Asked by Ray
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?
I tried using the m1v1=m2v2 equation but it doesn't look like that equation would work for this problem??
I tried using the m1v1=m2v2 equation but it doesn't look like that equation would work for this problem??
Answers
Answered by
DrBob222
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!
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!
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