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Topic: VIS absorption  (Read 4929 times)

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Offline jlachapelle

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VIS absorption
« on: April 05, 2007, 02:15:14 PM »
I am trying to find the absorption coefficient of CCl4 liquid at 532 nanometers and 25 degrees C. It is expected to be in the 10^-4---10^-6 range so I need a very precise measurement. Does anyone know a possible data source or, better yet, the coefficient?

Offline ARGOS++

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Re: VIS absorption
« Reply #1 on: September 30, 2007, 12:39:02 PM »

Dear JLachapelle,

Let’s give you the sad news first:
I don’t believe that there will be any solution for your problem! – Sorry!

I even DON’T believe that your value of: ε = 1.0 * 10-4 L / mol can be of any true, too!
Contrary I believe that nobody is able to give you a true value for!

Let’s do some calculation together:  (But do self-control too!)
  • A.)   For the highest possible Concentration, pure CCl4 with a density of 1.596 kg/L and a molar weight of 0.1538 kg/mol results in a Molarity of: 10.38 mol/L
  • B.)   Now we can calculate the Absorbance in a 1.0 cm Cuvette, if you use only the “Special Beer-Lambert Law”:  A = ε * c * d
       A532 = 1.0 * 10-4 L / mol * 10.38 10.38 mol / L * 0.01 m

    and we result in a final Absorption of:  A532 ~ 1.038 * 10-5 AU’s

Not only I have to believe that NOBODY till jet was ever able to build a Spectrometer of such low noise, to bring your Signal significantly out of the measuring noise at all.

If you go the other way and bring your final Signal to 1.0 AU you can easy calculate, that it would require a Cuvette with a Pathlength of nearly ONE Kilo-Meter.
I think also an absolute impossibility!

You could finally use a so called “Multi-Reflection-Cuvette” to shorten the physical Dimension, but I don’t believe you are able to find anyone with a higher Factor then 100.
But even in the case you could find one, the Noise-Problem will still survive!!

How much higher will be the Problem in case of even lower Concentration of CCl4?

I can only hope that there will be any other Solution to your Problem, but as I don’t know it in more details, no hint can be given.

I hope it may be anyway of help to you!

Good Luck!

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