The limiting molar conductivities $\left({\wedge }^0\right)$ for $NaCl,KBr$ and $KCl$ are 126, 152 and $150Sc{m}^{2}mo{l}^{-1}$ respectively. Then ${\wedge }^0$ for $NaBr$ is

The limiting molar conductivity, ${\wedge }^0$ for NaBr is calculated as: 

For any Electrolyte from   of individual ions

$$\begin{gathered} {\wedge }_{{ }_{\mathrm{NaBr}}}^0={\wedge }_{{ }_{\mathrm{NaCl}}}^0+{\wedge }_{{ }_{\mathrm{KBr}}}^0-{\wedge }_{{ }_{\mathrm{KCl}}}^0 \\ {\wedge }_{{ }_{\mathrm{NaBr}}}^0=\left(126+152-150\right) \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1} \\ =128 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1} \end{gathered}$$