Question

A solution containing $3.92mg/100mL$of $A(335g/mol)$has a transmittance of$64.1\%$ in a $1.50-cm$ cell at $425nm$. Calculate the molar absorptivity of $A$ at this wavelength.

Short answer

The molar absorptivity is$1.10\times {10}^3\mathrm{L}{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1}$

Step-by-step solution

Step 1. Given Information

Percent transmittance is$64.1\%$at$425nm$

Step 2. Formula used 

The Beer's law, also known as the Lambert-Beers law, asserts that the absorbance (A) of a solution is proportional to the thickness or path length (b) of the solution and the concentration (c) of the absorbing species for monochromatic light. This law is mathematically expressed as follows when the concentration is in moles per litre (M) and the route length is in centimetres:

$A=\epsilon bc.....\left(1\right)$

is a proportionality constant called molar absorptivity.

$A=-{\log }_{10}\left(T\right)....\left(2\right)$

Step 3. Conversion of concentration

substitute the value of Transmittance in the formula of absorbance,

$$\begin{gathered} A=-{\log }_{10}(0.641) \\ =0.193 \end{gathered}$$

Solution contains $3.92\mathrm{mg}/100\mathrm{mL}$of $\mathrm{A}(335\mathrm{g}/\mathrm{mol})$

Convert it into mol/L as,

$$\begin{gathered} 3.92\mathrm{mg}/100\mathrm{ml}=(3.92\mathrm{mg}/100\mathrm{ml})\left({10}^{-3}\frac{\mathrm{g}}{\mathrm{mg}}\times \frac{1000\mathrm{ml}}{\mathrm{L}}\right)\left(\frac{1\mathrm{mol}}{335\mathrm{g}}\right) \\ =1.170\times {10}^{-4}\mathrm{mol}{\mathrm{L}}^{-1} \end{gathered}$$

Step 4. Finding molar absorptivity

Substitute the values in equation $\left(2\right)$,

$$\begin{gathered} \epsilon =\frac{0.193}{1.50\mathrm{cm}\times 1.170\times {10}^{-4}\mathrm{M}} \\ =1.10\times {10}^3\mathrm{L}{\mathrm{mol}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$