Certainly! Let's work through the problem step-by-step.
We are given the following information:
- Concentration of the analyte solution, [tex]\(c\)[/tex] = [tex]\(10^{-4} \, M\)[/tex]
- Path length of the sample cell, [tex]\(l\)[/tex] = [tex]\(1 \, \text{cm}\)[/tex]
- Absorbance of the solution, [tex]\(A\)[/tex] = [tex]\(0.139\)[/tex]
We need to determine the molar absorptivity, [tex]\(\varepsilon\)[/tex], of the analyte at 350 nm. To do this, we will use Beer's Law, which is expressed by the formula:
[tex]\[ A = \varepsilon \cdot c \cdot l \][/tex]
Where:
- [tex]\(A\)[/tex] is the absorbance,
- [tex]\(\varepsilon\)[/tex] is the molar absorptivity (L·mol[tex]\(^{-1}\)[/tex]·cm[tex]\(^{-1}\)[/tex]),
- [tex]\(c\)[/tex] is the concentration (mol/L),
- [tex]\(l\)[/tex] is the path length (cm).
Rearranging the formula to solve for molar absorptivity, [tex]\(\varepsilon\)[/tex]:
[tex]\[ \varepsilon = \frac{A}{c \cdot l} \][/tex]
Now, let's substitute the given values into the formula:
[tex]\[ \varepsilon = \frac{0.139}{(10^{-4} \, \text{M}) \cdot (1 \, \text{cm})} \][/tex]
[tex]\[ \varepsilon = \frac{0.139}{10^{-4}} \][/tex]
[tex]\[ \varepsilon = 0.139 \times 10^{4} \][/tex]
[tex]\[ \varepsilon = 1390.0 \, \text{L} \cdot \text{mol}^{-1} \cdot \text{cm}^{-1} \][/tex]
So, the molar absorptivity of the analyte at 350 nm is:
[tex]\[ \varepsilon = 1390.0 \, \text{L} \cdot \text{mol}^{-1} \cdot \text{cm}^{-1} \][/tex]
This concludes our calculation.
