If the drinking water supply for a municipality has a reported concentration of lead at 14.0...
Homework Answers
volume of water = 1.80 L = 1800 mL
assuming density of water = 1.00 g/mL
mass of water = (volume of water) * (density of water)
mass of water = (1800 mL) * (1.00 g/mL)
mass of water = 1800 g
mass of lead = (concentration of lead / 109) * (mass of water)
mass of lead = (14.0 ppb / 109) * (1800 g)
mass of lead = 2.52 x 10-5 g
If an individual drank 1.80 liters of water per day, then exposure to lead will be 2.52 x 10-5 g/day
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