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23 July, 08:38

What volume (in L) of a 2.46 M.

magnessium nitrate (Mg (NO3) 2)

solution would be needed to make

275 mL of a 0.758 M solution by

dilution?

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Answers (1)
  1. 23 July, 08:49
    0
    0.0847L

    Explanation:

    The following data were obtained from the question:

    C1 = 2.46 M

    V1 = ?

    C2 = 0.758 M

    V2 = 275 mL = 275/1000 = 0.275L

    Using the dilution formula C1V1 = C2V2, we can confidently calculate the volume of the original solution as follows:

    C1V1 = C2V2

    2.46 x V1 = 0.758 x 0.275

    Divide both side by 2.46

    V1 = (0.758 x 0.275) / 2.46

    V1 = 0.0847L

    Therefore, 0.0847L is needed
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