The limiting molar conductivities of NaI, NaNO _ 3 and AgNO _ 3 are...

(1) ${\lambda}_{m}^{\infty}\left(NaI\right) = 12 . 7{mSm}^{2}{mol}^{- 1}$

(2) ${\lambda}_{m}^{\infty}\left({NaNO}_{3}\right) = 12 . 0{mSm}^{2}{mol}^{- 1}$

(3) ${\lambda}_{m}^{\infty}\left({AgNO}_{3}\right) = 13 . 3{mSm}^{2}{mol}^{- 1}$

${\lambda}_{m}^{\infty}\left(AgI\right) = \left(1\right) + \left(3\right) - \left(2\right)$

$= 12 . 7 + 13 . 3 - 12 . 0$

$= 26 . 0 - 12 . 0$

${\lambda}_{m}^{\infty}\left(AgI\right) = 14 . 0$