Question

A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The volume of water treated is 25,000,000 gal/day. What is the mass of chlorine needed by this plant per day

Answer 1

Answer: The mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

Explanation:

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, $\frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L$

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for $9.4625\times 10^7L$ of water, the amount of chlorine applied will be $\frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg$

Hence, the mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$