Question

The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0-ppb solution. What assumption did you have to make in your calculation? (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 m3 of water?

Answer 1

Answer:

0.54 g of Pb in 60 m³ of [Pb] = 9 ppb

Explanation:

We need to know, that ppb is a sort of concentration that indicates (in value of volume) ng / mL (ppb = parts per billion)

9 ppb means that 9 ng are contained in 1mL of solution

We know that 1mL = 1 cm³

Let's convert 60 m³ to cm³ → 60 m³ . 1×10⁶ cm³ / 1m³ = 6×10⁷ cm³

Then, we can make a rule of three:

1 cm³ has 9 ng of Pb

Therefore in 6×10⁷ cm³ we must have ( 6×10⁷ cm³ . 9ng) / 1 cm³ =

5.4×10⁸ ng

We convert ng to g → 5.4×10⁸ ng . 1 g / 1×10⁻⁹ ng = 0.54 g