Fluoridation is the process of adding fluorine to drinking water to avoid tooth decay.
The question is incomplete; the complete question is;
Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose. (1 ppm means one part per million, or 1 g of fluorine per 1 million g of water.) The compound normally chosen for fluoridation is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gallons.
(Sodium fluoride is 45.0 percent fluorine by mass. 1 gallon = 3.79 L; 1 year = 365 days; 1 ton = 2000 lb; I lb = 453.6 g; density of water = 1.0 g/mL.)
The quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gallons is [tex]9.32 * 10^9 Kg[/tex].
From the question, we know that;
Daily consumption of water per person = 150 gallons.
Population of the city = 100,000 people
Total daily city consumption =[tex]100,000 * 150 gallons = 15 * 10^6 gallons[/tex]
Annual water consumption = [tex]15 * 10^6 gallons * 365 = 5.475 * 10^9 gallons[/tex]
1 gallon = 3.785 L = 3.785 kg
[tex]5.475 * 10^9 gallons = 5.475 * 10^9 gallons * 3.785 kg/ 1 gallon = 2.1 * 10^10 Kg[/tex]
Since sodium fluoride accounts for 45% by mass of water
Mass of sodium fluoride= [tex]0.45 * 2.1 * 10^10 Kg= 9.32 * 10^9 Kg[/tex]
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