The maximum concentration set by the U.S. Environmental Protection Agency for lead in drinking water is 15 ppb.

What is this concentration in milligrams per liter?

How many liters of water contaminated at this maximum level must you drink to consume 1.0 ug of lead?

1 mg/L is 1 ppm.

Therefore, 1 ppb is 0.001 mg/L. Check my thinking.

To convert parts per billion (ppb) to milligrams per liter (mg/L), you need to know the molecular weight of the substance you are converting. For lead, the molecular weight is approximately 207.2 grams per mole.

Here's how you can convert ppb to mg/L for lead in drinking water:

1. Convert the ppb concentration to micrograms per liter (µg/L):
15 ppb = 15 µg/L

2. Convert micrograms (µg) to milligrams (mg):
15 µg = 0.015 mg

Therefore, the lead concentration of 15 ppb is equivalent to 0.015 mg/L.

Now, let's calculate the number of liters needed to consume 1.0 µg of lead contaminated at the maximum level:

1 µg = 0.001 mg

To determine the number of liters needed, we'll set up a proportion:

0.015 mg 1 L
--------- = ---------
x L 0.001 mg

Cross multiplying and solving for x, we get:

0.015 × 0.001 = x × 1

0.000015 = x

Therefore, you would need to consume approximately 0.000015 liters (15 µL) of water contaminated at the maximum lead level to consume 1.0 µg of lead.