Calculate how to prepare 100.0mL of a 1.25x10^-4 M aqueous solution of Fe(NO3)3(aq) by dilution of stock of 0.0025M Fe(NO3)3(aq) in a 100.0mL volumetric flask?

Question

I tried using the m1v1=m2v2 equation but it doesn't look like that equation would work for this problem??

Answer 1

Don't know why.

1.25E-4M x 100 mL = 0.0025M*mL
Solve for mL of the 0.0025 M solution, add that to a 100 mL volumetric flask, dilute to the mark, mix thoroughly, stopper, voila!