Question

1) A solution containing 3.75 mg/100 ml of A (MW - 220 g/mol) has a transmittance of 39.6% in a 1.50 cm cell at 480 nm. Calculate (a) the molar absorptivity of A, and (b) the transmittance of a solution of A which is twice as concentrated as the solution described above.

Answer 1

a) Given T = 39.6%, I=1.5cm, C=3.75mg/100ml, Mwt. = 220g/mol

No. of moles = 3.75mg/ 220g / mol =0.375 g /220g/mol = 0.0017 moles

Absorbance A = ɛ x l x c, where A is the amount of light absorbed by the sample for a given wavelength, ɛ is the molar absorptivity, l is the distance that the light travels through the solution, and c is the concentration of the absorbing species per unit volume.

T=39.6%, Absorbance = 2 – log(%T) = 2 - log (39.6) =0.41

60.4= ɛ x 1.5 x 0.0017

ɛ = 0.41 / (1.5cm x0.0017mole) =160 cm -1 mole-1

b) C= 0.0034 moles, I=1.5cm, ɛ = 160 cm -1 mole-1

A = ɛ x l x c = 160 cm -1 mole-1x 1.5 cm x 0.0034moles = 0.816

Transmittance = Antilog (2-absorbance) = Antilog (2- 0.816) =15.275%