Answer : The final molarity of iodide anion in the solution is 0.0508 M.
Explanation :
First we have to calculate the moles of
and
.

Molar mass of KI = 166 g/mole

and,

Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:

From the balanced reaction we conclude that
As, 1 mole of KI react with 1 mole of 
So, 0.0178 mole of KI react with 0.0178 mole of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of 
From the reaction, we conclude that
As, 1 mole of
react to give 1 mole of 
So, 0.0178 moles of
react to give 0.0178 moles of 
Thus,
Moles of AgI = Moles of
anion = Moles of
cation = 0.0178 moles
Now we have to calculate the molarity of iodide anion in the solution.


Therefore, the final molarity of iodide anion in the solution is 0.0508 M.