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)

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

( 1L = 1000 mL)

Mass of water = m

Density of water = d = 1.0 g/mL

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:

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

Percentage of fluorine in NaF = 45.0 %

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)

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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

a electronegativity decreases beacuse the size of nucleus is bigger.