Calculate molar conductivity at infinite dilution for $\mathrm{NaBr}$ if molar conductivity at infinite…

Calculate molar conductivity at infinite dilution for $\mathrm{NaBr}$ if molar conductivity at infinite dilution for $\mathrm{NaCl}, \mathrm{KBr}$ and $\mathrm{KCl}$ are 126, 152 and $150 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively
  1. $128 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  2. $302 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  3. $278 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  4. $176 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \wedge{ }^0 \mathrm{NaBr}=\wedge{ }^0 \mathrm{NaCl}+\wedge^{\circ} \mathrm{KBr}-\wedge^{\circ} \mathrm{KCl} \\ & =126+152-150 \\ & =128 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mole}^{-1}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 1)

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