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23 April, 08:26

how many grams of naf must be added to 1500 liters of water to fluoridate it at a level of 0.7 mg f - / l?

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  1. 23 April, 08:34
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    Given:

    Volume of water = 1500 L

    Amount of F - ion = 0.7 mg/L

    To determine:

    Amount of NaF that must be added to obtain the given levels of F - ions

    Explanation:

    Atomic weight of F = 19 g/mol

    # moles of F - in 1 L of water = 0.7*10⁻³ g/19 g mol⁻¹ = 3.684*10⁻⁵ moles

    Therefore, # moles of F - in 1500 L of water = 3.684*10⁻⁵ * 1500 = 0.0553 moles

    Based on the stoichiometry of NaF:

    1 mole of NaF has 1 mole of Na + and 1 mole of F-

    Therefore, # moles of NaF required = 0.0553 moles

    Molar mass of NaF = 23 + 19 = 42 g/mol

    Mass of NaF required = 0.0553 moles * 42 g/mol = 2.323 g

    Ans : NaF = 2.323 g
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