The limiting molar conductivity of \(AgI\),\(Λ^0_m(AgI) = Λ^0_m(NaI)+Λ^0_m(AgNO_3)-Λ^0m(NaNO_3)\)\(Λ^0_m(AgI) = 12.7 + 13.3 – 12.0\)\(Λ^0_m(AgI) = 26 – 12\)\(Λ^0_m(AgI) =14\ mS m^2 mol^{–1}\)
So, the answer is \(14\ mS m^2 mol^{–1}\).