Convert the accompanying transmittance data to absorbance: 53.86% 0.567% Express
ID: 898793 • Letter: C
Question
Convert the accompanying transmittance data to absorbance: 53.86% 0.567% Express the following absorbances in terms of percent transmittance: What is the relationship between percentage transmittance and absorbance? If the percent transmittance for a sample is 100 at a given wavelength, what is the value of the absorbance? Explain why it was necessary to add ortho-phenanthroline to the solutions. What is the purpose of adding hydroxylamine hydrochloride to the solution? Calculate the molar concentration of a solution if the absorbance (A) is 0.244, the molar absorptivity (A) is 3.39 Times 10^4 L/mole cm and the path length of the cell is 2.00 cm.Explanation / Answer
1) convert transmittance (%T) data to absorbance (A)
Solution-
According to beer’s law absorbance can be shown as below
A = 2 - log10 %T
a) %T = 53.86%
Plug the given value in above equation
A = 2 - log10 (56.86) = 2 – 1.7548
A = 0.245
b) %T = 0.567%
Plug the given value in above equation
A = 2 - log10 (0.567) = 2 – (-0.246)
A = 2.246
2) Express following absorbance in terms of (%T) percent transmittance
a) A = 0.918
By using the beers law
A = 2 - log10 %T
0.918 = 2 - log10 %T
1.082 = log10 %T
Taking antilog of both the side
%T = 12.07%
b) A = 0.072
By using the beers law
A = 2 - log10 %T
0.072 – 2 = - log10 %T
1.928 = log10 %T
Taking antilog of both the side
%T = 84.72%
3) What is the relation between %T and absorbance? If the %T of sample is 100 at the given wavelength what is the value of absorbance.
Solution-
Absorbance (A) is inversely proportion to (% T) percent transmittance
Transmittance, T = P / P0
P = radiant power
P0 = Incident radiant power
% Transmittance, %T = 100 T
Absorbance,
A = log10 P0 / P
A = log10 1 / T
A = log10 100 / %T
A = 2 - log10 %T
Given - %T 100
Plug the value in above equation
A = 2 - log10 %T = 2- log10(100) = 2 -2
A = 0
5) calculate the molar concentration of solution . If the absorbance is 0.244, molar absorbtivity 3.39 x 104 L/mol cm and path length of cell (b) is 2.0 cm.
Solution-
Given-
a = 0.244
Molar absorptivity (A) = 3.39 x 104 L/mol cm
Path length (b) = 2.0 cm
Molar concentration (c) =?
By using beer’s law molar absorbtivityis as below
A = a/bc
C = a/Ab
Plug the value in equation
3.39 x 104 L/mol cm = 0.244 /(2*C)
C = 0.244/(2*3.39 x 104 L/mol cm) =
C =3.59 X 10-6 mol L-1
molar concentration of solution = 3.59 X 10-6 mol L-1
Transmittance, T = P / P0
P = radiant power
P0 = Incident radiant power
% Transmittance, %T = 100 T
Absorbance,
A = log10 P0 / P
A = log10 1 / T
A = log10 100 / %T
A = 2 - log10 %T