Respuesta :
Answer: The mass of fluoride ions that must be added will be [tex]8.943\times 10^6g[/tex]
Explanation:
The equation used to calculate the volume of cylinder is:
[tex]V=\pi r^2h[/tex]
where,
r = radius of the reservoir= [tex]\frac{d}{2}=\frac{4.30\times 10^2m}{2}=215m[/tex]
h = height of the reservoir = 56.03 m
Putting values in above equation, we get:
[tex]\text{Volume of reservoir}=(3.14)\times (215)^2\times 56.03\\\\\text{Volume of reservoir}=8.13\times 10^6m^3[/tex]
- Converting this into liters, we use the conversion factor:
[tex]1m^3=1000L[/tex]
So, [tex]8.13\times 10^6m^3=8.13\times 10^9L[/tex]
Mass of water reservoir = [tex]8.13\times 10^9kg[/tex] (Density of water = 1 kg/L )
We are given:
Concentration of fluoride ion in the drinking water = 1.10 ppm
This means that 1.10 mg of fluoride ion is present in 1 kg of drinking water
Calculating the mass of fluoride ion in given amount water reservoir, we use unitary method:
In 1 kg of drinking water, the amount of fluoride ion present is 1.10 mg
So, in [tex]8.13\times 10^9kg[/tex] of drinking water, the amount of fluoride ion present will be = [tex]\frac{1.10mg}{1kg}\times 8.13\times 10^9kg=8.943\times 10^9mg[/tex]
- Converting this into grams, we use the conversion factor:
1 g = 1000 mg
So, [tex]8.943\times 10^9mg\times \frac{1g}{1000mg}=8.943\times 10^6g[/tex]
Hence, the mass of fluoride ions that must be added will be [tex]8.943\times 10^6g[/tex]
