The limiting molar conductivities of NaI, $$NaNO_3$$ and $$AgNO_3$$ are $$12.7, 12.0$$ and $$13.3$$ mS m$$^2$$ mol$$^{-1}$$, respectively (all at $$25°$$C). The limiting molar conductivity of AgI at this temperature is ______ mS m$$^2$$ mol$$^{-1}$$.

We use Kohlrausch's Law of Independent Migration of Ions, which states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the limiting molar conductivities of its constituent ions.

We need to find $$\Lambda^\circ_{AgI}$$.

We can write:

$$\Lambda^\circ_{AgI} = \Lambda^\circ_{Ag^+} + \Lambda^\circ_{I^-}$$

Using the given data, we construct this from known electrolytes:

$$\Lambda^\circ_{AgI} = \Lambda^\circ_{AgNO_3} + \Lambda^\circ_{NaI} - \Lambda^\circ_{NaNO_3}$$

This works because:

$$\Lambda^\circ_{AgNO_3} = \Lambda^\circ_{Ag^+} + \Lambda^\circ_{NO_3^-}$$

$$\Lambda^\circ_{NaI} = \Lambda^\circ_{Na^+} + \Lambda^\circ_{I^-}$$

$$\Lambda^\circ_{NaNO_3} = \Lambda^\circ_{Na^+} + \Lambda^\circ_{NO_3^-}$$

So: $$\Lambda^\circ_{AgNO_3} + \Lambda^\circ_{NaI} - \Lambda^\circ_{NaNO_3} = \Lambda^\circ_{Ag^+} + \Lambda^\circ_{NO_3^-} + \Lambda^\circ_{Na^+} + \Lambda^\circ_{I^-} - \Lambda^\circ_{Na^+} - \Lambda^\circ_{NO_3^-}$$

$$= \Lambda^\circ_{Ag^+} + \Lambda^\circ_{I^-} = \Lambda^\circ_{AgI}$$

Substituting the values:

$$\Lambda^\circ_{AgI} = 13.3 + 12.7 - 12.0$$

$$= 14.0 \text{ mS m}^2 \text{ mol}^{-1}$$

Therefore, the limiting molar conductivity of AgI is 14 mS m$$^2$$ mol$$^{-1}$$.