At the time of the 2010 census, St. Paul, MN had a population of 285,068 people.
ID: 850474 • Letter: A
Question
At the time of the 2010 census, St. Paul, MN had a population of 285,068 people. If the city consumes 170 gallons of water per person per day, what mass of sodium fluoride (NaF), in Kilograms, must be added to the water supply each year (365) to produce a fluoride ion concentration (not NaF concentration) of 1 ppm (part per million)? NaF is 45% fluoride by mass.
This was a test question and my professor said I went about it the wrong way.... I always thought ppm was mg/L in water chemistry. I suppose I was wrong. If you could work it all the way through that would be great. I got something like 1.48*10^5 KG but obviously thats wrong. Thank you in advance!
Explanation / Answer
Population = 285065
Water consumed = 170 gallons per person per day
Total water consumed per day = 170 * 285065 = 48461050 gallons per day
Total days in a year = 365
Total water consumed in one year = 365 * 48461050 = 17688283250 gallons
We use 1 gallon = 3.78 litres
Volume of water = 17688283250 * 3.78 = 66861710685 litres
We use 1 ppm = 1 mg / L
Concentration = Mass of Flouride ion (in mg) / Volume (in Litre)
or
Mass of Flouride ion (in mg) = Concentration * Volume (in Litre)
Mass of Flouride ion (in mg) = 1 * 66861710685 = 66861710685 mg
We know 1 g = 1000 mg and 1 kg = 1000 g therefore 1 kg = 10^6 mg
Mass of Flouride ion = 66861710685 mg = 66861.710685 kg
Let mass of NaF = M kg
Flouride ion is 45 percent by mass
Mass of Flouride = 45/100 * M = 0.45 * M kg
0.45 * M = 66861.710685 kg
M = 66861.710685 kg / 0.45 = 148581.5793 kg (Answer)
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