The molar conductivity of $\text{KNO}_3\text{, NaCl and KCl}$ are 111, 128 and $152\text{ S cm}^2\text{ mol}^{-1}$, respectively. The molar conductivity of $\text{NaNO}_3$ will be:

$ 87\ \text{S}\ \text{cm}^2\ \text{mol}^{-1} $

The correct answer is Option (4) → $ 87\ \text{S}\ \text{cm}^2\ \text{mol}^{-1} $

Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte

Given in question,

Let:

$\Lambda_{\text{m}}(\text{KNO}_3)=\lambda_{\text{K}^{+}}+\lambda_{\text{NO}_3^{-}}=111 $

$\Lambda_{\text{m}}(\text{NaCl})=\lambda_{\text{Na}^{+}}+\lambda_{\text{Cl}^{-}}=128 $

$\Lambda_{\text{m}}(\text{KCl})=\lambda_{\text{K}^{+}}+\lambda_{\text{Cl}^{-}}=152 $

Molar conductivity of:

$\text{NaNO}_3=\lambda_{\text{Na}^{+}}+\lambda_{\text{NO}_3^{-}}=\text{Equation }1+2-3=111+128-152=87 $

Molar conductivity of: $NaNO_3$

$ 87\ \text{S}\ \text{cm}^2\ \text{mol}^{-1} $