The molar conductances of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm² mol^-1 respectively. The molar conductance of CH₃COOH at infinite dilution is. Choose the right option for your answer.

To find the molar conductance of acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) at infinite dilution, we can apply Kohlrausch's Law of Independent Migration of Ions.

Kohlrausch's Law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of its constituent anions and cations. Therefore, for acetic acid, we can write:
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\lambda }^{\circ }\left({{\mathrm{CH}}_3\mathrm{COO}}^{-}\right)+{\lambda }^{\circ }\left(H^{+}\right)$

We are given the molar conductances at infinite dilution (${\Lambda }_m^{\circ }$) for the following strong electrolytes:
1. Sodium chloride (NaCl):
${\Lambda }_m^{\circ }\left(\mathrm{NaCl}\right)={\lambda }^{\circ }\left({\mathrm{Na}}^{+}\right)+{\lambda }^{\circ }\left({\mathrm{Cl}}^{-}\right)=126.45{\text{S cm}}^2{\text{mol}}^{-1}$
2. Hydrochloric acid (HCl):
${\Lambda }_m^{\circ }\left(\mathrm{HCl}\right)={\lambda }^{\circ }\left(H^{+}\right)+{\lambda }^{\circ }\left({\mathrm{Cl}}^{-}\right)=426.16{\text{S cm}}^2{\text{mol}}^{-1}$
3. Sodium acetate (${\mathrm{CH}}_3\mathrm{COONa}$):
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COONa}\right)={\lambda }^{\circ }\left({{\mathrm{CH}}_3\mathrm{COO}}^{-}\right)+{\lambda }^{\circ }\left({\mathrm{Na}}^{+}\right)=91.0{\text{S cm}}^2{\text{mol}}^{-1}$

To obtain the expression for ${\mathrm{CH}}_3\mathrm{COOH}$, we can combine the equations for the strong electrolytes. We want to keep the acetate ion (${{\mathrm{CH}}_3\mathrm{COO}}^{-}$) and the hydrogen ion ($H^{+}$), while canceling out the sodium (${\mathrm{Na}}^{+}$) and chloride (${\mathrm{Cl}}^{-}$) ions. This can be achieved as follows:
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COONa}\right)+{\Lambda }_m^{\circ }\left(\mathrm{HCl}\right)-{\Lambda }_m^{\circ }\left(\mathrm{NaCl}\right)$

Substituting the given numerical values into the equation:
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)=91.0+426.16-126.45$

Performing the addition and subtraction:
$91.0+426.16=517.16$
$517.16-126.45=390.71{\text{S cm}}^2{\text{mol}}^{-1}$

Thus, the molar conductance of acetic acid at infinite dilution is 390.71 S cm² mol⁻¹.