How many grams of f– must be added to a cylindrical water reservoir having a diameter of 7.96 × 102 m and a depth of 45.54 m? (b) how many grams of sodium fluoride, naf, contain this much fluoride?

(a) The diameter of cylindrical water reservoir is and its height is 45.54 m.

Radius of cylinder is half of its diameter thus,

Volume of cylinder can be calculated as follows:

Here, r is radius and h is height, putting the values,

Since,

Thus, volume will be

The concentration of fluoride is approximately 1 ppm.

Since, 1 ppm=1 mg/L

thus,

Mass can be calculated as follows:

m=C×V=

Therefore, mass of fluoride will be .

(b) In NaF, sodium and florine are present in 1:1 ratio thus, 1 mole of fluoride gives 1 mole of NaF.

Mass of fluoride is and its molar mass is 19 g/mol thus, number of moles will be:

Thus, number of moles of NAF formed will be

Molar mass of NaF is 41.98 g/mol, mass can be calculated as follows:

.

Therefore, grams of NaF containing this much of fluoride is .

(a) 1.812×10⁷ grams of F⁻ must be added.

(b) 4.0068×10⁷ grams of sodium fluoride, NaF, contain this much fluoride.

Explanation:

You didn't show it in your question, but assuming drinking water contain 0.800 ppm fluoride.

Since, Volume of cylinder, V = πr²h

radius, r = Diameter / 2 ⇒ 7.96×10² / 2 ⇒ 398

depth, h = 45.54 m

V = πr²h ⇒ 3.14 × (398)² × 45.54

  ⇒ 2.265×10⁷ m³ × (100cm/m)³

  ⇒ 2.265×10¹³ cm³

  ⇒ 2.265×10¹³ mL

V ⇒ 2.265×10¹⁰ L

(a)

Mass of F⁻ = (0.8 mg/L) × (2.265×10¹⁰ L)

                 = 1.812×10¹⁰ mg

or               = 1.812×10⁷ g of F⁻ to be added

(b)

Mass of NaF = 1.812×10⁷ g F  ÷  19 g/mole F

                 = 953684.21

                 = 9.54×10⁵ moles F present

1 mole F    :    1 mole NaF

⇒ 9.54×10⁵ moles NaF × 42 g/mol

     = 4.0068×10⁷ /g NaF

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explanation:

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