1 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 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. 2 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

beer lamberts law +6 more

by VEDANT MILIND ATHAVALE

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1 Attachment 3 years ago