New York City’s 7.9 million people in 1978 had a daily per capita consumption of 676
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-strengthening dose of 1
part (by weight) fluorine per million parts water? The density of water is 1.000 g/cm3, or
1.000 kg/L.
How many liters.
7.9E6 people x 676 L/person/yr = 5.340E9 L
1 ppm = 1 mg F^-/L = 1 mg x (1/0.45)/L so for 5.34E9L you will need
1 mg x 1/0.45 x 5.34E9 L = total mg
Convert total mg to metric tons (10^3 kg) you will need of NaF.
Well, that's quite a tricky question! Let's break it down with a sprinkle of humor.
First, we need to find the total water consumption per day in liters. So, 7.9 million people consuming 676 liters of water per day would result in a total of... *drumroll*... a whopping number! Yes, it's 5,324,000,000 liters of water per day!
Now, onto our goal of achieving a tooth-strengthening dose of 1 part fluorine per million parts of water. If we look at this ratio, it's like trying to find a needle in a haystack!
To calculate the amount of fluorine required, we need to convert the liters of water into metric tons (103 kg). Using the given density of water (1.000 kg/L), we can divide the total liters by 1,000 to get the total water in metric tons.
So, 5,324,000,000 / 1,000 = 5,324,000 metric tons of water per day!
Now, we need to find the amount of fluorine needed to reach our desired concentration of 1 part per million. Since sodium fluoride is 45% fluorine by weight, we can calculate the amount required by multiplying the total water in metric tons by the desired concentration.
5,324,000 x (1/1,000,000) = 5.324 metric tons of fluorine!
Voila! To give New York City's 7.9 million residents a tooth-strengthening dose of fluorine, we would need approximately 5.324 metric tons of sodium fluoride per year.
Remember, though, this is all just a playful calculation, and if you have any specific requirements for water intake or fluoridation, it's always best to consult with a dental professional or an expert.
To calculate the amount of sodium fluoride required, we need to consider the per capita consumption of water and the desired concentration of fluorine. Here is the step-by-step solution:
Step 1: Calculate the total annual water consumption
Total water consumption per year = per capita consumption x population x number of days in a year
Given:
Per capita consumption = 676 liters
Population = 7.9 million people
Number of days in a year = 365 days
Total water consumption per year = 676 L/person x 7.9 million people x 365 = 2,074,462,000,000 liters
Step 2: Calculate the desired concentration of fluorine
Desired concentration of fluorine = 1 part per million (ppm) = 1 mg/L = 1 mg/1,000,000 mg
Step 3: Calculate the amount of fluorine required per year
Amount of fluorine required per year = Total water consumption per year x desired concentration of fluorine
Given:
Total water consumption per year = 2,074,462,000,000 liters
Desired concentration of fluorine = 1 mg/1,000,000 mg
Amount of fluorine required per year = 2,074,462,000,000 L x 1 mg/1,000,000 mg = 2,074,462 mg
Step 4: Convert milligrams to metric tons
1 metric ton = 1,000,000 grams = 1,000,000,000 milligrams
Amount of fluorine required per year in metric tons = Amount of fluorine required per year / 1,000,000,000
Amount of fluorine required per year in metric tons = 2,074,462 mg / 1,000,000,000 = 0.002074462 metric tons
Therefore, approximately 0.0021 metric tons (or 2.074462 kg) of sodium fluoride would be required per year to give New York City's water a tooth-strengthening dose of 1 part fluorine per million parts water.
To calculate the amount of sodium fluoride required, we need to go through a series of steps.
Step 1: Calculate the total water consumption per day
To find the total water consumption per day, multiply the per capita consumption (676 liters) by the population (7.9 million):
Total water consumption per day = Per capita consumption × Population
Total water consumption per day = 676 liters × 7.9 million
Total water consumption per day = 5,346.4 million liters
Step 2: Convert liters of water to kilograms
Since the density of water is 1.000 kg/L, we can convert liters to kilograms by using this conversion factor:
Total water consumption per day (in kg) = Total water consumption per day (in liters) × Density of water
Total water consumption per day (in kg) = 5,346.4 million liters × 1.000 kg/L
Total water consumption per day (in kg) = 5,346.4 million kg
Step 3: Calculate the amount of fluorine required per day
The desired dose is 1 part of fluorine per million parts of water. So, the amount of fluorine required per day is:
Fluorine required per day = Total water consumption per day (in kg) / 1 million
Fluorine required per day = 5,346.4 million kg / 1 million
Fluorine required per day = 5,346.4 kg
Step 4: Calculate the amount of sodium fluoride required per day
The sodium fluoride contains 45% fluorine by weight. So, to find the amount of sodium fluoride required per day, we divide the fluorine required by the weight percentage of fluorine in sodium fluoride:
Sodium fluoride required per day = Fluorine required per day / Weight percentage of fluorine in sodium fluoride
Sodium fluoride required per day = 5,346.4 kg / 0.45
Sodium fluoride required per day = 11,881.8 kg
Step 5: Convert kilograms to metric tons
To convert kilograms to metric tons, divide the sodium fluoride required per day by 1000:
Sodium fluoride required per day (in metric tons) = Sodium fluoride required per day (kg) / 1000
Sodium fluoride required per day (in metric tons) = 11,881.8 kg / 1000
Sodium fluoride required per day (in metric tons) = 11.8818 metric tons
Therefore, approximately 11.8818 metric tons of sodium fluoride would be required per year to give New York City’s population a tooth-strengthening dose of 1 part fluorine per million parts of water.