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1. A solution of Tryptophan has an absorbance at 280 nm of 0.54 in a 0.5 cm
length cuvette. Given the absorbance coefficient of trp is 6.4 × 103
LMol-1cm-1
What is the concentration of a solution?
Solution:
As A = ε l c l= 0.5 cm A=
0.54
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ε = 6.4 × 103 LMol-1cm-1
C=?
So c = A/ε l
= 0.54 / 6.4 × 103 × 0.5
Answer = 0.000168 M
2. Calculate the molar absorptivity of a 1 x 10-4 M solution, which has an
absorbance of 0.20, when the path length is 2.5 cm.
Solution:
A = ε l c
l= 2.5 cm
A= 0.20
C= 1 x 10 – 4 M
ε =?
So ε = A / l c
= 0.20/ 2.5 ×1 x 10-4
Answer = 800 dm3
/mol/cm.
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3. The molar absorptivity of a 0.5 x 10-3 M solution is 261.53 dm3
/mol/cm,
If it has an absorbance of 0.17, Calculate the path length.
Solution:
A = ε l c
ε = 261.53 dm3
/mol/cm
A= 0.17
C= 0.5 x 10-3 M
l = ?
So l = A / ε c
= 0.17/ (261.53 × 0.5 x 10-3)
Answer = 1.3 cm.
4. A 1.00 × 10–4 M solution of an analyte is placed in a sample cell with a path
length of 1.00 cm. When measured at a wavelength of 350 nm, the solution’s
absorbance is 0.139. What is the analyte’s molar absorptivity at this
wavelength?
l = 1.00 cm
c = 1.00 × 10–4 M
A=0.139
ε =?
So
A = ε l c
ε = A / l c
= 0.139/ 1.0 × 1.00 x 10-4
Answer = 1390 cm−1 M−1
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