The limiting equivalent conductance of NaCI, KCI and KBr are 126.5, 150.0 and 152.0 S cm² eq^-1 respectively. The limiting equivalent ionic conductance of Br^- is 76 S cm² eq^-1. The limiting equivalent ionic conductance of Na⁺ is

$$\begin{gathered} {\wedge }_{KBr}^{\infty }={\mathrm{\lambda }}_{{\mathrm{Br}}^{-}}^{°}+{\mathrm{\lambda }}_{{\mathrm{K}}^{+}}^{°}=152 ........ \left(1\right) \\ {\wedge }_{\mathrm{KCl}}^{°}={\mathrm{\lambda }}_{{\mathrm{K}}^{+}}^{°}+{\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^{°}=150 ........ \left(2\right) \\ {\wedge }_{\mathrm{NaCl}}^{\infty }={\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^{°}+{\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^{°}=126.5 ......... \left(3\right) \end{gathered}$$

Therefore (1)+(3) - (2)  ⇒ ${\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^{°}+{\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^{°} + {\mathrm{\lambda }}_{{\mathrm{Br}}^{-}}^{°}+{\mathrm{\lambda }}_{{\mathrm{K}}^{+}}^{°}-{\mathrm{\lambda }}_{{\mathrm{K}}^{+}}^{°}-{\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^{°}={\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^{°}+ {\mathrm{\lambda }}_{{\mathrm{Br}}^{-}}^{°}$ 

⇒ ( 126.5 + 152 - 150 ) = 76 + ${\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^{°}$

⇒ ${\mathrm{\lambda }}_{{\mathrm{Na}}^{+}}^{°}$ = 52.5

$\therefore$Equivalent ionic conductance for Na⁺ is 52.5 cm² eq^-1.