Question

To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g of F- must be added to a cylindrical water reservoir having a diameter of 3.36x10^2 m and a depth of 21.80 m? B) How many grams of sodium fluoride, NaF, contain this much fluoride?

Answer 1

a) 1.740 g of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Step-by-step explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

`\pi  r^(2)h` ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

`r=(d)/(2)`

`=(3.36x10^(2))/(2)`

= 168 m

Volume =

`=(22* 168^(2)* 21.80)/(7)`

`=1.93* 10^(6) m^(3)`

2.Convert ppm to g/m3 and Solve for mass of F-

`1ppm = 1g/m^(3)`

`0.9ppm = 0.9g/m^(3)`

Because both ppm and g/m3 are same quantity .

`g/m^(3) =(mass\ of\ F-(g))/(Volume\ m^(3))* 10^(6)`

`0.9 =(mass\ of\ F-)/(1.93* 10^(6) m^(3))* 10^(6)`

`mass\ of\ F- =1.740g`

mass of F- required = 1.740 g

3. Apply mole concept to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

`2Na+F_(2)\rightarrow 2NaF`

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

`\frac {84}{38}* 1.74`

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced