I need help with answering the following question in Instrumental Analysis. Plea

Question

I need help with answering the following question in Instrumental Analysis. Please Show me the work step by step.

Teflon capillary tubes up to 5 m have been constructed to detect dilute amounts of material using absorbance spectroscopy. For these systems the path length is 5 meters. For a dye that has an absorption max at 590 nm draw a diagram of an absorbance that incorporates the 5 m cell. You want to maximize your sensitivity and keep your cost low. For getting your signal into and out of the Teflon capillary just draw a lens and say coupling optic. If 5.05 x 10-5 M of an analyte can be detected with a 1 cm path length, what concentration could be detected with a 5 m path length?

Explanation / Answer

FRom Beer's law,

Absorbance A = ecl

e = molar extinction coefficient

c = concentration

l = path length

So,

   c1l1 = c2l2

Given that

c1 = 5.05 x 10-5 M

l1= 1 cm

c2 = ?

l2 = 5m = 500 cm

c1l1 = c2l2

c2 =c1l1 / l2

= (5.05 x 10-5 M) (1 cm) / (500 cm)

= 0.0101 x 10-5 M

c2 = 0.0101 x 10-5 M

Therefore,

0.0101 x 10-5 M could be detected with a 5 m path length.

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