Question

The percentage transmittance of light at 250 nm through a certain aqueous solution is 20.0% at 25°C. The experimental cell length is 1.0 cm, and the concentration of the solution is. Calculate the absorbance. Calculate the molar absorption coefficient.

Short answer

The absorbance and molar absorption coefficient were calculated as 0.698,$1397.94{\text{Lmol}}^{-1}{\text{cm}}^{-1}$ respectively.

Step-by-step solution

Definitions of Absorption and Transmittance

Absorption: The amount of incident radiation absorbed in the solution is referred to as absorption. Absorption is also known as optical density. The absorption is calculated using the Lambert-Beer law.

Transmittance: The amount of incident radiation transmitted through the solution is known as transmittance.

Explanation Absorption and Transmittance

The absorbance and transmittance are linked together through the Lambert-Beer law. Absorbance is directly proportional to the concentration of the solution and cell length. After removing the proportionality sign the constant value added to the solution is later known as the molar absorption coefficient.

Calculations for absorbance and molar absorption coefficient

Lambert-Beer law is stated in form of an equation.

$A=\log \left(\frac{I_0}{I}\right)=\epsilon cl$, where A: absorbance, I_˳is the incident intensity of radiation, I is transmitted intensity of radiation, ε is the molar absorption coefficient, c is the concentration of the solution, and l is the length of the cell.

First, calculate absorbance:

$\begin{array}{rcl}\text{\% Transmission(T)}& =& \frac{\text{I}}{{\text{I}}_{\circ }},\\ & & \text{T =}\frac{20}{100}\\ & =& \frac{\text{I}}{{\text{I}}_{\circ }}\end{array}$

Solve further.

$$\begin{gathered} \text{A}=\log \left(\frac{{\text{I}}_{\circ }}{\text{I}}\right) \\ =\text{log}\left(\frac{100}{20}\right) \\ =\text{log}\left(5\right) \\ =0.698 \end{gathered}$$

Calculate for Molar absorption coefficient:

$$\begin{gathered} A=\epsilon cl \\ 0.698=\epsilon 5\times {10}^{-4}\times 1 \\ \Rightarrow \epsilon =1397.94{\text{Lmol}}^{-1}{\text{cm}}^{-1} \end{gathered}$$