At infinite dilution, molar conductivity of ${\mathrm{Ba}}^{2+}$ and ${\mathrm{Cl}}^{-}$ions are =127.32 $\mathrm{S} {\mathrm{cm}}^2/\mathrm{mol}$ and 76.34$\mathrm{S} {\mathrm{cm}}^2/\mathrm{mol}$ respectively. What is  ${\wedge }_m^{\infty }$ for ${\mathrm{Bacl}}_2$ at same dilution ?

Here given is :
Molar conductivity of ${\mathrm{Ba}}^{2+}=127.32 \mathrm{s} {\mathrm{cm}}^2/{\mathrm{mol}}^{-1}$
Molar conductivity of${\mathrm{Cl}}^{-}=76.34 \mathrm{S} {\mathrm{cm}}^2/{\mathrm{mol}}^{-1}$
Following is reaction 

${\mathrm{BaCl}}_2\to {\mathrm{Ba}}^{2+}+2{\mathrm{Cl}}^{-}$
According to Kohlrausch's law,

${\wedge }_{\mathrm{m},B{\mathrm{aCl}}_2}^{\infty }={\wedge }_{\mathrm{m},{\mathrm{Ba}}^{2+}}^{\infty }+2\times {\wedge }_{\mathrm{m},{\mathrm{Cl}}^{-}}^{\infty }$
By putting given values in equation,

${\wedge }_{\mathrm{m},{\mathrm{Bacl}}_2}^{\infty }=127.32+2\times 76.34=280 \mathrm{S} {\mathrm{cm}}^2/{\mathrm{mol}}^{-}$

Therefore option A is right.