Given the molar conductances at infinite dilution: Calculate $$\Lambda^\infty_{HOAc}$$ using appropriate molar conductances of the electrolytes listed above at infinite dilution in H$$_2$$O at $$25^\circ$$C

The value of $$\Lambda^\infty_{HOAc}$$ is obtained from Kohlrausch’s law of independent ionic migration.

Kohlrausch’s law
For any strong electrolyte AB that ionises as $$A^+ + B^-$$ at infinite dilution, the molar conductance is the sum of the ionic conductances of the cation and the anion:$$\Lambda^\infty_{AB} = \lambda^\infty_{A^+} + \lambda^\infty_{B^-}$$

For the four electrolytes involved, write the expressions:

$$\Lambda^\infty_{HCl} = \lambda^\infty_{H^+} + \lambda^\infty_{Cl^-}$$ $$-(1)$$
$$\Lambda^\infty_{NaCl} = \lambda^\infty_{Na^+} + \lambda^\infty_{Cl^-}$$ $$-(2)$$
$$\Lambda^\infty_{NaOAc} = \lambda^\infty_{Na^+} + \lambda^\infty_{OAc^-}$$ $$-(3)$$
$$\Lambda^\infty_{HOAc} = \lambda^\infty_{H^+} + \lambda^\infty_{OAc^-}$$ $$-(4)$$

Add equations $$-(1)$$ and $$-(3)$$, then subtract $$-(2)$$:

$$\left[\lambda^\infty_{H^+} + \lambda^\infty_{Cl^-}\right] + \left[\lambda^\infty_{Na^+} + \lambda^\infty_{OAc^-}\right] - \left[\lambda^\infty_{Na^+} + \lambda^\infty_{Cl^-}\right]$$
$$= \lambda^\infty_{H^+} + \lambda^\infty_{OAc^-} = \Lambda^\infty_{HOAc}$$

Hence, algebraically,

$$\boxed{\;\Lambda^\infty_{HOAc} = \Lambda^\infty_{HCl} + \Lambda^\infty_{NaOAc} - \Lambda^\infty_{NaCl}\;}$$

Insert the experimental values at $$25^\circ\text{C}$$:
$$\Lambda^\infty_{HCl} = 426.2\; \text{S cm}^2\ \text{mol}^{-1}$$
$$\Lambda^\infty_{NaOAc} = 91.0\; \text{S cm}^2\ \text{mol}^{-1}$$
$$\Lambda^\infty_{NaCl} = 126.5\; \text{S cm}^2\ \text{mol}^{-1}$$

$$\Lambda^\infty_{HOAc} = 426.2 + 91.0 - 126.5 = 390.7\; \text{S cm}^2\ \text{mol}^{-1}$$

Therefore, the molar conductance at infinite dilution for acetic acid is

Option C which is: $$390.7\; \text{S cm}^2\ \text{mol}^{-1}$$