Asked by Gabby
How would a student make 500.0 mL of a 0.08500 M solution of iron(III) using a stock solution that is 0.3180 M in iron(III)? Explain how this would be done.
Answers
Answered by
DrBob222
I know not how a student would do it but I would use c1v1 = c2v2
0.08500M*500 mL = 0.3180M*v2
v2 = 133.6 mL
The way you do this is to pipet or measure accurately 133.6 mL of the stock solution and transfer quantitatively to a 500 mL volumetric flask. Then add distilled water to the mark of the volumetric flask.
0.08500M*500 mL = 0.3180M*v2
v2 = 133.6 mL
The way you do this is to pipet or measure accurately 133.6 mL of the stock solution and transfer quantitatively to a 500 mL volumetric flask. Then add distilled water to the mark of the volumetric flask.
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