Answer to Question #4329 in Chemistry for sarah

Question

You are the public health officer in the water treatment facility for a city of 50,000 people. A concentration of 1.0 ppm of flourine in the drinking water is sufficient for the purpose of helping to prevent tooth decay. (1 ppm means "one part per million," or 1 g of flourine per one million grams of water). The compound normally chosen for flouridation is the same as is found in some toohtpaste's sodium flouride. You know that sodium flouride is 45% flourine by mass. Calculate how many kilograms of sodium flouride you will need to have in order to flouridate the city's water supply for one year, based upon your estimate that the average daily consumption of water is 150 gallons per person. (1 gallon = 3.79 L; 1 year = 365 days; 1 ton = 2000 lb; 1 lb = 456.3 g; density of water = 1.0 g/mL )

Answer 1

V_general(H₂O) = 50 000*150*3.79*365 = 1.4*10¹⁰ L
m _general(H₂O) = 1.4*10¹⁰*1 = 1.4*10¹⁰ kg
m(F)_H2O = m(F)/(m(F) + m(H₂O))(10^-6) = m(F)/(m(F) + 1.4*10^10)*(10^-6)
m(F) = 1.4*10^4 kg
(NaF) = 1.4*10^4/0.45 = 2.3*10^4 kg