Chapter 12, Problem 12.6QAP

Interpretation Introduction

(a)

Interpretation:

Mass absorption coefficient of a solution containing $11.00$ g of KI and $89.00$ g of water should be calculated.

Concept introduction:

Mass absorption coefficient ( ${\mu }_M$ ) is the absorption cross section per unit mass (units are ${\text{cm}}^{\text{2}}\text{/g)}\text{.}$ Mass absorption coefficient is independent of chemical and physical nature of elements.

Mass absorption coefficient ( ${\mu }_M$ ) of a sample can be calculated by adding the contributions of the mass absorption coefficients of each element present in the sample.

Contribution of “element A” to the mass absorption coefficient of a sample can be written as $W_A{\mu }_A$

Where, $W_A=\text{ weight fraction of the element A and }{\mu }_A=\text{mass absorption coefficient of the element}$

Therefore, the mass absorption coefficient ( ${\mu }_M$ ) of a sample can be written as,

${\mu }_M=W_A{\mu }_A+W_B{\mu }_B+W_C{\mu }_C+\mathrm{...}$

In order to calculate mass fractions of each element in a sample, the atomic mass of each element and the molar mass of each compound should be known.

$\text{Weight fraction (}{W}_A\text{) = }\frac{\text{atomic mass of the element}}{\text{molar mass of the compound}}\times \text{ weight of the compound}\times \frac{\text{1}}{\text{weight of the sample}}$

$\begin{array}{l}\begin{array}{l}\text{Atomic/molar masses of }\\ \text{K =39}\text{.10 g/mol}\end{array}\hfill \\ \text{I =126}\text{.90 g/mol}\hfill \\ \text{K =39}\text{.10 g/mol}\hfill \\ \text{H =1}\text{.01 g/mol}\hfill \\ \text{O =16}\text{.0 g/mol}\hfill \\ \text{KI =166}\text{.0 g/mol}\hfill \\ {\text{H}}_{\text{2}}\text{O =18}\text{.02 g/mol}\hfill \end{array}$

Interpretation Introduction

(b)

Interpretation:

Transmitted fraction of the radiation when the Mo $K\alpha$ source radiation is passed through a $0.40\text{ cm}$ layer of the solution should be calculated.

Concept introduction:

Beer’s law can be written as follows, in which, mass absorption coefficient ( ${\mu }_M$ ) is included.

$\begin{array}{l}\ln \frac{p_o}{p}={\mu }_M\rho x\\ P_o\text{= power of incident beam}\\ p=\text{ power of transmitted beam}\\ \rho \text{ = density of the sample}\\ x=\text{ sample thickness in cm}\end{array}$

Transmittance (T) of a medium is the fraction of incident radiation transmitted by the medium.

$\text{T=}\frac{P}{P_o}$

Therefore, transmittance can be rewritten as:

$\frac{p}{p_o}=\frac{1}{e^{{\mu }_M\rho x}}$