Chapter 13: Problem 39

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

Short Answer

Expert verified

The mass of chlorine that must be added to 100.0 L of water to achieve a level of 0.5 ppm is 50 mg.

Step by step solution

Understanding Units

First, understand that 1 ppm is equivalent to 1 mg/L. This is because 1 ppm means 1 part in 1 million, and in the context of water solutions, this is often measured as milligrams per liter since 1 liter of water weighs approximately 1 million milligrams. Therefore, in this exercise, when it is said that the concentration of chlorine is 0.5 ppm, it is equivalent to saying that the concentration is 0.5 mg/L.

Applying the Concept of Concentration

Now, as the volume of water is given to be 100.0 L, we need to find out how much chlorine (in mg) is needed. Since 1 L of water requires 0.5 mg of chlorine (from step 1), for 100.0 L of water, you need to multiply 100.0 L by 0.5 mg/L.

Calculation

Carry out the multiplication: 100.0 L * 0.5 mg/L = 50 mg. So, 50 mg of chlorine is required to achieve a concentration of 0.5 ppm in 100.0 L of water.

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

Short Answer

Step by step solution

Understanding Units

Applying the Concept of Concentration

Calculation

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