Question

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?

Answer 1

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

Answer 2

The mass of NaF needed to be added to 1500 L of water to fluoridate it at a level of 0.7 mg/L F ¯ is 2.32 g

How to determine the mole of F¯ in the solution

We'll begin by calculating the mass of F¯. This can be obtained as follow:

• Molarity of F¯ = 0.7 mg/L = 0.7×10¯³ g/L
• Volume = 1500 L

• Mass of F¯ =?

Mass of F¯ = Molarity (g/L) × Volume

Mass of F¯ = 0.7×10¯³ × 1500

Mass of F¯ = 1.05 g

Thus, the mole of F¯ can be obtained as follow:

• Mass of F¯ = 1.05 g
• Molar mass of F¯ = 19 g/mol
• Mole of F¯ =?

Mole = mass / molar mass

Mole of F¯ = 1.05 / 19

Mole of F¯ =  0.0553 mole

How to determine the mass of NaF needed

Balanced equation

NaF(aq) —> Na⁺(aq) + F¯(aq)

From the balanced equation above,

1 mole of NaF produced 1 mole of F¯.

Therefore,

0.0553 mole of NaF will also produce 0.0553 mole of F¯

Thus, the mass of NaF needed can be obtained as follow:

• Mole of NaF = 0.0553 mole
• Molar mass of NaF = 23 + 19 = 42 g/mol

• Mass of NaF =?

Mass = mole × molar mass

Mass of NaF = 0.0553 × 42

Mass of NaF = 2.32 g

Learn more about stoichiometry:

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