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 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 115.0 gallons. (Sodium fluoride is 45.0 percent fluorine by mass. 1 gallon = 3.79 L; 1 ton=2000lb; 1 lb= 453.6 g; density of water =1.0 g/mL)

Answer 1

Answer:

The quantity of sodium fluoride is 17,676,14 kilograms is needed per year for a city of 50,000 people.

Explanation:

Population of the city = 50,000

Volume of water consumed by an individual in day= 115.0 gallons

Volume of water consumed by 50,000 people in day: V

V = 115.0 gallons × 50,000 = 5,750,000 gallons

1 gallon = 3.79 L

$V=5,750,000 gallons=5,750,000\times 3.79 L = 21,792,500 L$

$V=21,792,500 L=21,792,500,000 mL$

( 1L = 1000 mL)

Mass of water = m

Density of water = d = 1.0 g/mL

$m=d\times V=1.0 g/mL\times 21,792,500,000 mL=21,792,500,000 g$

Concentration of Fluorine in water = 1 ppm = 1 gram of F /1 million grams of water

Then mass of fluorine present in 21,792,500,000 g of water:

$\frac{1}{10^6}\times 21,792,500,000 g=21,792.5 g$

Mass of sodium fluoride with 21,792.5 g of F = M

Percentage of fluorine in NaF = 45.0 %

$45\%=\frac{21,792.5 g}{M}\times 100$

M = 48,427.78 g = 48.427 kg ( 1g = 0.001 kg)

48.427 kg of NaF should be added to water in day for city population.

Amount of NaF needed per year for city with 50,000 population :

(1 year = 365 days)

$48.427 kg\times 365=17,676.14 kg$

The quantity of sodium fluoride is 17,676,14 kilograms is needed per year for a city of 50,000 people.