Homework Answers
a) Given T = 39.6%, I=1.5cm, C=3.75mg/100ml, Mwt. = 220g/mol
No. of moles = 3.75mg/ 220g / mol =0.375 g /220g/mol = 0.0017 moles
Absorbance A = ɛ x l x c, where A is the amount of light absorbed by the sample for a given wavelength, ɛ is the molar absorptivity, l is the distance that the light travels through the solution, and c is the concentration of the absorbing species per unit volume.
T=39.6%, Absorbance = 2 – log(%T) = 2 - log (39.6) =0.41
60.4= ɛ x 1.5 x 0.0017
ɛ = 0.41 / (1.5cm x0.0017mole) =160 cm -1 mole-1
b) C= 0.0034 moles, I=1.5cm, ɛ = 160 cm -1 mole-1
A = ɛ x l x c = 160 cm -1 mole-1x 1.5 cm x 0.0034moles = 0.816
Transmittance = Antilog (2-absorbance) = Antilog (2- 0.816) =15.275%
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