Use the Beer-Lambert relationship in quantitative spectrophotometric analysis. Calculate the following (express your answer to 3 significant figures): (a) The concentration \(\left(\mu \mathrm{g} \mathrm{mL}^{-1}\right)\) of pentachlorophenol in a test solution giving an absorbance at \(300 \mathrm{~nm}\left(A_{300}\right)\) of \(0.57\) in a cuvette of path length \(5 \mathrm{~mm}\), based on an absorptivity of \(20 \mathrm{~L} \mathrm{~g}^{-1} \mathrm{~cm}^{-1}\). (b) The amount (ng) of pentachlorophenol in a \(50 \mu \mathrm{L}\) sub-sample from a test solution where \(A_{300}=0.31\) in a cuvette of path length \(1 \mathrm{~cm}\), based on an absorptivity of \(20 \mathrm{~L} \mathrm{~g}^{-1} \mathrm{~cm}^{-1}\).

Step by step solution

01

Calculate the concentration of pentachlorophenol

Rearrange the Beer-Lambert equation to solve for the concentration, \(c = \frac{A}{\varepsilon \times l}\). Substituting the given values, we get \(c = \frac{0.57}{20 \times 0.5} = 0.057 \mu g mL^{-1}\).

02

Calculate the amount of pentachlorophenol in the sub-sample

First calculate the concentration of the sub-sample similarly, \(c = \frac{0.31}{20 \times 1} = 0.0155 \mu g mL^{-1}\). Since we know 1 mL = 1000 \mu L, the test solution volume in L is \(50 \mu L \times \frac{1 mL}{1000 \mu L} = 0.05 mL = 0.00005 L\). The amount of pentachlorophenol in the sample is then found by multiplying the volume (in L) by the concentration (in \mu g/mL), and converting to ng: \(0.0155 \mu g/ mL \times 0.00005 L \times \frac{1 mL}{1 L} \times \frac{1000 ng}{1 \mu g} = 0.775 ng\)

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