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.
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.
To calculate the amount of sodium fluoride required per year to achieve a tooth-strengthening dose of 1 part fluorine per million parts water, we need to follow these steps:
Step 1: Calculate the total water consumption per day.
Given that the daily per capita consumption of water in 1978 was 656 liters and the population of New York City was 7.9 million, we can calculate the total water consumption per day as follows:
Total water consumption per day = Daily per capita consumption * Population
Total water consumption per day = 656 liters * 7.9 million
Total water consumption per day = 5,170,400,000 liters
Step 2: Convert the water consumption to metric tons per year.
Since the density of water is 1000 kg/L or 1000 tons/L, we can convert the total water consumption per day to metric tons per year:
Total water consumption per year = Total water consumption per day * 365 days * (1/1000)
Total water consumption per year = 5,170,400,000 liters * 365 * (1/1000)
Total water consumption per year = 1,887,196,000 tons
Step 3: Calculate the amount of fluorine required per year.
To achieve 1 part fluorine per million parts water, we need to convert parts per million (ppm) to kilograms of fluorine per million kilograms of water (kg/kg):
1 ppm = 1 kg/1,000,000 kg = 0.001 kg/kg
Amount of fluorine required per year = Total water consumption per year * Fluorine concentration
Amount of fluorine required per year = 1,887,196,000 tons * 0.001 kg/kg
Amount of fluorine required per year = 1,887,196 tons
Step 4: Calculate the mass of sodium fluoride required.
Given that sodium fluoride is 45% fluorine by weight, we can calculate the mass of sodium fluoride required:
Mass of sodium fluoride required = Amount of fluorine required per year / Fluorine content in sodium fluoride
Mass of sodium fluoride required = 1,887,196 tons / 0.45
Mass of sodium fluoride required = 4,193,103 tons (approximately)
Therefore, approximately 4,193,103 metric tons of sodium fluoride would be required per year to achieve a tooth-strengthening dose of 1 part fluorine per million parts water for New York City's population of 7.9 million in 1978.