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18 January, 21:44

The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0-ppb solution. What assumption did you have to make in your calculation? (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 m3 of water?

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  1. 19 January, 01:11
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    0.54 g of Pb in 60 m³ of [Pb] = 9 ppb

    Explanation:

    We need to know, that ppb is a sort of concentration that indicates (in value of volume) ng / mL (ppb = parts per billion)

    9 ppb means that 9 ng are contained in 1mL of solution

    We know that 1mL = 1 cm³

    Let's convert 60 m³ to cm³ → 60 m³. 1*10⁶ cm³ / 1m³ = 6*10⁷ cm³

    Then, we can make a rule of three:

    1 cm³ has 9 ng of Pb

    Therefore in 6*10⁷ cm³ we must have (6*10⁷ cm³. 9ng) / 1 cm³ =

    5.4*10⁸ ng

    We convert ng to g → 5.4*10⁸ ng. 1 g / 1*10⁻⁹ ng = 0.54 g
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