Question

The transmittance of a 0.010 M solution of a compound in a 0.100-cm-path length cell is .$\mathbf{T}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{23}\mathbf{\%}$Find the absorbance (A) and the molar absorptive (ε).

Short answer

The value of absorbance is$\approx 1.08$

The value of molar absorptive is$1.08\mathrm{x}1{10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}$

Step-by-step solution

Calculate absorbance:

Equation for absorbance is

$\mathrm{A}=-\log \mathrm{T}$ →(1)

To find transmittance the equation is$\mathrm{T}=\frac{\%\mathrm{T}}{100\%}$

Substitute $8.23\%\mathrm{for}\%$in above equation,

$$\begin{gathered} \mathrm{T}=\frac{\%\mathrm{T}}{100\%} \\ =0.0823 \end{gathered}$$

Substitute the value of transmittance in (1),

$$\begin{gathered} \mathrm{A}=-\log \mathrm{T} \\ \mathrm{A}=-\log \left(0.0823\right) \\ \approx 1.08 \end{gathered}$$

The value of absorbance is$\approx 1.08$

Calculate molar absorptive:

Equation to calculate ε is,

$\mathrm{\epsilon }=\frac{\mathrm{A}}{\mathrm{bc}}$ →(2)

Given,

$$\begin{gathered} \mathrm{c}=0.010\mathrm{M} \\ \mathrm{b}=0.100\mathrm{cm} \\ \mathrm{A}=1.08 \end{gathered}$$

Substitute in equation (2),

$$\begin{gathered} \mathrm{\epsilon }=\frac{1.08}{\left(0.0100\mathrm{cm}\right)\left(0.010\mathrm{M}\right)} \\ \mathrm{\epsilon }=1.08\mathrm{x}{10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

The value of molar absorptive is$1.08\mathrm{x}{10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}$