Posted by TROUBLED on .
a portion (10.0 ml)of a solution containing the chloride anion was treated with silver nitrate to yield a precipitate (0.1713 g) of the silver chloride. calculate the molar concentration of the chloride anion in the solution. first write the equation.
Convert 0.1713 g AgCl to g Cl
moles AgCl = 0.1713/molar mass AgCl
moles Cl = moles Ag since there is 1 mole Cl in 1 mole AgCl.
Then M = moles Cl/L soln.
AgNO3 + Cl^ ==> AgCl + NO3^-