/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Fluoridation is the process of a... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(=1.0 \mathrm{~g} / \mathrm{mL} .)\)

Short Answer

Expert verified
The city requires 23,066.94 kg of sodium fluoride annually, with 98.944% being wasted.

Step by step solution

01

Convert Gallons to Liters

First, we will convert the daily water consumption per person from gallons to liters. Given that 1 gallon = 3.79 L, for 150 gallons:\[ 150 ext{ gallons/person} \times 3.79 ext{ L/gallon} = 568.5 ext{ L/person} \]
02

Daily Water Consumption for the City

Next, calculate the daily water consumption for the entire city. With 50,000 people each consuming 568.5 L:\[ 50,000 ext{ people} \times 568.5 ext{ L/person} = 28,425,000 ext{ L/day} \]
03

Annual Water Consumption

Calculate the annual water consumption by multiplying the daily consumption by the number of days in a year (365 days):\[ 28,425,000 ext{ L/day} \times 365 ext{ days/year} = 10,379,625,000 ext{ L/year} \]
04

Convert Liters to Grams

Since the density of water is 1.0 g/mL, each liter is equivalent to 1 kg or 1000 grams. Therefore, convert the annual water consumption from liters to grams:\[ 10,379,625,000 ext{ L/year} \times 1000 ext{ g/L} = 10,379,625,000,000 ext{ g/year} \]
05

Calculate Fluorine Required for Water

Determine how much fluorine is needed at a concentration of 1 ppm (1 g/million g of water):\[ \frac{10,379,625,000,000 ext{ g water}}{1,000,000} = 10,379,625 ext{ g of fluorine/year} \]
06

Calculate Sodium Fluoride Required

Since sodium fluoride is 45% fluorine by mass, calculate the total sodium fluoride needed using the formula:\[ \frac{10,379,625 ext{ g fluorine/year}}{0.45} = 23,066,944.44 ext{ g of sodium fluoride/year} \]Convert this to kilograms:\[ \frac{23,066,944.44 ext{ g}}{1000} = 23,066.94 ext{ kg/year} \]
07

Calculate Water Usage for Drinking/Cooking

Determine the amount of water used for drinking and cooking, considering only 6 L per person a day is used for these purposes:\[ 6 ext{ L/person} \times 50,000 ext{ people} = 300,000 ext{ L/day} \]
08

Annual Water Used for Drinking/Cooking

Convert the daily drinking/cooking water consumption to annual consumption:\[ 300,000 ext{ L/day} \times 365 ext{ days/year} = 109,500,000 ext{ L/year} \]
09

Calculate Percent Sodium Fluoride Wasted

Calculate the percentage of sodium fluoride wasted, knowing that the fluorine intended is used only for drinking/cooking:\[ \frac{109,500,000}{10,379,625,000} \times 100 = 1.056\% \]Thus, \[ 100\% - 1.056\% = 98.944\% \text{ of sodium fluoride is wasted.} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Fluoride
Sodium fluoride is a chemical compound often used in public health to help prevent tooth decay. This is because it provides fluoride ions, which strengthen the enamel on our teeth. Sodium fluoride is usually added to water as part of a fluoridation process. It is also a common ingredient in many toothpastes.
One key property of sodium fluoride is that it is 45% fluorine by mass. This means for every 100 grams of sodium fluoride, 45 grams are actually fluorine. This is important when calculating how much sodium fluoride to use in water treatment processes. For example, if a certain amount of fluoride is needed in a water supply, you must use the proportion to find out how much sodium fluoride will meet this requirement.
In water fluoridation, sodium fluoride is typically dosed to achieve a concentration of 1 part per million (ppm) of fluoride, which is considered safe and effective for reducing the incidence of dental cavities in the population.
ppm (parts per million)
The term ppm stands for "parts per million" and is a unit of measurement used to describe the concentration of one substance within another. In the context of water fluoridation, ppm helps denote how much fluoride is present compared to the amount of water.
For instance, 1 ppm of fluoride signifies that there is 1 gram of fluoride for every 1 million grams (or liters, since water's density is 1 g/mL) of water. This measurement is relatively straightforward, as 1 liter of water equals 1 kilogram, making it easier to calculate the amount of fluoride required for large water supplies.
Understanding ppm is crucial for adjustments in municipal water systems, ensuring that the amount of fluoride added is both effective for dental health and compliant with health regulations.
Water Consumption Calculation
Calculating water consumption accurately is essential in determining the appropriate amount of fluoridation. In our example, each person in the city consumes 150 gallons of water every day. To work with a consistent unit, this amount is converted into liters. Using the conversion factor (1 gallon = 3.79 liters), 150 gallons equal 568.5 liters per person per day.
Next, the entire city's daily consumption is calculated by multiplying the per-person use by the total population (50,000 people), leading to a daily consumption of 28,425,000 liters.
To find the annual water consumption, one multiplies the daily consumption by 365 (the number of days in a year), resulting in about 10.38 billion liters of water consumed per year. Knowing this annual consumption helps in calculating how much sodium fluoride needs to be added to achieve the desired fluoridation level of 1 ppm, tailored to ensure community dental health is maintained.

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Most popular questions from this chapter

Carry out these conversions: (a) A 6.0-ft person weighs 168 lb. Express this person's height in meters and weight in kilograms. \((1 \mathrm{lb}=453.6 \mathrm{~g} ; 1 \mathrm{~m}=\) \(3.28 \mathrm{ft} .\) ) (b) The current speed limit in some states in the United States is 55 miles per hour. What is the speed limit in kilometers per hour? (c) The speed of light is \(3.0 \times 10^{10} \mathrm{~cm} / \mathrm{s} .\) How many miles does light travel in 1 hour? (d) Lead is a toxic substance. The "normal" lead content in human blood is about 0.40 part per million (that is, \(0.40 \mathrm{~g}\) of lead per million grams of blood). A value of 0.80 part per million (ppm) is considered to be dangerous. How many grams of lead are contained in \(6.0 \times 10^{3} \mathrm{~g}\) of blood (the amount in an average adult) if the lead content is \(0.62 \mathrm{ppm} ?\)

The surface area and average depth of the Pacific Ocean are \(1.8 \times 10^{8} \mathrm{~km}^{2}\) and \(3.9 \times 10^{3} \mathrm{~m},\) respectively. Calculate the volume of water in the ocean in liters.

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Define these terms: (a) element, (b) compound.

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