# Answer to Question #4329 in Other Chemistry for sarah

Question #4329

You are the public health officer in the water treatment facility for a city of 50,000 people. A concentration of 1.0 ppm of flourine in the drinking water is sufficient for the purpose of helping to prevent tooth decay. (1 ppm means "one part per million," or 1 g of flourine per one million grams of water). The compound normally chosen for flouridation is the same as is found in some toohtpaste's sodium flouride. You know that sodium flouride is 45% flourine by mass. Calculate how many kilograms of sodium flouride you will need to have in order to flouridate the city's water supply for one year, based upon your estimate that the average daily consumption of water is 150 gallons per person. (1 gallon = 3.79 L; 1 year = 365 days; 1 ton = 2000 lb; 1 lb = 456.3 g; density of water = 1.0 g/mL )

Expert's answer

V

m

m(F)

m(F) = 1.4*10^4 kg

(NaF) = 1.4*10^4/0.45 = 2.3*10^4 kg

_{general}(H_{2}O) = 50 000*150*3.79*365 = 1.4*10^{10 }Lm

_{general}(H_{2}O) = 1.4*10^{10}*1 = 1.4*10^{10}kgm(F)

_{H}_{2}_{O}= m(F)/(m(F) + m_{}(H_{2}O))(10^-6) = m(F)/(m(F) + 1.4*10^10)*(10^-6)m(F) = 1.4*10^4 kg

(NaF) = 1.4*10^4/0.45 = 2.3*10^4 kg

## Comments

Assignment Expert05.10.15, 12:37Dear Kira,

1) mass = volume × density. Density of water equals 1 kg/L. That's why m = 1.4×10

^{10}L × 1 kg/L = 1.4×10^{10}kg;2) mass of F can be 1 kg per 1000000 kg of water, or 1 g per 1000000 g of water (thanks, there was a mistake in the previous comment);

3) to find the mass of F we need to find million part of water mass (million part of 1.4×10

^{10}). That's why we divide 1.4×10^{10}by 10^{6}(1000000).Kira03.10.15, 06:45Thank you for your quick reply! I still have few questions. How did you get the mass of water is 1.4x10^10 kg? And why the mass of F is 1g per 1000000 kg of water? Isn't "1g of flourine per million GRAMS of water"?And why do you have to divided 100000 by 1.4x10^10?

Thanks!!!

Assignment Expert02.10.15, 10:59Dear Kira,

1.Calculate the volume and mass of water needed to supply a city: 1.4×10

^{10}kg.

2.Calculate the mass of fluorine (F) needed for preventing tooth decay: 1 kg per

1000000 kg of water, mass (F) = 1.4×10

^{10}/10^{6}= 1.4×10^{4 }kg.3. Calculatethe amount of NaF needed (NaF consists of 45% of F): 1.4×10

^{4}/0.45 = 2.3×10^{4}kg.

Kira02.10.15, 08:45Still comfused!! would you mind explain?

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