High levels of Pb+2 ions made Flint's water corrosive. Suppose Flint has 100 milligrams of Pb+2 ions per fluid ounce of water. If you were to drink eight cups of Flint's water in one day, how many grams of Pb+2 ions would you consume? (One cup = 0.237 liters. Density of water = 1000g/L. 1 fluid ounces = 29.574 mililiters) Please help me solve this problem.
one cup = 0.237 liter then 8 cup = 8 0.237 = 1.896
liter
we drink 1.896 liter water
1 liter = 1000 mililiter then 1.896 liter = 1896 mililiter
1 ounce = 29.574 mililiter
1 once contain 100 milligam of Pb2+ that mean 29.574 mililiter water contain 100 miligram of Pb2+
29.574 mililiter water contain 100 miligram of Pb2+ then 1896 mililiter water contain
100 1896 / 29.574 =
6411 miligram of Pb2+
6411 miligram = 6.411 gram
we consume 6.411 gram of Pb2+
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