Chemical Forums: Units of molar absorptivity. Is L.mol^-1.cm^-1 = mL.mmol^-1.cm^-1?

This is more of a mathematical question but here goes.

The Beer-Lambert equation rearranged is: ε=A/cl

Where c = concentration (mol.L^-1), A = absorbance (AU) and l = path length (cm)

Since “mol.L^-1” and “cm” are in the denominator, molar absorptivity has the units L.mol^-1.cm^-1.

Edit (Thanks to chenbeier):

The prefix milli- (m) means multiplication by 10^-3 to convert to the unit without a prefix eg. 350 mL = 350*10^-3 L = 0.35 L.

Similarly, the prefix kilo- (kg) means multiplication by 10^3 to convert to the unit without a prefix eg. 1.2 kg = 1.2*10^3 g = 1200 g

And similarly, the prefix micro- (μ) means multiplication by 10^-6 to convert to the unit without a prefix eg. 250000 μm = 250000*10^-6 m = 0.25 m

The question is,

is molar absorptivity in L.mol^-1.cm^-1 numerically the same as molar absorptivity in mL.mmol^-1.cm^-1 given that 1 mol.L^-1 is equals 1 mmol.mL^-1? Here is my attempt at answering my own question.

Suppose you have a solution that has a concentration of 0.00025 mol.L^-1. That’s 0.00025 moles per 1 litre. Converting to mmol per litre, the concentration is 0.25 mmol.L^-1. So now that’s 0.25 mmol per 1 litre which is the same is 0.25 mmol per 1000 mL. Expressing the concentration in mmol per mL (mmol.mL^-1), we return to the number 0.00025 once more (0.00025 mmol.mL^-1). Therefore, the concentration in mol.L^-1 is numerically the same as the concentration in mmol.mL^-1.

Suppose we are trying to work out molar absorptivity and A = 0.5 , c = 0.00025 mol.L^-1 and l = 1 cm.

We established earlier that mol.L^-1 = mmol.mL^-1. Redoing the calculation above but using mmol.L^-1 instead as the unit for concentration, A = 0.5 , c = 0.00025 mmol.mL^-1 and l = 1 cm.

To conclude, am I right in saying that numerically, the molar absorptivity in L.mol^-1.cm^-1 is equal to the molar absorptivity in mL.mmol^-1.cm^-1?