you want to determine the amount of chlorophyll(Molecular weight=893.51g/mol)in a 25ml marine sample.before analysis by UV spectoscopy,2ml of sollution was centrifuged.0.5ml of the supernatant was diluted to 150ml with DI water.the diluted sample was analysed using UV spectroscopy and had an absobance of 0.540 at 670.what was the amount in mg of chlorophll in the original sample.

Work:
A=ɛlc(beers lambarts law)
c=0.54 ÷ 0.984=0.549
C1V1=C2V2
C1.0.5=0.549.150=164.163
m=n×Mm (164.163×893.51)=1.47×10^-5

1. I don't see a molar absorptivity constant nor any information to calculate same.

2. Please clarity if that A = 0.540(you have 0.549 in your calculation) is the amount in 0.5 mL or the amount in the 2mL. The way the problem reads to me is confusing; i.e., I don't know if we are looking at a 2 mL sample or a 0.5 mL sample.