Soal 2. Fluoridasi adalah proses penambahan senyawa fluor ke dalam air minum untuk membantu mencegah kerusakan gigi. Konsentrasi 1 ppm fluorin sudah cukup untuk tujuan tersebut. (1 ppm berarti satu bagian per sejuta, atau 1 g fluor per 1 juta g air). Senyawa yang biasanya dipilih untuk fluoridasi adalah natrium fluorida, yang juga ditambahkan ke beberapa pasta gigi. Hitunglah jumlah natrium fluorida dalam kilogram yang dibutuhkan per tahun untuk sebuah kota berpenduduk 50.000 orang jika konsumsi air per orang per hari adalah 150 galon. Berapa persen natrium fluorida yang "terbuang" jika setiap orang hanya menggunakan \( 6,0 \mathrm{~L} \) air per hari untuk minum dan memasak? (Natrium fluorida mengandung 45,0 persen fluor berdasarkan massa. 1 galon \( =3,79 \mathrm{~L} ; 1 \) tahun \( =365 \) hari; massa jenis air \( =1,0 \mathrm{~g} / \mathrm{mL}) \).
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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 one part per million, or $1 \mathrm{~g}$ 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. gallon $=3.79 \mathrm{~L} ; 1$ year $=365$ days; density of water $=1.0 \mathrm{~g} / \mathrm{mL}$.)
Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of $1 \mathrm{ppm}$ of fluorine is sufficient for the purpose (1 ppm means one part per million, or 1 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 gal. 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 gal $=3.79$ L; 1 year $=365$ days; 1 ton $=2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{g} ; \text { density of water }=1.0 \mathrm{g} / \mathrm{mL}) ?$
The fluoridation of city water supplies has been practiced in the United States for several decades because it is believed that fluoride prevents tooth decay, especially in young children. This is done by continuously adding sodium fluoride to water as it comes from a reservoir. Assume you live in a medium-sized city of 150,000 people and that each person uses 175 gal water per day. How many tons of sodium fluoride must you add to the water supply each year ( 365 days) to have the required fluoride concentration of 1 part per million (that is, 1 ton of fluoride per million tons of water)? (Sodium fluoride is $45.0 \%$ fluoride, and 1 U.S. gallon of water has a mass of $8.34 \mathrm{lb}$.)
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