The absorbance of #2# x #10^-4# #M# solution was found to be #0.123# when placed in a cell of #1# #cm# length. Calculate the transmittance, percent transmittance and molar absorptivity of a cell of length #2# #cm#?

Question

Wyatt Atkins · Accepted Answer

#T=0.568#
#%T=56.8%#
#epsilon=615M^(-1)*cm^(-1)#

#C=2xx10^(-4)M#
#A=0.123#
#l=2cm#
#?T#, #?%T#, #?epsilon#

The relationship between transmittance #T# and absorbance is:

#A=-logT=>T=10^(-A)#

Since the length of the cell was increased to #l=2cm#, the absorbance will double and therefore, #A_2=2xx0.123=0.246#

#=>T=10^(-0.246)=0.568#

Therefore, #%T=100xxT=100xx0.568=56.8%#

In order to solve for the molar absorptivity #epsilon#, we will use the Beer-Lambert law:

#A=epsilonxxlxxC=>epsilon=A/(lxxC)#

#epsilon=(0.123)/(1cmxx2xx10^(-4)M)=615M^(-1)*cm^(-1)#

Here is a video that explains Beer-Lambert law and it use experimentally:
AP Chemistry Investigation #1: Beer-Lambert Law.

Wyatt Atkins · Answer

undefined Transmittance (T) = 10^(-absorbance) = 10^(-0.123) ≈ 0.792

Percent transmittance = (Transmittance × 100) = 0.792 × 100 ≈ 79.2%

Molar absorptivity (ε) = (absorbance / concentration) × cell path length = (0.123 / 2 × 10^(-4)) / 1 ≈ 615 L/(mol·cm)