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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.

  • Chemistry -

    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.

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