Question 743335
A solution containing 10,000 mg (10g) of something per liter (1000 milliliters) of solution has a concentration of
10g/1000mL = 1g/100mL = 1%(w/v)
We call it 1% weight in volume because we are expressing the amount of dissolved material by weight, in grams, and the amount of solution by volume, in milliliters. Knowing that a milliliter of a reasonably dilute aqueous solution weighs about one gram, we pair milliliters with grams.
Unless this is a problem for Chemical Engineering class, we can ignore the fact that 1.00 gram of sodium hypochlorite is not exactly equivalent to 1.00 g of chlorine gas, as well as ignore issues about the purity of the available solutions.
 
We say that 100 liters of 2% solution contain 2,000 grams chlorine
1% had 10,000mg (10g) per liter, so
2% must have 20g/L, and 100 x 20g = 2000g in 100L.
 
The available 12% solution contains
12 x 10g = 120 g per liter, so we need to use
2000g ÷ (120g/L) = {{{highlight(16.7L)}}} of 12% solution
 
If for each liter of feed water we need to add enough chlorine to meet the 2.2mg demand and have an extra 0.3mg residual chlorine, we need to add
2.2mg + 0.3mg = 2.5mg chlorine per liter of feed water.
If we are using that 2% solution containing
20g/L = 20,000mg/1000mL = 20mg/mL,
we need to add
2.5mg ÷ (20mg/mL) = 0.125mL of solution per liter of feed water
That would be 1.25mL/10L or 10mL/80L.