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
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.
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