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The standard for arsenic content in public water supplies is less than or equal to 10 parts per billion (ppb) arsenic. Assuming that this arsenic is present as arsenate ASO43-,
What mass of sodium arsenate would be present in a 1 L sample of drinking water that just meets the stadard?
The strings:
S7P1A15 (containership - mass).
The math:
Pj Problem of Interest is of type containership (mass).
1 ppb = 1μg/L = (1 x 10-6) g /L
So, 10 ppb = 10 x 10-6 g/L
So, mass of arsenic in 1 L sample of drinking water = 10-5 g.
Molar mass of arsenic = 74.92 g
So mol of arsenic in 1 L = (1/74.92)(10-5) = 0.01335 x 10-5 mol
Na2AsO4 -------> 3Na+ + AsO43-
Mole ratio Na3AsO4:AsO43- = 1:1
So, mol of Na3AsO4 in 1 L sample of drinking water = 0.01335 x 10-5 mol.
Molar mass of Na3AsO4 =3x23 +74.92 + 4x16 = 69 + 74.92 + 64 = 207.92 g
So, mass of Na2AsO4 in 1 L sample of drinking water = 207.92(0.01335)(10-5) = 2.8 x 10-5 g of Na2AsO4.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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