1. The problem statement, all variables and given/known data The new standard for arsenate in drinking water, mandated by the Safe Drinking Water Act, requires that by January 2006, public water supplies must contain no greater than 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 1.00-L sample of drinking water that just meets the standard? 2. Relevant equations molar mass of sodium arsenate, Na3AsO4, = 207.89 g 3. The attempt at a solution (10 moles of Na3AsO4/109moles of H2O) * (1 mole H2O/18 g) * (1000 g/ 1 L) * (207.89 g/ 1 mole Na3AsO4) = 1.2 * 10-4 g of Na3AsO4 in 1 L of water I did conversions from the ratio of moles of sodium arsenate and water to grams of sodium arsenate per liter of water. However, this answer is incorrect according to my textbook's answer, which is 2.8 * 10-5 g of Na3AsO4 in 1 liter of water. Please help me understand what I am doing wrong, or what I'm misinterpreting.