The limiting molar conductivities of NaI , NaNO_3 and AgNO_3 are 12.7, 12.0 and 13.3 mS m2 mol - 1, respectively (all at 25 ^\circ C). The limiting molar conductivity of AgI at this temperature is \underline\hspace2cm mS m2 mol - 1.

Electrochemistry PYQ — JEE Main 2022 · 27 Jun · Shift 1 · Q14 with Solution

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Correct answer: 14

$\Lambda_m^\circ(\text{electrolyte}) = \lambda_+^\circ + \lambda_-^\circ$

So,

$\Lambda_m^\circ(\text{NaI}) = \lambda_{\text{Na}^+}^\circ + \lambda_{\text{I}^-}^\circ = 12.7$

$\Lambda_m^\circ(\text{NaNO}_3) = \lambda_{\text{Na}^+}^\circ + \lambda_{\text{NO}_3^-}^\circ = 12.0$

$\Lambda_m^\circ(\text{AgNO}_3) = \lambda_{\text{Ag}^+}^\circ + \lambda_{\text{NO}_3^-}^\circ = 13.3$

We need:

$\Lambda_m^\circ(\text{AgI}) = \lambda_{\text{Ag}^+}^\circ + \lambda_{\text{I}^-}^\circ$

$\Lambda_m^\circ(\text{AgI}) = \Lambda_m^\circ(\text{AgNO}_3) + \Lambda_m^\circ(\text{NaI}) - \Lambda_m^\circ(\text{NaNO}_3)$

Substitute values:

$\Lambda_m^\circ(\text{AgI}) = 13.3 + 12.7 - 12.0$

$\Lambda_m^\circ(\text{AgI}) = 14.0$

$\boxed{14\ \text{mS m}^2\text{ mol}^{-1}}$