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. Choose the right option for your answer.

To find the molar conductance of acetic acid (CH₃COOH) at infinite dilution, we use the fact that the molar conductance of an electrolyte is the sum of the contributions of its constituent ions.

Let

$a\; =\; \lambda ₀(Na⁺)$

$b\; =\; \lambda ₀(Cl⁻)$

$c\; =\; \lambda ₀(H⁺)$

$d\; =\; \lambda ₀(CH₃COO⁻)$

From the given data we have three equations:

$a\; +\; b\; =\; 126.45$     (1)  (molar conductance of NaCl)

$c\; +\; b\; =\; 426.16$     (2)  (molar conductance of HCl)

$a\; +\; d\; =\; 91.0$      (3)  (molar conductance of CH₃COONa)

We need the value of $c\; +\; d$, which is the molar conductance of CH₃COOH at infinite dilution.

Subtract (1) from (2) to eliminate $b$:

$(c\; +\; b)\; –\; (a\; +\; b)\; =\; 426.16\; –\; 126.45$

$c\; –\; a\; =\; 299.71$

Hence

$c\; =\; a\; +\; 299.71$

From (3) we solve for $d$:

$d\; =\; 91.0\; –\; a$

Now add $c$ and $d$:

$c\; +\; d\; =\; (a\; +\; 299.71)\; +\; (91.0\; –\; a)$

The $a$ terms cancel:

$c\; +\; d\; =\; 299.71\; +\; 91.0\; =\; 390.71$

Therefore, the molar conductance of CH₃COOH at infinite dilution is

$390.71 S cm2 mol-1$