Molar conductivities `(Lambda_(m)^(@))` at infinite dilution of `NaCl, HCl` and `CH_(3)COONa` arc `126.4, 425.9` and `91.0 S cm^(2) mol^(-1)` respectively. `Lambda_(m)^(@)` for `CH_(3)COOH` will be

Explanation

Molar conductivity at infinite dilution for acetic acid can be calculated as :
`Lambda^(@) (CH_(3) COOH) = lambda_(H)^(@) + lambda_(CH_(3) COO^(-))^(@) = lambda_(CH_(3) COO^(-))^(@) + lambda_(Na^(+))^(@) + lambda_(H^(+))^(@) + lambda_(Cl^(-))^(@) - lambda_(Na^(+))^(@) - lambda_(Cl^(-))^(@)`
`Lambda^(@) (CH_(3) COOH) = Lambda^(@) (CH_(3) COONa) + Lambda^(@) (HCl ) - Lambda^(@) (NaCl)`
`Lambda_(@) (CH_(3) COONa) = 91.0 S cm^(2) mol^(-1)`
`Lambda^(@) (HCl) = 425.9 S cm^(2) mol^(-1)`
`Lambda^(@) (NaCl) = 126.4 S cm^(2) mol^(-1)`
`therefore Lambda^(@) (CH_(3) COOH) = 91.0 + 425.9 - 126.4`
= `390.5 S cm^(2) mol^(-1)`