Question

A solution 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?

Answer 1

Answer:

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 = $\frac{\textup{68,137,470}}{\textup{ 1,000,000}}$

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 = $\frac{\textup{Mass}}{\textup{Density}}$

or

Volume of chlorine = $\frac{\textup{68.13747 g}}{\textup{1.10 g/mL}}$

or

Volume of chlorine = 61.943 mL