Final answer:
To prepare a 50 PPM chlorine-based sanitizing solution, calculate the required mass of chlorine for the desired concentration in liters. In a 5,000-gallon pool, to achieve a concentration of 3.0 ppm, you must add 63.6 grams of chlorine.
Step-by-step explanation:
To prepare a 50 PPM chlorine-based sanitizing solution, you must calculate the necessary mass of chlorine needed for the desired concentration in a given volume of water. The concentration PPM (parts per million) is equivalent to milligrams of solute per liter of solution.
For the scenario where water in pools should be chlorinated to a maximum level of 3.0 ppm, in a typical 5,000-gallon pool that contains 21,200 kg of water, you would need to add a certain amount of chlorine to achieve this concentration. Since 1 ppm is equal to 1 mg per liter, for a 3.0 ppm concentration, you will need 3.0 mg of chlorine per liter of water. Therefore, the total mass of chlorine needed would be:
Mass of chlorine = Volume of water (liters) x Desired concentration (mg/L)
The volume of the pool in liters is 21,200 kg of water (assuming the density of water is about 1 kg/L), which converts to 21,200 liters. Multiplying this volume by the concentration gives:
Mass of chlorine = 21,200 liters x 3.0 mg/L = 63,600 mg of chlorine
63,600 mg is equivalent to 63.6 grams of chlorine. So, to achieve a chlorination level of 3.0 ppm in a 5,000-gallon pool, you would need to add 63.6 grams of chlorine.