Directions: Show all work for this problem. You may want to write your answer on
ID: 852820 • Letter: D
Question
Directions: Show all work for this problem. You may want to write your answer on a piece of notebook paper. Answer all questions by showing all mathematical work and/or interpreting all major peaks in the IR and all peaks in the H - NMR spectra. You may wish to do complete your assignment an other sheet of paper. A solution of ethanol has been contaminated with benzene a technique mp eyed that makes ethanol unfit to drink. Benzene has a molar absorptive of 230 M-1 'cm-1 at 260 nm in ethanol, and ethanol shows no absorbance at 260 nm. How could the concentration of benzene in the solution be determined? A 4.0 times 10-5 M solution of a compound in hexane shows an absorbance of 0.40 at 252 nm in a cell with a 1-cm light path. What is the molar absorptive of the compound in hexane at 252 nm? The visible spectrum of (3-carotene (C40H56, log epsilon (463 nm) = 5.10, the orange pigment in carrots) dissolved in hexane shows intense absorption maxima at 463 nm and 494 nm, both in the blue-green region. Because light of these wavelengths is absorbed by beta -carotene, we perceive the color of this compound as that of the complement to blue-green, namely red-orange. Calculate the concentration in milligrams per milliliter of beta -carotene that gives an absorbance of 1.8 at 463 nm.Explanation / Answer
Q#1 UV/Visible spectroscopy can be used to determine the impurity or the contamination of ethanol with benzezene.
we can make standard solutions of varous concentration (in the range in which our contaminated sample may fall in) of benzene in ethanol and detrmining their absobance of each solution at given wavelength. then drawing a calibration curve between absorbance and concentration. the concentration or contamination of unknown benzene concentration can be detrmined by comparing its value from the calibration curve.
Q#2
concentration=0.00004M
absorbance=0.4
path length=1cm
molar absorptivity=?
by beers law
absorbance=molar absorptivity*path length*concentration
0.4/(path length*concentration)=molar absorptivity
molar absorptivity=0.4/(1cm*0.00004moles/L)
molar absorptivity=10000L/mol-cm
Q#3
log(apsilon)=5.1
antilog(log(apsilon)=molar absorptivity==antilog(5.1)=125892.5L/mol-cm
absorbance=1.8
path length=1cm
concentration=absorbance/(molar absorptivity*path length)
=1.8/(125892.5L/mole-cm*1cm)
=1.43x10-5M
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