Question

6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/(mol cm). The depth of the cell is 5 mm. (b) What is the %T? (7) Calculate the absorbance of the solution if the transmitted light intensity is 70% of the initial light beam intensity

Answer 1

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

$A = \epsilon cl\\A = \text{35.9 L&\cdot mol ^{-1} cm ^{-1} } \times 0.0278 mol \cdot L ^{-1} \times 0.5 cm = \mathbf{0.499}$

(b) Percent transmission

$A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}$

7) Absorbance

$A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}$