The conductivity of 2.5 × 10−4 M methanoic acid is 5.25 × 10−5 S cm−1…

The correct answer is Option (3) → Molar conductivity = 210 S cm² mol⁻¹; degree of dissociation = 0.525

The molar conductivity is the conductivity per mole of electrolyte. With κ in S cm⁻¹ and the concentration c in mol L⁻¹:

    \( \displaystyle \Lambda_m=\frac{\kappa\times1000}{c} \)

For a weak electrolyte such as methanoic acid, the degree of dissociation is the ratio of the molar conductivity to the limiting molar conductivity:

    \( \displaystyle \alpha=\frac{\Lambda_m}{\Lambda^{\circ}_m} \)

Molar conductivity:

    \( \displaystyle \Lambda_m=\frac{5.25\times10^{-5}\times1000}{2.5\times10^{-4}} \)

    \( \displaystyle \Lambda_m=210\,\mathrm{S\,cm^{2}\,mol^{-1}} \)

Limiting molar conductivity, from the ions \( \ce{H^+} \) and \( \ce{HCOO^-} \) (Kohlrausch’s law):

    \( \displaystyle \Lambda^{\circ}_m=\lambda^{\circ}_{\ce{H^+}}+\lambda^{\circ}_{\ce{HCOO^-}}=349.5+50.5=400\,\mathrm{S\,cm^{2}\,mol^{-1}} \)

Degree of dissociation:

    \( \displaystyle \alpha=\frac{210}{400}=0.525 \)

So the acid has a molar conductivity of 210 S cm² mol⁻¹ and is 52.5% dissociated.