The molar conductance 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

The correct option is 390.71 S cm² mol⁻¹.

Step-by-Step Explanation:
According to Kohlrausch's Law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is the sum of the limiting molar conductivities of its constituent ions.

We want to find the molar conductance of acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) at infinite dilution:

${\Lambda }_m^{\infty }\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\lambda }_{H^{+}}^{\infty }+{\lambda }_{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}^{\infty }$

We are given the molar conductances at infinite dilution for the following strong electrolytes:
1. For sodium chloride ($\mathrm{NaCl}$):
${\Lambda }_m^{\infty }\left(\mathrm{NaCl}\right)={\lambda }_{{\mathrm{Na}}^{+}}^{���}+{\lambda }_{{\mathrm{Cl}}^{-}}^{\infty }=126.45{\text{S cm}}^2{\text{mol}}^{-1}$ (Equation 1)

2. For hydrochloric acid ($\mathrm{HCl}$):
${\Lambda }_m^{\infty }\left(\mathrm{HCl}\right)={\lambda }_{H^{+}}^{\infty }+{\lambda }_{{\mathrm{Cl}}^{-}}^{\infty }=426.16{\text{S cm}}^2{\text{mol}}^{-1}$ (Equation 2)

3. For sodium acetate (${\mathrm{CH}}_3\mathrm{COONa}$):
${\Lambda }_m^{\infty }\left({\mathrm{CH}}_3\mathrm{COONa}\right)={\lambda }_{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}^{\infty }+{\lambda }_{{\mathrm{Na}}^{+}}^{\infty }=91.0{\text{S cm}}^2{\text{mol}}^{-1}$ (Equation 3)

To obtain the expression for ${\Lambda }_m^{\infty }\left({\mathrm{CH}}_3\mathrm{COOH}\right)$, we can add Equation 2 and Equation 3, and then subtract Equation 1:

${\Lambda }_m^{\infty }\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\Lambda }_m^{\infty }\left({\mathrm{CH}}_3\mathrm{COONa}\right)+{\Lambda }_m^{\infty }\left(\mathrm{HCl}\right)-{\Lambda }_m^{\infty }\left(\mathrm{NaCl}\right)$

Let us verify the algebraic combination of the ions:
$({\lambda }_{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}^{\infty }+{\lambda }_{{\mathrm{Na}}^{+}}^{\infty })+({\lambda }_{H^{+}}^{\infty }+{\lambda }_{{\mathrm{Cl}}^{-}}^{\infty })-({\lambda }_{{\mathrm{Na}}^{+}}^{\infty }+{\lambda }_{{\mathrm{Cl}}^{-}}^{\infty })={\lambda }_{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}^{\infty }+{\lambda }_{H^{+}}^{\infty }$

Substitute the given values into the equation:

${\Lambda }_m^{\infty }\left({\mathrm{CH}}_3\mathrm{COOH}\right)=91.0+426.16-126.45$

Let's perform the calculation:
$91.0+426.16=517.16$
$517.16-126.45=390.71{\text{S cm}}^2{\text{mol}}^{-1}$

Therefore, the molar conductance of ${\mathrm{CH}}_3\mathrm{COOH}$ at infinite dilution is 390.71 S cm² mol⁻¹.