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Answer :
Final answer:
The absorbance of the solution with the given molar extinction coefficient, concentration, and path length is approximately 0.5, and the percent transmittance is approximately 31.6%. When the transmitted light is 70% of the initial light beam intensity, the absorbance is approximately 0.523.
Explanation:
The subject matter is related to the concept of absorbance in Chemistry. In this specific case, you are required to know and apply the Beer-Lambert law, which states that the absorbance (A) of a solution is directly proportional to its concentration (c) and the path length (l). The formula used is A = εcl, where ε is the molar extinction coefficient.
For part (a), plug the given values into the above formula to get: A = 35.9 L/(mol cm) * 0.0278 mol/L * 0.5 cm = 0.4987. So, the absorbance of the solution is approximately 0.5.
Now, for part (b), the percent transmittance (%T) can be calculated using the relationship A = 2 - log(%T). Solving for %T gives: %T = 10^(2-A) = 10^(2-0.4987) = 31.6%. So the %T is approximately 31.6%.
For part (7), the decrease in the intensity of light by 70% means the transmitted light is 30% of the initial intensity. The absorbance in this case can be calculated directly from the formula A = -log(I/I0) where I/I0 = 0.30. Hence, A = -log(0.30) = 0.523. So the absorbance, when the transmitted light intensity is 70% of the initial light beam intensity, is approximately 0.523.
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Rewritten by : Jeany
Answer:
6) (a) 0.499; (b) 31.7 %
7) 0.15
Explanation:
6) (a) Absorbance
Beer's Law is
[tex]A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}[/tex]
(b) Percent transmission
[tex]A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}[/tex]
7) Absorbance
[tex]A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}[/tex]
