Asolution used to chlorinate a home swimming pool contains 7% chlorine by mass. an ideal chlorine level for the pool is one part per million (1 ppm). (think of 1 ppm as being 1 g chlorine per million grams of water). if you assume densities of 1.10 g/ml for the chlorine solution and 1.00 g/ml for the swimming pool water, what volume of the chlorine solution, in liters, is required to produce a chlorine level of 1 ppm in an 18,000 gal swimming pool?

Volume of chlorine = 61.943 mL

Explanation:

Given:

Volume of the water in the Pool = 18,000 gal

also,

1 gal = 3785.412 mL

thus,

Volume of water in pool = 18,000 × 3785.412 = 68,137,470 mL

Density of water = 1.00 g/mL

Therefore,

The mass of water in the pool = Volume × Density

or

The mass of water in the pool = 68,137,470 mL × 1.00 g/mL = 68,137,470 g

in terms of million =

or

= 68.13747 g

also,

1 g of chlorine is present per million grams of water

thus,

chlorine present is 68.13747 g

Now,

volume =

or

Volume of chlorine =

or

Volume of chlorine = 61.943 mL

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