Fluoridation is the process of adding fluorine compounds to drinking water to he

Question

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 tooth pastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 1,500,000 people if the daily consumption of water per person is 75 gallons. What percentage of the sodium fluoride is wasted if each person uses only 4.5 L of water a day for drinking and cooking? (Sodium fluoride is 45.0 % fluorine by mass. 1 gallon = 3.79 L, 1 year = 365 days; 1 ton = 2000 pounds, 1 pound = 453.6 g and the density of water is assumed 1.0 g/mL). 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 tooth pastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 1,500,000 people if the daily consumption of water per person is 75 gallons. What percentage of the sodium fluoride is wasted if each person uses only 4.5 L of water a day for drinking and cooking? (Sodium fluoride is 45.0 % fluorine by mass. 1 gallon = 3.79 L, 1 year = 365 days; 1 ton = 2000 pounds, 1 pound = 453.6 g and the density of water is assumed 1.0 g/mL).

Explanation / Answer

daily consumption of water per person= 75 gallons= 75*3.79 L =284.25 L

density =1 g/ml =1000g/L

dailiy consumption in terms of mass = 284.25*1000 g=284250 gms

total consumption of water for 1,500,000 people = 1.500.000*284250 =4.26*1011 gms

1ppm of fluoride correspond to 1 g /106 gm of water

106 gm of water contains 1 gm of fluoride

4.26*1011 gm contains   4.26*1011/106 =426000 mgs = 426 kg of fluoride

Fluoride is supplimented through NaF which contains 45 fluoride by mass, so NaF= 426*45/100=191.7 kgs

when 4.5 L of water is wasted, fluoride wasted= (4.5/75)*191.7 =11.502 kg

percentage of fluoride wasted= 100*11.502/191.7=6%

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