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The drinking water in Saplingville was found to contain 13 ppb (parts per billion) of lead. What is the concentration of lead in molarity?

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Final answer:

To convert 13 ppb of lead in drinking water to its molarity, we first convert ppb to g/L and then divide by the molar mass of lead, leading to a molarity of 6.27×10-11 M.

Step-by-step explanation:

The question asks about converting the concentration of lead in drinking water from parts per billion (ppb) to molarity. To calculate the molarity of a 13 ppb lead solution, we first convert ppb to grams per liter (g/L), then use the molar mass of lead to find the molarity. The process involves the following conversions and calculations:

  1. Convert ppb to g/L: since 1 ppb corresponds to 1×10-9 g per gram of solution, for water, we can assume a density of approximately 1 g/mL. Hence, 13 ppb is 13×10-9 g/mL, which is also 13×10-9 g/g of water since 1 mL of water weighs approximately 1 g.
  2. To convert to grams per liter, multiply by 1000, because there are 1000 mL in one liter. Therefore, 13 ppb is equivalent to 13×10-6 g/L.
  3. To find the molarity, divide the mass of lead in grams by its molar mass (207.2 g/mol for Pb) and the volume in liters. So, the molarity is (13×10-6 g/L) / (207.2 g/mol), which equals 6.27×10-11 M.

Therefore, the molarity of lead in the Saplingville water supply is 6.27×10-11 M.

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