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?

Drinking water with a concentration of Pb of 9.0 ppb, has a molarity of Pb of 4.3 × 10⁻⁸ M. There are 0.54 g of Pb in a 60 m³ pool.

The maximum allowable concentration of lead in drinking water is 9.0 ppb, that is, 9.0 μg of Pb per liter of water.

We want to calculate the molarity (M) of the solution. We will use the following expression, assuming that the volume of water is equal to the volume of the solution.

M = mass Pb / molar mass Pb × liters of solution

M = (9.0 × 10⁻⁶ g) / (207.2 g/mol) × 1 L = 4.3 × 10⁻⁸ M

To calculate the mass of Pb in a 60 m³ pool, we need to consider that 1 m³ = 1000 L.

60 m³ Water × (1000 L Water/ 1 m³ Water) = 60,000 L Water

60,000 L Water × (9.0 × 10⁻⁶ g Pb/1 L Water) = 0.54 g Pb

Drinking water with a concentration of Pb of 9.0 ppb, has a molarity of Pb of 4.3 × 10⁻⁸ M. There are 0.54 g of Pb in a 60 m³ pool.

Learn more about parts per billion here: link

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

your answer is c) plasma

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