The formula of the complex ion present in the solution is found by adding
aqueous silver nitrate to the solution. This only reacts with the free chloride ions
to form a precipitate of silver chloride.
The precipitate is then filtered, washed, dried and weighed.
In an experiment, 0.012 mol of one of the forms of chromium(III) chloride was
used and 3.44 g of silver chloride was formed.
Deduce the formula of the complex ion. You must show your working.
