A water sample has 20ppm (by mass) Cl2 present in it. Calculate the wquantity of Cl2 present in 100mL water(density of water= 1g/mL

To determine the quantity of chlorine gas (Cl2) present in a 100 mL water sample with a concentration of 20 ppm (parts per million), we can follow a straightforward calculation. The ppm value indicates how many grams of solute are present in one million grams of solution. In this case, since we are dealing with water, we can use the density of water to convert the volume into mass.

Step-by-Step Calculation

1. Convert Volume to Mass

Given that the density of water is 1 g/mL, we can easily convert the volume of water into mass. For 100 mL of water:

• Mass of water = Volume × Density
• Mass of water = 100 mL × 1 g/mL = 100 g

2. Calculate the Mass of Cl2 in the Sample

Now that we have the mass of the water, we can calculate the mass of Cl2 present using the ppm concentration. Since 20 ppm means 20 grams of Cl2 per 1,000,000 grams of water, we can set up a proportion:

• Mass of Cl2 = (Concentration in ppm × Mass of water) / 1,000,000
• Mass of Cl2 = (20 ppm × 100 g) / 1,000,000
• Mass of Cl2 = 2,000 / 1,000,000 = 0.002 g

3. Convert Grams to Milligrams (if needed)

Since it might be more practical to express this in milligrams, we can convert grams to milligrams:

• 0.002 g = 2 mg

Final Result

Thus, in a 100 mL water sample with a concentration of 20 ppm Cl2, there are 2 milligrams of chlorine gas present.

Understanding ppm in Context

This calculation illustrates how ppm works in practical scenarios. It’s a useful way to express very dilute concentrations, especially in environmental science and chemistry, where even small amounts of substances can have significant effects. For example, in water quality testing, knowing the concentration of pollutants in ppm helps in assessing safety and compliance with health standards.