A substance in an aqueous solution at a concentration of 0.01M shows an optical transmittance of 28% with a path length of 2mm calculate the molar absorption coefficient of the solute? What would be the transmittance in a cell of 1cm thick?

The absorption of radiation

Many compounds absorb ultraviolet (UV) or visible (VIS) light as it passes through a solution.

If a beam of radiation of power #P_0# passes through a solution, absorption takes place. The radiation leaving the sample has power #P#.

We can express the amount of radiation absorbed as transmittance #T#.

#color(blue)(bar(ul(|color(white)(a/a)T = P/P_0color(white)(a/a)|)))" "#

We can express the amount of radiation absorbed as absorbance #A#

#A = log(P_0/P) = log(1/T)#

#color(blue)(bar(ul(|color(white)(a/a)A = "-log"Tcolor(white)(a/a)|)))" "#

The Beer-Lambert Law

The Beer-Lambert Law states that the absorbance #A# of a solution is directly proportional to the molar concentration #"c"# of the solute and the path length #l# (usually expressed in centimetres).

#color(blue)(bar(ul(|color(white)(a/a)A = epsilonclcolor(white)(a/a)|)))" "#

#epsilon# is a proportionality constant called the molar absorption coefficient.

Now, we can write

#logT = -epsiloncl#

or

#epsilon = -logT/(cl)#

Part (a). Calculate the molar absorption coefficient

#epsilon = -logT/(cl) = -log0.28/("0.01 mol·L"^"-1" × "0.2 cm") = "280 L·mol"^"-1""cm"^"-1"#

Part (b). Calculate the transmittance

#logT = -epsiloncl = "-280" color(red)(cancel(color(black)("L·mol"^"-1""cm"^"-1")))× 0.01 color(red)(cancel(color(black)("mol·L"^"-1"))) × 1 color(red)(cancel(color(black)("cm"))) = "-2.8"#

#T = 10^"-2.8" = 0.0017#