Question

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.700 ppm fluoride (F–).

a) How many grams of F– must be added to a cylindrical water reservoir having a diameter of 3.01 × 102 m and a depth of 75.43 m? (b) How many grams of sodium fluoride, NaF, contain this much fluoride?

Answer 1

Solution :-

Lets first calculate the volume of the reservoir

V= pi* r^2 * h

V= 3.14 * (301 m/2)^2* 75.43 m

V= 5364716 m3

Now lets convert m3 to L

5364716 m3 * 1000 L / 1 m3 = 5.365*10^9 L

Now lets calculate the mass of the F- needed to get 0.700 ppm concentration

0.700 ppm means 0.700 mg per L

So

5.365*10^9 L * 0.700 mg/ L = 3.755*10^9 mg F-

Now lets convert the mg to gram

3.755*10^9 mg * 1 g / 1000 mg = 3.755*10^6 g F-

So mass of the F- needed = 3.755*10^6 g

Now lets calculate the mass of the NaF

1mol NaF = 41.988 g

In 1 mole NaF mass of F- = 18.988 g

So

3.755*10^6 F- * 41.988 g NaF / 18.988 g F- = 8.304*10^6 g NaF

So the mass of sodium fluoride NaF needed = 8.304*10^6 g