Most community water supplies have 0.50 ppm of chlorine added for purification.

What mass of chlorine must be added to 134.0 L of water to achieve this level?

anwser grams

How many grams of chlorine gas are needed to make 5.00×106 g of a solution that is 1.50 ppm chlorine by mass?

0.50 ppm means 0.50 mg Cl2/L

So for 134.0 L you need .....?

To find the mass of chlorine needed to achieve a concentration of 0.50 ppm in 134.0 L of water, we can use the formula:

Mass = Concentration × Volume

First, let's convert the concentration from parts per million (ppm) to grams per liter (g/L):

1 ppm = 1 mg/L = 0.001 g/L

So, 0.50 ppm = 0.50 × 0.001 g/L = 0.0005 g/L

Now, we can calculate the mass of chlorine:

Mass = 0.0005 g/L × 134.0 L

Mass = 0.067 g

Therefore, approximately 0.067 grams of chlorine must be added to 134.0 L of water to achieve a concentration of 0.50 ppm.

To find the mass of chlorine that must be added to 134.0 L of water, we need to use the information given, which is that most community water supplies have 0.50 ppm (parts per million) of chlorine added for purification.

First, we need to convert the volume of water from liters to milliliters since the concentration of chlorine is given in parts per million (ppm), which is equivalent to milligrams per liter (mg/L).

1 L = 1000 mL

So, 134.0 L of water is equal to 134.0 × 1000 mL = 134,000 mL

Next, we can calculate the mass of chlorine using the concentration in ppm.

0.50 ppm means 0.50 mg of chlorine per liter (mg/L).

Since we have 134,000 mL of water, we can set up a proportion to find the mass of chlorine:

(0.50 mg / 1 L) = (x mg / 134,000 mL)

To solve for x, we can cross-multiply and divide:

x mg = (0.50 mg / 1 L) * 134,000 mL

x mg = 67,000 mg

Finally, we need to convert milligrams to grams:

1 g = 1000 mg

67,000 mg is equal to 67,000 ÷ 1000 = 67 g

Therefore, the mass of chlorine that must be added to 134.0 L of water to achieve a level of 0.50 ppm is 67 grams.