A water sample has 20.0 ppm (by mass) Cl2 present in it. Calculate the

quantity of Cl2 present in 100 ml water. (Density of water = 1.0 g ml−1.)

To find the quantity of chlorine gas (Cl2) present in a 100 ml water sample with a concentration of 20.0 ppm (parts per million), we can break down the calculation step by step. The term "ppm" indicates that there are 20 grams of Cl2 in one million grams of water. Since the density of water is 1.0 g/ml, we can easily convert the volume of water into mass.

Step-by-Step Calculation

1. Convert Volume to Mass

First, we need to convert the volume of water (100 ml) into mass. Given that the density of water is 1.0 g/ml, the mass of 100 ml of water can be calculated as follows:

• Mass of water = Volume × Density
• Mass of water = 100 ml × 1.0 g/ml = 100 g

2. Understanding ppm Concentration

Next, we need to interpret what 20.0 ppm means. This concentration tells us that there are 20 grams of Cl2 in 1,000,000 grams of water. To find out how much Cl2 is in our 100 g of water, we can set up a proportion:

• 20 g Cl2 / 1,000,000 g water = x g Cl2 / 100 g water

3. Setting Up the Equation

Now, we can cross-multiply to solve for x:

• 20 g Cl2 × 100 g water = x g Cl2 × 1,000,000 g water
• 2000 = 1,000,000x

4. Solving for x

To isolate x, divide both sides by 1,000,000:

• x = 2000 / 1,000,000
• x = 0.002 g

Final Result

The quantity of Cl2 present in 100 ml of water is 0.002 grams, or 2.0 mg. This means that in a relatively small volume of water, we can still detect a measurable amount of chlorine gas, which is important in various applications, including water treatment and sanitation.

Real-World Implications

Understanding concentrations like ppm is crucial in fields such as environmental science, chemistry, and public health. For instance, knowing the concentration of chlorine in drinking water helps ensure that it is safe for consumption, as chlorine is often used to disinfect water but can be harmful in excessive amounts.