To assay silver content in an alloy, a 13.38 g sample of alloy was dissolved in nitric acid. The addition of 0.53 molar sodium chloride resulted in the precipitation of silver chloride. If the alloy were pure silver, what volume of sodium chloride is needed to precipitate the silver?

Answer 1

Convert 13.38 g Of Ag to moles by dividing by the gram-atomic mass of Ag.

1 mole of Ag produces 1 mole of AgNO3
1 mole of AgNO3 + 1 mole NaCl ---> 1 mole AgCl
so.....
Moles of Ag = Moles of AgNO3 = moles of NaCl

(Moles of NaCl) / (molarity of NaCl) = liters NaCl

Answer 2

Just a quick note here to have you realize that this is a ridiculous problem. An alloy is at least two somethings together so it can't be pure silver as the problem states.