To find the amount of sodium fluoride required per year to give the water a tooth-strengthening dose of 1 part fluorine per million parts water, we need to calculate the total amount of water consumed per year and then determine the amount of sodium fluoride needed to achieve the desired concentration.
Step 1: Calculate the total amount of water consumed per year.
Given:
- New York City's population in 1978: 7.9 million people
- Daily per capita consumption of water: 656 liters
Total water consumed per year = (Daily per capita consumption) * (Number of people) * (Days in a year)
Total water consumed per year = 656 liters * 7.9 million * 365 days
Step 2: Calculate the amount of sodium fluoride needed to achieve the desired fluorine concentration.
Given:
- Sodium fluoride contains 45% fluorine by weight.
- Desired fluorine concentration: 1 part per million parts water.
To convert the desired concentration to a weight ratio, we use the density of water:
1 part per million parts water = 1 gram per 1 million grams water = 1 gram per 1 million milliliters water = 1 gram per 1 million liters water.
Weight of sodium fluoride needed per year = Total water consumed per year * (Desired fluorine concentration / 1 million)
Weight of sodium fluoride needed per year = (Total water consumed per year) * (1 / 1 million)
Step 3: Convert the calculated weight to metric tons.
Weight of sodium fluoride needed per year = Weight of sodium fluoride needed per year / (1000 kg/ton) = Weight of sodium fluoride needed per year / 1000 metric tons/kg
Combining the steps and given information, we can calculate the amount of sodium fluoride needed per year:
Total water consumed per year = 656 liters * 7.9 million * 365 days
Total water consumed per year = 1,750,668,000,000 liters
Weight of sodium fluoride needed per year = (Total water consumed per year) * (1 / 1 million)
Weight of sodium fluoride needed per year = 1,750,668,000,000 * (1 / 1,000,000) grams
Converting grams to metric tons:
Weight of sodium fluoride needed per year = 1,750,668,000,000 / 1,000,000 metric tons
Weight of sodium fluoride needed per year = 1,750.668 metric tons
Therefore, approximately 1,750.668 metric tons of sodium fluoride would be required per year to give New York City's 7.9 million people in 1978 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_ strengthing dose of 1 part(by weight) fluorine per million parts water? The density of water is 1000g/cm^3, or 1000kg/L
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