[Zoloft]

I'm not as good in math as I thought I was, apparently. So here I post a petty problem that should be in theory basic but that I haven't figured out?

Problem
1)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 )

F = 1.00 ppm

1 L = 1 dm³

1 Lb = 456.2 g / 1 lb

Per year = (150 gallons / person ) x (365 days / 1 year ) x (5.0 x 10³ / city ) = city's total water supply / year = 27.375 x 10⁸ gallons of water.

Do I convert 1 g of flourine to kilograms?

And then divide by the amount of water (but convert to liters? maybe)

Any help is appreciated. 

[enahs]

The amount (gallons) looks ok.

What is the mass of that amount of water?
So you need 1 PPM of that mass, what is it?
Now knowing your sodium fluoride is only 45% fluorine, then how much sodium fluoride is required?
 

[Zoloft]

What is the mass of that amount of water?

I convert it to kg?

So you need 1 PPM of that mass, what is it?

So a millionth of that?

Now knowing your sodium fluoride is only 45% fluorine, then how much sodium fluoride is required?

55%?!

Hopefully this will steer me in the right direction.

[JGK]

I'm not as good in math as I thought I was, apparently. So here I post a petty problem that should be in theory basic but that I haven't figured out?

Problem
1)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 )

F = 1.00 ppm

1 L = 1 dm³

1 Lb = 456.2 g / 1 lb

Per year = (150 gallons / person ) x (365 days / 1 year ) x (5.0 x 10³ / city ) = city's total water supply / year = 27.375 x 10⁸ gallons of water.

Do I convert 1 g of flourine to kilograms?

And then divide by the amount of water (but convert to liters? maybe)

Any help is appreciated. 

there is an error in your city population size 5.0 x 10³ =5000 not 50,000


once you have a total volume in gallons convert to litres then to grams then calculate how many grams of fluorine gives you the 1 ppm needed then divide by the % fluorine in NaF to get the amount of NaF you need.

[Riley_5000]

Your final answer is about 23 tons / day of NaF. 1 ton = 1000kg

Work towards that.