Question

Page 9 Questions 1. A substance exhibits an absorbance of 0.47 in a 1.00 cm cell at 525 nm. (a) What will be its absorbance in a 2.7 cm cell? (b) If the substance concentration is 2.1 x 10-5 mol dm, compute its molar absorptivity. (c) What absorbance will it have in a 5.0 cm cell at 2. = 405 nm (think about it!)? 2. You are setting up an absorption spectrometer to determine the concentration of a gas which is present in very small amounts in the atmosphere. During initial testing between two tall buildings, you measure a 46% transmittance when the concentration of the particular gas is known to be 8 x 10 mol m-. What is the distance between the two buildings? You may assume that the molar absorptivity, E, of the gas in question is 2 x 10 m conditions. mol-'m-' under these

Answer 1

SOLN 1

Absorbance (A) = e.c.l

e = molar absorptivity

c = concentration

l =path length

(a) hence A is directly proportional to path length

so in 1 cm if A = 0.47

so in 2.7 it will be 0.47 * 2.7 = 1.269

(b) A=e.c.l

We have A = 0.47, b = 1.00 cm = 0.1 dm, C = 2.1 x 10^-5 mol.dm^-3. $\epsilon$ = ?

Using these values in eq.(1)

0.47 = $\epsilon$ x 0.1 x 2.1 x 10^-5.

$\epsilon$ = 0.47 / (2.1 x 10^-6)

$\epsilon$ = 223810 dm²/mol.

(c) Absorbance is measured at particular value of  $\lambda$max only.

hence this question is not directly solved for Absorbance.

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