Question

It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir with a diameter of 4.50 X 10' m and a depth of 10.0 m. (The volume is rh, where r is the radius and h is the height.) How many grams of F should be added to give 1.6 ppm? Fluoride is provided by hydrogen hexafluorosilicate, H. SiF. How many grams of H Sif contain this much F?

Answer 1

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Step-by-step explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =
`4.50* 10^1 m=45 m`

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

`V=\pi r^2 h`

`=3.14* (22.5 m)^2* 10.0 m=15,896.25 m^3=15,896,250 L`

`1 m^3=1000 l`

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

`m=d* V=1 kg/L* 15,896,250 L=15,896,250 kg`

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then
`15,896,250 kg` of water have x mass of fluorine:

`\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}`

`x=(1.6 kg)/(1000,000)* 15,896,250 kg=25.434 kg`

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

`F\%=(6* 19 g/mol)/(144 g/mol)* 100=79.16\%`

Total mass of hydrogen hexafluorosilicate = m'

`79.16\%=(25,434 g)/(m')* 100`

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.