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 find the amount of sodium fluoride required per year, we need to calculate the total consumption of water in liters per year and then convert it to metric tons.
Given:
Population in 1978 = 7.9 million people
Daily per capita consumption of water = 656 liters
Total water consumption per day = Population × Daily per capita consumption
Total water consumption per day = 7.9 million × 656 liters
Total water consumption per day = 5,164,000,000 liters
Total water consumption per year = Total water consumption per day × 365 days
Total water consumption per year = 5,164,000,000 liters × 365
Total water consumption per year = 1,884,760,000,000 liters
Now, we need to calculate the amount of sodium fluoride required to achieve 1 part fluorine per million parts of water.
1 part fluorine per million parts of water means the concentration of fluorine is 1 ppm.
In other words, we need 1 gram of fluorine for every 1 million grams (or liters) of water.
Total amount of fluorine required per year = Total water consumption per year × Fluorine concentration in ppm
Total amount of fluorine required per year = 1,884,760,000,000 liters × 1 ppm (1 gram per million grams)
Total amount of fluorine required per year = 1,884,760,000,000 grams
Since the density of water is 1000 kg/L, we can convert the amount of fluorine required to metric tons.
Total amount of fluorine required per year (in metric tons) = Total amount of fluorine required per year (in grams) / 1,000,000 (conversion factor)
Total amount of fluorine required per year (in metric tons) = 1,884,760,000,000 grams / 1,000,000
Total amount of fluorine required per year (in metric tons) = 1,884,760 metric tons
To calculate the amount of sodium fluoride required, we need to consider that sodium fluoride is 45% fluorine by weight.
Amount of sodium fluoride required per year (in metric tons) = Total amount of fluorine required per year (in metric tons) / Fluorine content in sodium fluoride
Amount of sodium fluoride required per year (in metric tons) = 1,884,760 metric tons / 0.45
Amount of sodium fluoride required per year (in metric tons) = 4,188,244 metric tons
Therefore, approximately 4,188,244 metric tons of sodium fluoride would be required per year to give New York City's 7.9 million people a tooth-strengthening dose of 1 part fluorine per million parts water.
To calculate the amount of sodium fluoride required per year, we need to consider the following information:
1. Population of New York City in 1978: 7.9 million people
2. Daily per capita water consumption: 656 liters
3. Sodium fluoride dose: 1 part fluorine per million parts water
4. Fluorine concentration in sodium fluoride: 45% fluorine by weight
5. Density of water: 1000 kg/L
Let's calculate the annual amount of water consumed by the population first:
Annual water consumption = Daily per capita water consumption * Population * 365 days
= 656 L/person/day * 7.9 million people * 365 days
= 1,885,684,400,000 L
Next, we need to calculate the fluorine concentration in the water:
Fluorine concentration = 1 part per million = 1 mg/L
Since the density of water is 1000 kg/L, the concentration can also be expressed as 1 mg/kg.
Now, let's determine the amount of sodium fluoride required to achieve this concentration:
Amount of sodium fluoride (in kg) = Annual water consumption * Fluorine concentration
= 1,885,684,400,000 L * 1 mg/kg
= 1,885,684,400 kg
Lastly, we convert the result to metric tons by dividing by 1000:
Amount of sodium fluoride (in metric tons) = 1,885,684,400 kg / 1000
= 1,885,684.4 metric tons
Therefore, approximately 1,885,684.4 metric tons of sodium fluoride would be required per year to give the population of New York City a tooth-strengthening dose of 1 part fluorine per million parts water.