Respuesta :
(a) The diameter of cylindrical water reservoir is [tex]7.96\times 10^{2}m[/tex] and its height is 45.54 m.
Radius of cylinder is half of its diameter thus,
[tex]r=\frac{d}{2}=\frac{7.96\times 10^{2} m}{2}=3.98\times 10^{2}m[/tex]
Volume of cylinder can be calculated as follows:
[tex]V=\pi r^{2}h[/tex]
Here, r is radius and h is height, putting the values,
[tex]V=(3.14)(3.98\times 10^{2}m)^{2}(45.54 m)=2.265\times 10^{7}m^{3}[/tex]
Since, [tex]1m^{3}=1000 L[/tex]
Thus, volume will be [tex]2.265\times 10^{10}L[/tex]
The concentration of fluoride [tex]F^{-}[/tex] is approximately 1 ppm.
Since, 1 ppm=1 mg/L
thus,
[tex]1 ppm=10^{-3}g/L[/tex]
Mass can be calculated as follows:
m=C×V=[tex]10^{-3}g/L\times 2.265\times 10^{10}L=2.265\times 10^{7}g[/tex]
Therefore, mass of fluoride will be [tex]2.265\times 10^{7}g[/tex].
(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 [tex]2.265\times 10^{7}g[/tex] and its molar mass is 19 g/mol thus, number of moles will be:
[tex]n=\frac{m}{M}=\frac{2.265\times 10^{7}g}{19 g/mol}=1.192\times 10^{6} mol[/tex]
Thus, number of moles of NAF formed will be [tex]1.192\times 10^{6} mol[/tex]
Molar mass of NaF is 41.98 g/mol, mass can be calculated as follows:
[tex]m=n\times M=1.192\times 10^{6} mol\times 41.98 g/mol\approx 5.0\times 10^{7} g[/tex].
Therefore, grams of NaF containing this much of fluoride is [tex]5.0\times 10^{7} g[/tex].
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
