

(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

Answer 16

(b)

Expert Solution

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 = 10−ε[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.mol−1.mm−1

Molar absorption coefficient is 5.48 dm3.mol−1.mm−1.

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

A = ε[J]L= (5.48 dm3.mol−1.mm−1)(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 = 10−A = 10( −1.315)= 0.048

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

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Ch. 11 - Prob. 11A.1ST Ch. 11 - Prob. 11A.2ST Ch. 11 - Prob. 11A.3ST Ch. 11 - Prob. 11B.1ST Ch. 11 - Prob. 11B.2ST Ch. 11 - Prob. 11B.3ST Ch. 11 - Prob. 11B.4ST Ch. 11 - Prob. 11B.5ST Ch. 11 - Prob. 11C.1ST Ch. 11 - Prob. 11C.2ST

Concept explainers

Explanation of Solution

Explanation of Solution

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Chapter 11 Solutions