Question

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?

Answer 1

Answer:

(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

Answer 2

(a) The diameter of cylindrical water reservoir is $7.96\times 10^{2}m$ and its height is 45.54 m.

Radius of cylinder is half of its diameter thus,

$r=\frac{d}{2}=\frac{7.96\times 10^{2} m}{2}=3.98\times 10^{2}m$

Volume of cylinder can be calculated as follows:

$V=\pi r^{2}h$

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

$V=(3.14)(3.98\times 10^{2}m)^{2}(45.54 m)=2.265\times 10^{7}m^{3}$

Since, $1m^{3}=1000 L$

Thus, volume will be $2.265\times 10^{10}L$

The concentration of fluoride $F^{-}$ is approximately 1 ppm.

Since, 1 ppm=1 mg/L

thus,

$1 ppm=10^{-3}g/L$

Mass can be calculated as follows:

m=C×V=$10^{-3}g/L\times 2.265\times 10^{10}L=2.265\times 10^{7}g$

Therefore, mass of fluoride will be $2.265\times 10^{7}g$.

(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 $2.265\times 10^{7}g$ and its molar mass is 19 g/mol thus, number of moles will be:

$n=\frac{m}{M}=\frac{2.265\times 10^{7}g}{19 g/mol}=1.192\times 10^{6} mol$

Thus, number of moles of NAF formed will be $1.192\times 10^{6} mol$

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

$m=n\times M=1.192\times 10^{6} mol\times 41.98 g/mol\approx 5.0\times 10^{7} g$.

Therefore, grams of NaF containing this much of fluoride is $5.0\times 10^{7} g$.