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26 May, 20:15

How many milliliters of 11.5 M HCl (aq) are needed to prepare 855.0 mL of 1.00 M HCl (aq) ?

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  1. 26 May, 20:21
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    The answer to your question is: V1 = 74.35 ml

    Explanation:

    Data

    Volume 1 = V1 = ?

    Concentration 1 = C1 = 11.5 M

    Volume 2 = V2 = 855 ml

    Concentration 2 = 1 M

    Formula

    C1V1 = C2V2

    V1 = C2V2 / C1

    Process

    V1 = (1 x 855) / 11.5

    V1 = 855 / 11.5

    V1 = 74.35 ml
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