Respuesta :
Answer : The percentage reduction in intensity is 79.80 %
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
[tex]A=\log \frac{I_o}{I}[/tex]
[tex]\log \frac{I_o}{I}=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution
C = concentration of solution = [tex]3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}[/tex]
l = path length = 2.5 mm = 0.25 cm
[tex]I_o[/tex] = incident light
[tex]I[/tex] = transmitted light
[tex]\epsilon[/tex] = molar absorptivity coefficient = [tex]855dm^3mol^{-1}cm^{-1}[/tex]
Now put all the given values in the above formula, we get:
[tex]\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)[/tex]
[tex]\log \frac{I_o}{I}=0.6947[/tex]
[tex]\frac{I_o}{I}=10^{0.6947}=4.951[/tex]
If we consider [tex]I_o[/tex] = 100
then, [tex]I=\frac{100}{4.951}=20.198[/tex]
Here 'I' intensity of transmitted light = 20.198
Thus, the intensity of absorbed light [tex]I_A[/tex] = 100 - 20.198 = 79.80
Now we have to calculate the percentage reduction in intensity.
[tex]\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100[/tex]
[tex]\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%[/tex]
Therefore, the percentage reduction in intensity is 79.80 %
