Beer Lambert Law solved problems

Beer Lambert Law solved problems

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  0 COMSATS Institute ofInformation Technology, Abbottabad Course title and code Analytical Techniques Assignment number 01 Assignment title 10 Numericals of Beer–Lambert law Submitted by Zohaib HUSSAIN Registration number Sp13-bty-001 Submitted To Dr. Murtazaa SAYED. Date of submission Sunday, March 13, 2016
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  1 1. A solutionof 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 × 10 3 LMol-1 cm-1 . What is the concentration of solution? Solution: As ε = A / l c l= 0.5 cm A= 0.54 ε = 6.4 × 10 3 LMol-1 cm-1 C=? So c = A/ε l = 0.54 / 6.4 × 10 3 × 0.5 Answer = 0.000168 M 2. A solution of thickness 2 cm transmits 40% incident light. Calculate the concentration of the solution, given that ε = 6000 dm3 /mol/cm. Solution: A = 2 - log10 %T = 2 - log10 40 = 2 – 1.6020 = 0.398 A = ε l c l= 2cm ε = 6000 dm3/mol/cm A=0.398 c=? So c = A/ ε l = 0.398/ 6000 × 2 Answer = 3.316 X 10 – 5 mol / dm3
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  2 3. A solutionshows a transmittance of 20%, when taken in a cell of 2.5 cm thickness. Calculate its concentration, if the molar absorption coefficient is 12000 dm3/mol/cm. Solution: A = 2 - log10 %T = 2 - log10 20 = 2 – 1.301 = 0.698 A = ε l c l= 2.5 cm ε = 12000 dm3/mol/cm A=0.698 c=? So c = A/ ε l = 0.698/ 12000 × 2.5 Answer = 2.33 X 10 – 5 mol / dm3 4. 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
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  3 = 0.20/ 2.5×1 x 10 -4 Answer = 800 dm3/mol/cm. 5. The concentration of yeast t-RNA in an aqueous solution is 10 M. The absorbance is found to be 0.209 when this Solution is placed in a 1.00 cm cuvette and 258 nm radiations are passed through it. a) Calculate the specific absorptivity, including units, of yeast t-RNA. b) What will be the absorbance if the solution is 5 M? c) What will be the absorbance if the path length of the original solution is increased to 5.00 cm? Solution 5a l = 1.00 cm c = 10.00 M A=0.209 So ε = A / l c = 0.209 / 1.00 cm X 10 M Answer = 0.0209 dm3/mol/cm. 5b ε = 0.0209 dm3/mol/cm. l = 1.00 cm c = 5.00 M A=? So A = ε l c
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  4 A = 0.0209dm3/mol/cm. X 1.00 cm X 5M Answer = 0.1045 5c ε = 0.0209 dm3/mol/cm. l = 5.00 cm c = 10.00 M A=? So A = ε l c A = 0.0209 dm3/mol/cm X 5.00 cm X 10.00 M Answer = 1.045 6. Calculate the molar absorptivity of a 0.5 x 10 -3 M solution, which has an absorbance of 0.17, when the path length is 1.3 cm. Solution: A = ε l c l= 1.3 cm A= 0.17 C= 0.5 x 10 -3 M ε =? So ε = A / l c = 0.17/ 1.3 × 0.5 x 10 -3 Answer = 261.53 dm3/mol/cm.
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  5 7. A CaCO3solution shows a transmittance of 90%, when taken in a cell of 1.9 cm thickness. Calculate its concentration, if the molar absorption coefficient is 9000 dm3/mol/cm. Solution: A = 2 - log10 %T = 2 - log10 90 = 2 – 1.954 = 0.045 A = ε l c l= 1.9 cm ε = 9000 dm3/mol/cm A=0.045 c=? So c = A/ ε l = 0.045/ 9000 × 1.9 Answer = 2.631 × 10 -6 mol / dm3 8. Extinction coefficient of NADH at 340 nm is 6440 L/mol/cm. whereas NAD does not absorb at 340nm. What absorbance will be observed when light at 340nm passes through a 1cm cuvette containing 10uM NADH and 10 uM NAD. Solution: ε = 6440 L/mol/cm. l = 1.00 cm c = 10.0 uM =10 X 10 -6 M A=? So A = ε l c A = 6440 L/mol/cm X 1.00 cm X 10 X 10 -6 M
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  6 Answer = 0.0644 Note:this absorbance is only for NADH because NAD do not absorb at 340nm. 9. 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 ε = A / l c = 0.139/ 1.0 × 1.00 x 10 -4 Answer = 1390 cm−1 M−1 10. The absorbance of a Cu sulphate solution containing 0.500 mg Cu/mL was reported as 0.3500 at 440 nm. a) Calculate the specific absorptivity, including units, of Cu sulphate on the assumption that a 1.00 cm cuvette was used. b) What will be the absorbance if the solution is diluted to twice its original volume Solution a) l = 1.00 cm c = 0.500 mg/ml A= 0.3500 ε =?
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  7 So ε =A / l c = 0.3500 / 1.0 × 0.500 Answer = 0.7 cm -1 (mg/mL) -1 b) c= 0.250 mg/ml ε= 0.7 cm -1 (mg/mL) -1 l = 1.00 cm A=? So A = ε l c A = 0.7 cm -1 (mg/mL) -1 X 1.00 cm X 0.250 mg/ml Answer = 0.175