Kohlrausch's Law

By Kohlrausch's law: $\Lambda^\circ_{m(CH_3COOH)} = \Lambda^\circ_{m(CH_3COONa)} + \Lambda^\circ_{m(HCl)} - \Lambda^\circ_{m(NaCl)}$ $= 91.0 + 426.16 - 126.45 = 390.71\ S\ cm^2\ mol^{-1}$

By Kohlrausch's law, $\lambda^0_{CH_3COOH}=\lambda^0_{CH_3COO^-}+\lambda^0_{H^+}=\lambda^0_{CH_3COOK}+\frac{1}{2}\lambda^0_{H_2SO_4}-\frac{1}{2}\lambda^0_{K_2SO_4}=z+\frac{x}{2}-\frac{y}{2}=\frac{(x-y)}{2}+z$.

By Kohlrausch's law, $\Lambda_m^\circ(NH_4OH) = \lambda^\circ(NH_4^+) + \lambda^\circ(OH^-)$
$\Lambda_m^\circ(NH_4Cl) = \lambda^\circ(NH_4^+) + \lambda^\circ(Cl^-)$
$\Lambda_m^\circ(NaOH) = \lambda^\circ(Na^+) + \lambda^\circ(OH^-)$
$\Lambda_m^\circ(NaCl) = \lambda^\circ(Na^+) + \lambda^\circ(Cl^-)$
Adding the first two and subtracting the third cancels $Na^+$ and $Cl^-$:
$\Lambda_m^\circ(NH_4OH) = \Lambda_m^\circ(NH_4Cl) + \Lambda_m^\circ(NaOH) - \Lambda_m^\circ(NaCl)$

By Kohlrausch's law: $\Lambda_m^0(CH_3COOH) = \Lambda_m^0(CH_3COONa) + \Lambda_m^0(HCl) - \Lambda_m^0(NaCl)$
(The $Na^+$ and $Cl^-$ contributions cancel, leaving $CH_3COO^-$ and $H^+$.)
$= 91.0 + 425.9 - 126.4 = 390.5$ S $cm^2$ $mol^{-1}$

Kohlrausch's law: at infinite dilution each ion makes its own contribution to the conductance, independent of the other ion.
In terms of equivalent conductances, one equivalent of any electrolyte contains one equivalent of the cation and one equivalent of the anion.
So $\Lambda^\circ_{eq}(\text{electrolyte}) = \Lambda^\circ_{eq}(\text{cation}) + \Lambda^\circ_{eq}(\text{anion})$, with no stoichiometric multipliers.
The ionic values given are already equivalent conductances, so: $\Lambda^\circ_{eq}[Al_2(SO_4)_3] = \Lambda^\circ_{Al^{3+}} + \Lambda^\circ_{SO_4^{2-}}$
(The form $2\Lambda^\circ + 3\Lambda^\circ$ would apply only if molar ionic conductances were given and the molar conductance were asked.)

Kohlrausch's law of independent migration of ions: at infinite dilution, when dissociation is complete, each ion makes a definite contribution to the equivalent (molar) conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
$\Lambda_m^\infty = \lambda_+^\infty + \lambda_-^\infty$
Here $\lambda_+^\infty$ and $\lambda_-^\infty$ are the ionic conductances at infinite dilution for the cation and the anion respectively.
The law holds only at infinite dilution, because only then are the ions far apart and free from inter-ionic attractions; so the statement with 'finite dilution' is wrong.
The contribution of an ion is independent of the other ion, so the statement 'depending on the nature of the other ion' is wrong.
The contribution is to the equivalent (or molar) conductance, not simply to the conductance.
The law is used to find $\Lambda^\infty$ of weak electrolytes such as acetic acid.

By Kohlrausch's law, the limiting equivalent conductance of an electrolyte is the sum of the limiting equivalent conductances of its ions.
The value 127 for barium corresponds to the whole $Ba^{2+}$ ion; per equivalent it is $\dfrac{1}{2} \times 127 = 63.5$.
$\Lambda^\circ_{eq}(BaCl_2) = \lambda^\circ\left(\dfrac{1}{2}Ba^{2+}\right) + \lambda^\circ(Cl^-) = 63.5 + 76$
$= 139.5$ ohm$^{-1}$ cm$^2$ eq$^{-1}$