Complete Question:
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Answer:
2.23x10⁶ g
Explanation:
The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:
0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³
0.7976 g/m³
The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:
A = πR², where R is the radius = 1.01x10² m (half of the diameter)
A = π*(1.01x10²)²
A = 32047 m²
The volume is then:
V = 32047 * 87.32
V = 2.7983x10⁶ m³
The mass of the F⁻ is the concentration multiplied by the volume:
m = 0.7976 * 2.7983x10⁶
m = 2.23x10⁶ g
