Question #848b4

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Question #848b4

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Answer 1 · 2017-12-19T09:50:15

$ε = 46.7 d m^{- 3} m o l^{- 1} c m^{- 1}$ $epsilon = 46.7 dm^(-3)mol^(-1)cm^(-1)$

Explanation:

To calculate this, we need to use the Beer-Lambert Law.
${log}_{10} (\frac{I_{0}}{I}) = A = ε C l$ $log_10(I_0/I) = A = epsilonCl$
Where $A$ $A$ = absorbance; $ε$ $epsilon$ = molar extinction coefficient ( $d m^{- 3} m o l^{- 1} c m^{- 1}$ $dm^(-3)mol^(-1)cm^(-1)$ ); $C$ $C$ = concentration (M); $l$ $l$ = path length (cm); $I_{0}$ $I_0$ = incident light; $I$ $I$ = transmitted light

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

The sample transmitted 20%, so we know that $I = 20 %$ $I = 20%$ Since the transmitted light is 20% of the incident light, we can take the incident light to be 100% $I_{0} = 100 %$ $I_0 = 100%$

$A = {log}_{10} (\frac{100}{20})$ $A = log_10(100/20)$
$A = 0.7$ $A = 0.7$

$ε = \frac{0.7}{0.01 \times 1.5}$ $epsilon = 0.7 / (0.01 xx 1.5)$

$ε = 46.7 d m^{- 3} m o l^{- 1} c m^{- 1}$ $epsilon = 46.7 dm^(-3)mol^(-1)cm^(-1)$

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Question #848b4

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