The disproportionation reaction of chlorine in alkaline medium can be balanced as follows. Chlorine (Cl2) is both reduced and oxidized:
Cl2 + 2OH− → Cl− + ClO− + H2O
Thus, the values of a, b, c, and d are 1, 2, 1, and 1 respectively.
So, the correct answer is: Option (1)
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$^{-1}$)