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