1. The problem is to calculate the absorbance $E$ using the Lambert-Beer law given by the formula: $$E = \varepsilon \cdot c \cdot l$$ where $\varepsilon$ is the molar absorptivity, $c$ is the concentration, and $l$ is the path length. 2. From the problem, we have: - $E = 1$ - $\varepsilon = 2.35 \times 10^{4}$ L/(mol·cm) - $l = 1$ cm 3. We need to find the concentration $c$: $$c = \frac{E}{\varepsilon \cdot l} = \frac{1}{2.35 \times 10^{4} \times 1} = 4.255 \times 10^{-5} \text{ mol/L}$$ 4. To find the mass concentration in g/L, multiply by the molar mass (452.36 g/mol): $$\text{mass concentration} = c \times \text{molar mass} = 4.255 \times 10^{-5} \times 452.36 = 0.01924 \text{ g/L}$$ 5. Final answer: The concentration is approximately $0.0192$ g/L.