Answer:
2835 kilograms of sodium hypochlorite must be added to the water supply each week to produce the required chlorine level of 1 ppm.
Explanation:
Volume of water used by 1 person = 750 L
Volume of water used by 1.8 million persons : V
[tex]V=1.8\times 10^6\times 750 L=1.35\times 10^{9} L[/tex]
Density of water,d = 1 kg/L
Mass of water used by 1.8 million persons = m
[tex]m=d\times V=1 kg/L\times 1.35\times 10^{9} L=1.35\times 10^{9} kg[/tex]
1 kilogram of chlorine per million kilograms of water. (Given)
Concentration of chlorine in water = 1 kg/ 1000,000 kg of water
In 1000,000 kg of water = 1 kg of chlorine
Then [tex]1.35\times 10^{9} kg[/tex] of water have x mass of chlorine:
[tex]\frac{x}{1.35\times 10^{9} \text{kg of water}}=\frac{1 kg}{1000,000 \text{kg of water}}[/tex]
[tex]x=\frac{1}{1000,000}\times 1.35\times 10^{9} kg=1.35\times 10^3 kg[/tex]
Mass of chlorine in water of mass [tex]1.35\times 10^{9} kg=1.35\times 10^{3} kg[/tex]
Percentage of chlorine in hypochlorite = 47.62%
[tex]47.62\%=\frac{1.35\times 10^{3} kg}{\text{Total mass of sodium hypochlorite}}\times 100[/tex]
Total mass of sodium hypochlorite = [tex]2834.94 kg\approx 2835 kg[/tex]
2835 kilograms of sodium hypochlorite must be added to the water supply each week to produce the required chlorine level of 1 ppm.