How to determine the concentration after a dilution with Beer's law?

Question

The molar extinction coefficient of pure antibody is $1.4$ absorbance units per milligram antibody at $\pu{280 nm}$. You have a concentrated antibody solution, so you make a $5$-fold dilution and measure at $\pu{280 nm}$, getting an absorbance of $0.845$. What is the concentration of that solution?

Using Beer's Law this is my work: $$0.845 = (1.4)\cdot (1)\cdot (c) \\ \therefore c = \pu{0.604 mol L^-1}$$

Then I think we need to factor in the dilution, therefore $c = (5) \cdot (0.604) = \pu{3.02 mol L^-1} $. Is this correct?

Answer 1

Beers law is

$$ E= \epsilon_{\lambda}\ \cdot c\ \cdot d$$

Where $c$ is the concentration of the solution. Then, you've found your answer once you've found $c$.

As Jerepierre points out, if you were to account for units in the answer, the question is actually missing some information. The pathlength $d$ is unaccounted for (but we're assuming its $1$), however if it was say $\mathrm{1\ cm}$, then the units for $\epsilon$ would need to be $\mathrm{AU\cdot mg^{-1} \cdot cm^{-1}}$. Which would mean that the units for your concentration would be in $\mathrm{mg}$. But, for concentration we would need to know antibody per some amount of volume. So taking units into account shows that something is missing.

What you're adding at the end is a step you would take to find the original concentration of the undiluted solution. Given that they diluted the concentration 5 times, but we don't know to what factor so we would have trouble solving for that.