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?

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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