Knowledge in Petro chemistry
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CHEMISTRY ASSIGNMENT
This pdf contains important mcq based exhaustive assignment of chemistry subject
priya mheta
CHEMISTRY
This pdf contains important mcq and question based on chemistry 1st year engineering
chemistry and petroleum
THIS PDF CONTAINS MCQ AND QUESTION BASED ON PETROLEUM CHEMISTRY
CHEMISTRY
This pdf contains simple mcq with answer based on chemistry(petroleum)
petroleum chemistry
This pdf contains simple but important mcq based on petroleum chemistry for 1st year engineering
KNOCKING OF FUELS , OCTANE AND CETANE VALUES AND LIQUID FUELS
These are the notes of the subject Petro Chemistry for engineering and are targeted to one of the difficult subtopics of KNOCKING OF FUELS, OCTANE, AND CETANE VALUES AND LIQUID FUELS. These will prove beneficial to FY Engineering students of all branches and help them to score good marks in their exams. Made by: A student at NIT Surat.
VEDANT MILIND ATHAVALE
BEER LAMBERTS LAW
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