Question #848b4

1 Answer
Dec 19, 2017

epsilon = 46.7 dm^(-3)mol^(-1)cm^(-1)ε=46.7dm3mol1cm1

Explanation:

To calculate this, we need to use the Beer-Lambert Law.
log_10(I_0/I) = A = epsilonCllog10(I0I)=A=εCl
Where AA = absorbance; epsilonε = molar extinction coefficient (dm^(-3)mol^(-1)cm^(-1)dm3mol1cm1); CC = concentration (M); ll = path length (cm); I_0I0 = incident light; II = transmitted light

http://6e.plantphys.net/topic07.01.html

We can rearrange to solve for the molar extinction coefficient:
epsilon = A / (Cl)ε=ACl

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%I0=100%

A = log_10(100/20)A=log10(10020)
A = 0.7A=0.7

epsilon = 0.7 / (0.01 xx 1.5)ε=0.70.01×1.5

epsilon = 46.7 dm^(-3)mol^(-1)cm^(-1)ε=46.7dm3mol1cm1