Equivalent conductivity of barium chloride at infinite dilution

Question

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.

Answer 1

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.