Chapter 13, Problem 13.14QAP

Interpretation Introduction

(a)

Interpretation:

Percent transmittance of the sample solution should be calculated.

Concept introduction:

$\begin{array}{l}\text{Transmittance (T) of a medium is the fraction of incident radiation }\\ \text{transmitted by the medium}\text{.}\\ \text{T=}\frac{P}{P_o}\\ \text{Where, }{P}_o\text{= power of incident beam and }P=\text{ power of transmitted beam}\\ \text{Percent Transmittance (T\%) = }\frac{P}{P_o}\times 100\%\end{array}$

Interpretation Introduction

(b)

Interpretation:

Absorbance of the sample solution should be calculated.

Concept introduction:

The relationship between absorbance ( $\text{A}$ ) and Transmittance (T) is as follows;

$\text{A = }-\text{ log(T)}$

Interpretation Introduction

(c)

Interpretation:

Transmittance of a solution having one third of absorber that of the original sample should be calculated.

Concept introduction:

Beer’s law states the relationship between absorbance ( $\text{A}$ ) and the concentration (c) of a sample:

$\begin{array}{l}\text{A = }\epsilon \text{bc}\\ \text{Where,}\\ \epsilon =\text{Molar absorptivity}\\ \text{b = path length of the sample/length of the cell in cm}\\ \text{c = concentration of the sample in moles per liter}\\ \text{Also, }\\ \text{A = }-\log (\text{T})\\ \text{Where, T = Transmittance}\end{array}$

Interpretation Introduction

(d)

Interpretation:

Transmittance of a solution having twice the absorber concentration that of the original sample should be calculated.

Concept introduction:

Beer’s law states the relationship between absorbance ( $\text{A}$ ) and the concentration (c) of a sample:

$\begin{array}{l}\text{A = }\epsilon \text{bc}\\ \text{Where,}\\ \epsilon =\text{Molar absorptivity}\\ \text{b = path length of the sample/length of the cell in cm}\\ \text{c = concentration of the sample in moles per liter}\\ \text{Also, }\\ \text{A = }-\log (\text{T})\\ \text{Where, T = Transmittance}\end{array}$