Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose. (1 ppm means \(1 \mathrm{~g}\) of fluorine per 1 million g of water.) The compound normally chosen for fluoridation is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium fluoride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gallons. What percent of the sodium fluoride is "wasted" if each person uses only \(6.0 \mathrm{~L}\) of water a day for drinking and cooking? (Sodium fluoride is 45.0 percent fluorine by mass. 1 gallon \(=3.79 \mathrm{~L} ; 1\) year \(=365\) days; density of water \(=1.0 \mathrm{~g} / \mathrm{mL} .)\)

Step by step solution

01

Calculate Total Water Consumption per Year

First, convert the daily consumption of water per person from gallons to liters. Multiply this consumption by the number of people and the number of days in a year to get the total annual water consumption for the city. \n\nGiven: \nDaily water consumption per person = 150 gallons\nNumber of people = 50,000\nNumber of days in a year = 365\n1 gallon = \(3.79 \mathrm{~L}\) \n\nTotal Water Consumption in Liters = Daily consumption per person * number of people * number of days in a year\n= (150 gallons/person/day * 3.79 L/gallon) * 50,000 people * 365 days/year

02

Calculate Total Amount of Sodium Fluoride Needed

Next, calculate the total amount of Sodium Fluoride to be added to the water. This is done by using the concentration of fluorine needed to fight tooth decay (1 ppm). This means \(1 \mathrm{~g}\) of fluorine per 1 million g of water. Remember that Sodium fluoride is 45.0 percent fluorine by mass. \n\nTotal amount of Sodium Fluoride = Total water consumption in liters * density of water * concentration of fluorine needed * proportion of sodium fluoride to fluorine\n= Total water consumption in L * \(1.0 \mathrm{~g/mL} * 1 \mathrm{~mL/L}\) * \(1 \mathrm{~g of fluorine / 10^6 ~g of water}\) * \(\frac{100}{45.0}\)

03

Calculate Amount of Sodium Fluoride for Drinking and Cooking

It's given that each person uses only \(6.0 \mathrm{~L}\) of water a day for drinking and cooking. Calculate the total amount of Sodium Fluoride that is consumed for drinking and cooking following the same approach as in Step 2. \n\nTotal amount of Sodium Fluoride consumed for drinking and cooking = Daily water consumption for drinking and cooking * number of people * number of days in a year * density of water * concentration of fluorine needed * proportion of sodium fluoride to fluorine\n= \(6.0 \mathrm{~L/person/day}\) * 50,000 people * 365 days/year * \(1.0 \mathrm{~g/mL} * 1 \mathrm{~mL/L}\) * \(1 \mathrm{~g of fluorine / 10^6 ~g of water}\) * \(\frac{100}{45.0}\)

04

Compute The Percentage of Wasted Sodium Fluoride

The percentage of wasted Sodium Fluoride can be calculated by subtracting the total amount of Sodium Fluoride used for drinking and cooking from the total amount of Sodium Fluoride added to the water, then dividing by the total amount of Sodium Fluoride added to the water, and finally multiplying by 100. \n\nWasted Sodium Fluoride % = \(\frac{Total ~amount ~of ~Sodium ~Fluoride ~- ~Sodium ~Fluoride ~used ~for ~drinking ~and ~cooking}{Total ~amount ~of ~Sodium ~Fluoride}\) * 100

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