Question

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 μg of lead?
Can someone just guide me through the steps for this!!! Thanks(:

Answer 1

since 1 L of water is roughly equal to 1 kg (at 4 celsius)

1 billionth of a kg is 1 \(\mu g\), so 1 part per billion means 1 \(\mu g\) in 1 L.
\((1 \;\mu g=0.000,000,000,1\;kg \;or\; 1.0*10^{-9}kg\)

so 15 ppb is \(15\mu g\) in 1 L, converting to mg it is: 0.015 mg/L

so the density is \(15\mu g/L\); use the formula \(\rho=\dfrac{m}{V}\) for the second part.

Answer 2

Thankkk you, it took me forever to finally figure that out since I had no help!
Thanks for responding though! You rock!!(: