Calculate the solubility of AgBr(s) in 0.15 M NaCN(aq). Ksp = 7.7 × 10-13 for AgBr Kf = 5.6 × 108 for Ag(CN)2-(aq)

Molar mass of    N    = 14 g/mol
Molar mass of   O2    = 32 g/mol
Adding both masses = 46 g/mol
Actual molar mass/ Empirical molar mass = 138.02 / 46 = 3
Now multiplying this co effecient with empirical fomula NO2 = 3(NO2)
                                                                                             = N3O6
So according to above explanation,
D) N3O6, is the correct answer.