What mass of cholrine must be added to 100.0 L of water to achieve this level? Most community water supplies have 0.5 ppm of cholrine added for purification.

0.5 ppm = 0.5 g Cl2 in 1,000,000 g solution. The density of water is essentially 1 g/mL and the mass of Cl2 is so small that 1 million grams of water + small amount Cl2 = 1 million grams solution or 1 million mL.

So we can think in terms of 0.5 g Cl2/10^6 mL water.
Convert that to grams Cl2/L. (that's 0.5 g Cl2/1000 mL), then to 100 L.

To determine the mass of chlorine that must be added to 100.0 L of water to achieve a certain level, we need to calculate based on the given concentration.

Given that most community water supplies have 0.5 ppm (parts per million) of chlorine added for purification, we know that 0.5 parts of chlorine are present in every 1 million parts of water.

To convert this to a more common concentration unit, we can convert parts per million to milligrams per liter (mg/L).

1 ppm = 1 mg/L

Thus, the concentration of chlorine that needs to be added is 0.5 mg/L.

To calculate the mass of chlorine that must be added, we need to multiply the concentration by the volume of water.

Mass of chlorine = concentration of chlorine × volume of water

In this case, the volume of water is given as 100.0 L.

Let's calculate the mass of chlorine:

Mass of chlorine = 0.5 mg/L × 100.0 L

Mass of chlorine = 50 mg

So, to achieve the desired level of chlorination, 50 milligrams of chlorine must be added to 100.0 liters of water.