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3 June, 08:15

Find the volume needed to prepare a 0.694 M Pb (NO3) 2 solution if you start with 110 g of Pb (NO3) 2

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  1. 3 June, 09:58
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    0.479 dm^3 (or 479 cm^3)

    Explanation:

    First, calculate the no. of moles in 110g of Pb (NO3) 2.

    According to the periodic table, molar mass of Pb is 207.2, N is 14,0, O is 16.0

    so, the molar mass of Pb (NO3) 2 = 207.2 + (14.0 + 16.0 x 3) x 2 = 331.2

    With the formula:

    no. of moles = mass/molar mass

    no. of moles in 110g of Pb (NO3) 2 = 110 / 331.2

    = 0.33213 mol

    Then, with the formula:

    molarity = no. of moles / volume

    0.694 = 0.33213 / v

    v = 0.479 dm^3 (or 479 cm^3)

    Therefore, the volume needed is 0.479 dm^3 (or 479 cm^3)
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