Question

A sample is subjected to UV-Vis spectroscopy in a 1 cm cell, and 23.2% of the incident light is transmitted through the sample to the detector. If the molar absorptivity of the analyte in the sample is 450 L/(mol*cm), what is the concentration of the analyte in the sample?

Answer 1

0.001410 mol/L is the concentration of the analyte in the sample.

Step-by-step explanation:

Using Beer-Lambert's law :

Formula used :

`A=\epsilon * C* l`

`A=\log (I_o)/(I)`

`\log (I_o)/(I)=\epsilon * C* l`

where,

A = absorbance of solution

C = concentration of solution =
`2.00* 10^(-3)M`

l = path length = 1.00 cm

`I_o` = incident light

`I` = transmitted light

`\epsilon` = molar absorptivity coefficient

Here we are given :
`\epsilon = 450 L/(mol cm)`

Transmittance of light = T = 23.2%

`T=(I)/(I_o)* 100`

`23.2\%=(I)/(I_o)* 100`

`0.232=(I)/(I_o)`

`I=0.232* I_o`

`A=\log (I_o)/(I)`

`A=\log (I_o)/(0.232* I_o)=0.6345`

`A=\epsilon * C* l`

`C=(A)/(\epsilon * l)=(0.6345)/(450 L/(mol cm)* 1 cm)`

C = 0.001410 mol/L

0.001410 mol/L is the concentration of the analyte in the sample.