Assume the % transmittance for a solution is 54.1% at a wavelength of 325nm. What is the absorbance?
Would I take the negative log of .541 to get the absorbance? Do I have to do anything with the wavelength in this problem?
2007-03-29 07:29:11 · 5 answers · asked by calculusgirl1979 2 in Science & Mathematics ➔ Chemistry
Yes, You are doing all right with the calculation.
No, You do not have to do anything with the wavelength. Your data just says that this absorbance is TRUE at 325 nm wavelength. At other wavelengths, the absorbance may or may not be the same.
2007-03-29 07:41:38 · answer #1 · answered by Aldo 5 · 0⤊ 0⤋
A = -log(T)
T = %T divided by 100
So yeah, just use the A = formula, wavelength doesn't have to do with anything in that problem. In fact, I have done a Beer's Law Lab and the only formula's used were the previous two a formula and A = bce, where
A= absorbance
b = measure of the spectrometer tube
c = concentration
e = a relationship between b and c (cm*mole/L)
But I doubt you need A = bce
2007-03-29 07:39:13 · answer #2 · answered by Roger 2 · 0⤊ 0⤋
No, the wavelength is only given because molar absorptivities are wavelength specific. In this case, you would take the -log of .541.
2007-03-29 07:38:57 · answer #3 · answered by orgchem72 3 · 0⤊ 0⤋
A = log I0/I
transmittance =I/I0
here transmittance 1/0.541 =1.848
you take the log 0.267= absorbance
you were correct
2007-03-29 07:37:45 · answer #4 · answered by maussy 7 · 0⤊ 0⤋
Yes, you 're right.
A = -log( I / Io)
The ratio (I / Io) is actually the percent transmittance.
2007-03-29 07:40:41 · answer #5 · answered by tuoidabuon 2 · 0⤊ 0⤋