Question

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.900 ppm fluoride, F−. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 8.34 × 102 m and a depth of 74.50 mm? mass: g F− How many grams of sodium fluoride, NaF, contain this amount of fluoride? mass:

Answer 1

Explanation:

The given data is as follows.

           Weight of $F^{-}$ = 0.9 ppm = 0.9 mg/kg

           Diameter = $8.34 \times 10^{2} m$

           radius = $\frac{Diameter}{2}$ = $\frac{8.34 \times 10^{2} m}{2}$

                       = $4.17 \times 10^{2} m$

           height = 74.50 mm = 0.0745 m   (as 1 mm = 0.001 m)

Hence, calculate the volume as follows.

                 Volume of tank = $\pi \times r^{2} \times h$

                                            = $3.14 \times (4.17 \times 10^{2} m)^{2} \times 0.0745 m$

                                             = $4.067 \times 10^{2} m^{3}$

                                             = $4067 \times 10^{2} l$   (as $1 m^{3}$ = 1000 L)

Therefore, weight of $F^{-}$ for $4067 \times 10^{2} kg$ water                                          

                           =  $0.9 mg/l \times 4067 \times 10^{2} l$

                           = $3.66 \times 10^{2} mg$

                           = $\frac{3.66 \times 10^{2} mg}{1000} g$

                           = 0.366 g

Hence, moles of $F^{-}$ will be calculated as follows.

                      No. of moles = $\frac{mass}{\text{molar mass}}$

                                             = $\frac{0.366 g}{19 g/mol}$

                                             = 0.0192 mol

Moles of $F^{-}$ = moles of NaF

Therefore, weight of NaF = Moles × Molecular weight

                                           = $0.0912 mol \times 42 g/mol$

                                           = 3.83 g

Thus, we can conclude that weight of NaF is 3.83 g.

Answer 2

A cylindrical water reservoir having a diameter of 8.34 × 10² m and a depth of 74.50 mm requires 7.25 g of NaF to contain 0.900 ppm F⁻.

A cylindrical water reservoir has a diameter (d) of 8.34 × 10² m and a depth of 74.50 mm (h). We can calculate the volume (V) of the reservoir using the following expression.

$V = h \times \pi \times (\frac{d}{2} )^{2} = (8.34 \times 10^{2}m ) \times \pi \times (\frac{74.50 \times 10^{-3}m }{2} )^{2} = 3.64 m^{3}$

We can convert 3.64 m³ to L using the conversion factor 1 m³ = 1000 L.

$3.64 m^{3} \times \frac{1000L}{m^{3} } = 3.64 \times 10^{3} L$

The required concentration of F⁻ is 0.900 ppm, that is, 0.900 mg of F⁻ per liter of water. The mass of F⁻ in 3.64 × 10³ L of water is:

$3.64 \times 10^{3} L \times \frac{0.900mgF^{-} }{L} = 3.28 \times 10^{3} mgF^{-} = 3.28 gF^{-}$

The mass ratio of F⁻ to NaF is 19.00:41.99. The mass of NaF that contains 3.28 g of F⁻ is:

$3.28 g F^{-} \times \frac{41.99gNaF}{19.00gF^{-} } = 7.25gNaF$

A cylindrical water reservoir having a diameter of 8.34 × 10² m and a depth of 74.50 mm requires 7.25 g of NaF to contain 0.900 ppm F⁻.

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