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 (103kg) 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 1000 g/cm^3, or 1000kg/L
To find the amount of sodium fluoride required, we need to calculate the total volume of water used per year and then determine the quantity of fluorine needed.
1. Total volume of water used per year:
We know that in 1978, the per capita daily consumption of water was 656 liters. Therefore, the total volume of water used per year is:
656 liters/day x 365 days/year = 239,240 liters/year
2. Converting the volume of water to mass:
We know that the density of water is 1000 kg/L. So, the mass of water used per year would be:
239,240 liters x 1000 kg/L = 239,240,000 kg/year
3. Calculating the amount of fluorine needed:
To achieve a tooth-strengthening dose of 1 part fluorine per million parts of water, we need to find the mass of fluorine required per year. One part per million (ppm) is equal to 1 milligram per liter.
1 ppm = 1 mg/L
Since water has a density of 1000 g/L (or 1 kg/L), 1 part per million (ppm) is also equal to 1 mg/kg. Therefore, we need 1 mg of fluorine per kilogram of water.
To calculate the amount of sodium fluoride required, we need to determine the mass of fluorine in sodium fluoride. Sodium fluoride is 45% fluorine by weight, so the mass of fluorine in one kilogram of sodium fluoride would be:
Mass of fluorine = 45% x 1 kg = 0.45 kg
Now, we can calculate the total mass of sodium fluoride needed per year:
Mass of sodium fluoride = (mass of fluorine / % fluorine) x mass of water
= (0.45 kg / 0.45) x 239,240,000 kg
= 239,240,000 kg
Therefore, approximately 239,240,000 kg (or 239,240 metric tons) of sodium fluoride would be required per year to give the water 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 follow these steps:
Step 1: Find the total water consumption per year.
Step 2: Calculate the fluorine content required per year.
Step 3: Convert the fluorine requirement to sodium fluoride.
Step 4: Determine the amount of sodium fluoride required.
Let's calculate it step by step:
Step 1: Find the total water consumption per year.
The daily water consumption per capita is given as 656 liters.
So, the total water consumption per year is:
Total water consumption per year = daily water consumption per capita * number of people * 365 days
Total water consumption per year = 656 L * 7.9 million * 365
Step 2: Calculate the fluorine content required per year.
We need 1 part fluorine per million parts of water.
So, the fluorine content required per year is:
Fluorine content = Total water consumption per year * (1/1,000,000)
Step 3: Convert the fluorine requirement to sodium fluoride.
Since sodium fluoride contains 45% fluorine by weight, we need to adjust our calculation:
Sodium fluoride content = Fluorine content / 45%
Step 4: Determine the amount of sodium fluoride required.
The density of water is given as 1000 kg/L, or 1 kg/L.
So, the amount of sodium fluoride required per year is:
Sodium fluoride required = Sodium fluoride content / density of water
Let's calculate the values:
Step 1: Total water consumption per year:
Total water consumption per year = 656 L * 7.9 million * 365
Step 2: Fluorine content required per year:
Fluorine content = Total water consumption per year * (1/1,000,000)
Step 3: Sodium fluoride content:
Sodium fluoride content = Fluorine content / 45%
Step 4: Amount of sodium fluoride required per year:
Sodium fluoride required = Sodium fluoride content / 1 kg/L
This calculation will give us the final answer.