Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction

Answer:
1.186L or 1186mL.
Explanation:
The volume of a 0.290 m NH4I solution that is required to react with 253 ml of a 0.680 m Pb(NO3)2 solution is 1.186 liters or 1186 ml.