An 81, 600 liter swimming pool has a chlorine concentration of 4.5 ppm (parts per million). How many liters must be replaced with chlorine-free water to reduce the chlorine concentration to 1.5 ppm?

You can handle this problem just as you would a percentage mixture problem. A ppm expresses a number out of a million whereas a percentage is a number expressed out of 100.

Let x = the amount of pool water in liters that needs replacing
with chlorine-free water.

Pool Mix | -Pool Mix | +Pure Water | Final Pool Mix | |
---|---|---|---|---|

ppm(chlorine) | 4.5 | 4.5 | 0 | 1.5 |

Vol. of Pool Mix | 81600 L | -x | +x | 81600 L |

The following equation describes volumes of chlorine. It states that the initial volume of chlorine in the pool minus the volume of chlorine removed plus the volume of chlorine added equals the final volume of chlorine in the pool:

81600 (4.5 ppm)- 4.5x + 0 x = 1.5 (81,600)

0.3672 - 0.0000045 x = 0.1224

Solving for x gives:

x = 54, 400 L