The reaction of lead chromate with NaOH (in hot, excess conditions) results in the formation of a soluble dianionic complex:
PbCrO4 + NaOH (hot, excess) → [Pb(OH)4]2- + Na2CrO4
The product [Pb(OH)4]2- is a dianionic complex with a coordination number of four.
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)