Equivalent conductivity of barium chloride at infinite dilution Find out the equivalent conductivity of barium chloride at infinite dilution given that the ionic conductivities of $\ce{Ba^{2+}, Cl-}$ are $\pu{127 ohm^-1 cm^2 equivalent^-1}, \pu{76 ohm^-1 cm^2 equivalent^{-1}}$ respectively.   


Equivalent conductivity is defined as.  


  The equivalent conductance of an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.   


One gram equivalent weight would be simply half of the entire compound so
$$\text{Equivalent Conductivity} = \dfrac{1}{2} \times 127 + \dfrac{1}{2} \times 2 \times 76$$   

This comes out to be 139.5 however the answer is 203.
One must be really careful with the units while dealing with conductivity, conductance problems. 

If you read the question carefully, the equivalent conductivity of $\ce{Ba^{2+}}$ and $\ce{Cl^{-1}}$ are provided to you. 

So the molar conductivity of $\ce{Ba^2+}$ is: $2 \times 127 ~\pu{ohm^-1 mole^-1} = 254 ~\pu{ohm^-1 mole^-1}$

and that of $\ce{Cl^{-1}} = 76~\pu{ohm^-1 cm^2 mole^-1} $

Now apply Kohlrausch's law of molar conductivity of solution at infinite dilution: 

$\lambda^o_{\ce{BaCl2}}= \lambda^o _{\ce{Ba^{2+}}} + 2\lambda^o_{\ce{Cl^{-1}}} \\ \implies  \lambda^o_{\ce{BaCl2}} = 254 + 2\times 76 = 406 ~\pu{ohm^-1 cm^2 mole^-1}$

Now, as you say, equivalent conductivity is $\dfrac{1}{2}$ times the molar conductivity for $\ce{BaCl_2}$ so equivalent conductivity of $\ce{BaCl_2} = \dfrac{1}{2}\times 406 = 203 ~ \pu{ohm^-1 cm^2 equivalent^-1}$ which is the correct answer. 

Alternatively, observe that $203 = 127 + 76$ so Kohlrausch's law for equivalent conductivity of strong electrolyte at infinite dilution may be stated as: 


  The equivalent conductivity of a strong electrolyte at infinite dilution is equal to the sum of the equivalent conductivities of the anions and cations.