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14 June, 14:51

If you wish to prepare a 1.13 M solution of NaNo3, to what volume (in liters) would you have to dilute 25.4 mL of 3.21 M NaNO3?

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  1. 14 June, 15:08
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    Answer is 72.15 or 0.07215

    Explanation:

    Given,

    Initial molarity M1 = 3.21M

    Initial volume V1 = 25.4mL

    Final molarity of M2 = 1.13 M

    Final volume V2 = ?

    M1*V1=M2*V2

    M1*V1/M2=V2

    3.21M*25.4/1.13=V2

    V2=72.15 mL. 1 mL=0.001 L

    V2 = 0.07215
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