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.

The correct option is 390.71 S cm² mol⁻¹.

Step-by-step Explanation:

To calculate the molar conductance of acetic acid (${\mathrm{CH}}_3\mathrm{COOH}$) at infinite dilution (${\Lambda }_m^{\circ }$), we can use Kohlrausch's Law of independent migration of ions. According to this law, the molar conductance of an electrolyte at infinite dilution is the sum of the individual molar conductivities of its constituent ions.

For the given electrolytes, we can write their molar conductances at infinite dilution as follows:

1) For Sodium Chloride ($\mathrm{NaCl}$):
${\Lambda }_m^{\circ }\left(\mathrm{NaCl}\right)={\lambda }^{\circ }\left({\mathrm{Na}}^{+}\right)+{\lambda }^{\circ }\left({\mathrm{Cl}}^{-}\right)=126.45{\text{S cm}}^2{\text{mol}}^{-1}$

2) For Hydrochloric Acid ($\mathrm{HCl}$):
${\Lambda }_m^{\circ }\left(\mathrm{HCl}\right)={\lambda }^{\circ }\left(H^{+}\right)+{\lambda }^{\circ }\left({\mathrm{Cl}}^{-}\right)=426.16{\text{S cm}}^2{\text{mol}}^{-1}$

3) For Sodium Acetate (${\mathrm{CH}}_3\mathrm{COONa}$):
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COONa}\right)={\lambda }^{\circ }\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)+{\lambda }^{\circ }\left({\mathrm{Na}}^{+}\right)=91.0{\text{S cm}}^2{\text{mol}}^{-1}$

We want to find the molar conductance of Acetic Acid (${\mathrm{CH}}_3\mathrm{COOH}$) at infinite dilution:
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\lambda }^{\circ }\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)+{\lambda }^{\circ }\left(H^{+}\right)$

We can obtain this target expression by combining the conductances of the three given salts 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 values given in the question:
${\Lambda }_m^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)=91.0+426.16-126.45$

Performing the addition first:
$91.0+426.16=517.16$

Next, subtracting the molar conductance of NaCl:
$517.16-126.45=390.71{\text{S cm}}^2{\text{mol}}^{-1}$

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