To calculate the amount of sodium fluoride required per year, we need to find the total volume of water consumed by New York City's population in 1978 and then determine the amount of sodium fluoride needed to achieve the desired concentration of fluorine.
Total volume of water consumed per day:
7.9 million people * 656 liters/person = 5,170,400,000 liters/day
Total volume of water consumed per year:
5,170,400,000 liters/day * 365 days/year = 1,886,856,000,000 liters/year
To convert liters to kg, we need to multiply by the density of water:
1,886,856,000,000 liters/year * 1000 kg/L = 1,886,856,000,000,000 kg/year
To achieve a concentration of 1 part fluorine per million parts water, we need 1 kg of sodium fluoride for every million kg of water.
Amount of sodium fluoride required per year:
1,886,856,000,000,000 kg/year / 1,000,000 = 1,886,856,000 kg/year
Since we want the amount of sodium fluoride in metric tons, we divide by 1000:
1,886,856,000 kg/year / 1000 = 1,886,856 metric tons/year
Since sodium fluoride is 45% fluorine by weight, the amount of fluoride in the required sodium fluoride is:
1,886,856 metric tons/year * 45% = 848,085.2 metric tons/year
Therefore, approximately 848,085.2 metric tons of sodium fluoride would be required per year to provide the New York City population in 1978 with a tooth-strengthening dose of 1 part fluorine per million parts water.
New York city's 7.9 million people in 1978 had a daily per Capita consumption of 656 liters of water. How many metric tons (103 kg) of sodium fluoride (45% fluorine by weight) would be required per year to give this water a tooth strenghting dose of 1 part ( by weight) fluorine per million parts water? The density of water is 1000 g/cm3, or 1000 kg/L.
1 answer