Question

fluoride is known to prevent tooth decay so it is added to the city water supply for drinking. A medium sized city is populated with 150,000 people. Each person appeoximately 660.0 L of water each day. How many kg of NaF must be added to the water supply each year to have the required fluoride concentration of 1.0 ppm? How to get the answer of 8.0×10^4 kgNaF/year?

Answer 1

Given the population of city = 150,000 people

Volume of water consumed by each person = 660.0 L

Total volume consumed in the city every year = $150,000 persons * \frac{660.0L}{1 person*1day}$*$\frac{365 days}{1 yr}$=$3.61*10^{10}L$/year

The required fluoride concentration is 1.0 ppm

1.0 ppm means 1.0 mg fluoride per liter of water

Calculating the mass of NaF that must be added in kg:

$3.61*10^{10} L/year*\frac{1 mg F^{-} }{1 L}*\frac{1g}{1000mg}*\frac{1molF^{-} }{19.00g}*\frac{1 mol NaF}{1 mol F^{-}}*\frac{42g}{1molNaF}*\frac{1kg}{1000g}$

                        = $8.0 * 10^{4} kg NaF/year$