Question

For Mo $\mathrm{K}\alpha$radiation $\left(0.711\stackrel{\mathrm{o}}{\mathrm{A}}\right)$the mass absorption coefficients for K, I, H, and O are $16.7,39.2,0.0,\mathrm{and}1.50{\mathrm{cm}}^2/\mathrm{g},$respectively.

(a) Calculate the mass absorption coefficient for a solution prepared by mixing $11.00\mathrm{g}$of KI with $89.00\mathrm{g}$of water.

(b) The density of the solution described in (a) is 1.086 g/cm³. What fraction of the radiation from a Mo $\mathrm{K}\alpha$source would be transmitted by a $0.40-\mathrm{cm}$layer of the solution?

Short answer

a. The mass absorption coefficient for a solution prepared by mixing 11.00 g of KI with 89.00 g of water is$4.92{\mathrm{cm}}^2/\mathrm{g}$.

b. Fraction of the radiation from a Mo source would be transmitted by a 0.40-cm layer of the solution is 0.118.

Step-by-step solution

Part (a) Step 1. Given information

For Mo $\mathrm{K}\alpha$radiation $\left(0.711\stackrel{\mathrm{o}}{\mathrm{A}}\right)$the mass absorption coefficients for K, I, H, and O are $16.7,39.2,0.0,$and $1.50{\mathrm{cm}}^2/\mathrm{g}$, respectively.

Determine the mass absorption coefficient for a solution prepared by mixing 11.00 g of KI with 89.00 g of water.

Part (a) Step 2. Explanation 

The mass absorption coefficient is the absorption cross-section per unit mass. The mass absorption coefficient is independent of the chemical and physical nature of elements. The mass absorption coefficient of a sample can be calculated by adding the contributions of the mass absorption coefficients of each element present in the sample.

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

Where,

$W_A$is the weight fraction of element A, and ${\mu }_A$is the mass absorption coefficient of element A.

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

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

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.

$\mathrm{Weight}\mathrm{fraction}=\frac{\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{element}}{\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{comound}}\times \mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{compound}\times \frac{1}{\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{sample}}$

Atomic/molar masses:

$$\begin{gathered} \mathrm{K}=39.10\mathrm{g}/\mathrm{mol} \\ \mathrm{I}=126.90\mathrm{g}/\mathrm{mol} \\ \mathrm{H}=1.01\mathrm{g}/\mathrm{mol} \\ \mathrm{O}=16.0\mathrm{g}/\mathrm{mol} \\ \mathrm{KI}=166.0\mathrm{g}/\mathrm{mol} \\ {\mathrm{H}}_2\mathrm{O}=18.02\mathrm{g}/\mathrm{mol} \end{gathered}$$

Part (a) Step 3. Weight fractions of each element in the solution

For I : $$\begin{gathered} \mathrm{Weight}\mathrm{fraction}\left(W_1\right)=\frac{\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{element}}{\mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{comound}}\times \mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{compound}\times \frac{1}{\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{sample}} \\ =\frac{126.90}{166.0}\times 11.0\times \frac{1}{100}=0.841 \end{gathered}$$

Similarly, weight fractions for other elements can be calculated.

$$\begin{gathered} W_K=\frac{39.10}{166.0}\times 11.0\times \frac{1}{100}=0.259 \\ {W}_{\mathrm{O}}=\frac{16.0}{18.02}\times 89.00\times \frac{1}{100}=0.791 \\ {W}_K=\frac{2.020}{18.02}\times 89.00\times \frac{1}{100}=0.099 \end{gathered}$$

Part (a) Step 4. The calculation for mass absorption coefficient 

Now, obtained weight fractions should be utilized to obtain the mass absorption coefficient of the solution as follows,

$$\begin{gathered} {\mu }_M=W_A{\mu }_A+W_B{\mu }_B+W_C{\mu }_C+...... \\ {\mu }_M=W_1{\mu }_1+W_{\mathrm{K}}{\mu }_{\mathrm{K}}+W_O{\mu }_{\mathrm{O}}+W_{\mathrm{H}}{\mu }_{\mathrm{H}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=0.841\times 39.2+0.259\times 16.7+0.791\times 1.5+0.099\times 0.0 \\ =4.92{\mathrm{cm}}^2/\mathrm{g} \end{gathered}$$

Part (b) Step 1. Given information 

The density of the solution described in (a) is 1.086 g/cm³. Calculate the transmitted fraction of the radiation when the Mo $\mathrm{K}\alpha$source radiation is passed through a 0.40 cm layer of the solution.

Part (b) Step 2. Beer’s law 

Beer’s law can be written as follows, in which, the mass absorption coefficient is included.

$$\begin{gathered} \ln \frac{P_o}{P}={\mu }_M\rho x \\ {P}_o=\mathrm{Power}\mathrm{of}\mathrm{incident}\mathrm{beam} \\ P=\mathrm{Power}\mathrm{of}\mathrm{transmittedt}\mathrm{beam} \\ \rho =\mathrm{Density}\mathrm{of}\mathrm{the}\mathrm{sample} \\ x=\mathrm{Sample}\mathrm{thickness}\mathrm{in}\mathrm{cm} \end{gathered}$$

Part (b) Step 3. Transmittance 

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

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

Therefore, transmittance can be rewritten as:

$\frac{P}{P_o}=\frac{1}{e^{{\mu }_M\rho x}}$

Part (b) Step 4. Fraction of the radiation 

Once the mass absorption coefficient is obtained, T (transmitted fraction of radiation) can be obtained by plugging the numerical values in the below equation.

$$\begin{gathered} \frac{P}{P_o}=\frac{1}{e^{{\mu }_M\rho x}} \\ =\frac{1}{e^{\left(4.92\times 1.086\times 0.40\right)}} \\ =0.118 \end{gathered}$$