The limiting equivalent conductance of NaCl, KCl and KBr are 126.5,150.0 and 152.0   ${\text{S\hspace{0.17em}\hspace{0.17em}cm}}^2{\text{\hspace{0.17em}eq}}^{\text{-1}}$respectively. The limiting equivalent ionic conductance of $B{r}^{-}$  is $76Sc{m}^{2}e{q}^{-1}$ . The limiting equivalent ionic conductance of $N{a}^{+}$  is  
 

${\wedge }_{NaCl}^{\infty }={\lambda }_{N{a}^{+}}^0+{\lambda }_{C{l}^{-}}^0=126.5\mathrm{.....}\left(i\right)$

${\wedge }_{KCl}^{\infty }={\lambda }_{K^{+}}^0+{\lambda }_{C{l}^{-}}^0=150\mathrm{.....}\left(ii\right)$

${\wedge }_{KBr}^{\infty }={\lambda }_{K^{+}}^0+{\lambda }_{B{r}^{-}}^0=152\mathrm{.....}\left(iii\right)$

Adding (i) & (iii)  subtract (ii)

${\lambda }_{Na}^0+{\lambda }_{C{l}^{-}}^0+{\lambda }_{K^{+}}^0+{\lambda }_{B{r}^{-}}^0-{\lambda }_{K^{+}}^0-{\lambda }_{C{l}^{-}}^0={\lambda }_{N{a}^{+}}^0+{\lambda }_{B{r}^{-}}^0$

$=(126.5+152-150)=\left(76\right)+{\lambda }_{N{a}^{+}}^0$
${\lambda }_{N{a}^{+}}^0=52.5$
 Equivalent ionic conductance for $N{a}^{+}$   is  52.5.