# Question 848b4

Dec 19, 2017

$\epsilon = 46.7 {\mathrm{dm}}^{- 3} m o {l}^{- 1} c {m}^{- 1}$

#### Explanation:

To calculate this, we need to use the Beer-Lambert Law.
${\log}_{10} \left({I}_{0} / I\right) = A = \epsilon C l$
Where $A$ = absorbance; $\epsilon$ = molar extinction coefficient (${\mathrm{dm}}^{- 3} m o {l}^{- 1} c {m}^{- 1}$); $C$ = concentration (M); $l$ = path length (cm); ${I}_{0}$ = incident light; $I$ = transmitted light

We can rearrange to solve for the molar extinction coefficient:
$\epsilon = \frac{A}{C l}$

The sample transmitted 20%, so we know that I = 20% Since the transmitted light is 20% of the incident light, we can take the incident light to be 100% I_0 = 100%#

$A = {\log}_{10} \left(\frac{100}{20}\right)$
$A = 0.7$

$\epsilon = \frac{0.7}{0.01 \times 1.5}$

$\epsilon = 46.7 {\mathrm{dm}}^{- 3} m o {l}^{- 1} c {m}^{- 1}$