Calculating Absorbance of Iron (III) Solution at 580nm for 2.50-cm Cell

[ahhppull]

Homework Statement

2. A 5.00 mL aliquot of a solution that contains 5.94 ppm iron (III) is treated with an appropriate excess of KSCN and is diluted to 50.0 mL.What is the absorbance of the resulting solution at 580 nm in 2.50-cm cell?

Homework Equations

The Attempt at a Solution

I'm having trouble finding the concentration of the solution. I'm not really good at using ppm. How do I convert this to concentration? If I have the concentration, I can finish the problem by plugging in ε,b and c.

 

[ahhppull]

Not sure if correct but here's what I did to find the concentration.

5.49 ppm = 5.49 micrograms/mL = 0.00549 grams/L

0.00549 grams/L /(55.845 grams/mole) = 0.000106 M

0.000106 M * 5 mL / 50 mL = 0.0000106 M

EDIT : Also I just realized that the molar absorptivity of FeSCN is not given in the book, so am I doing the problem wrong or do I need to find this online?

 

[Borek]

Shortcut - ppm usually means mg per L.

1.06e-4 M is OK.