Conductivity of 2.5 × 10^-4 M Methanoic acid is 5.25 × 10^-5 S cm^-1.What is its molar conductivity and degree of dissociation?
Given : λ⁰(H⁺) = 349.5 Scm² mol^-1
and λ⁰(HCOO^–) = 50.5 Scm² mol^-1. 

Molar Conductivity = 210 S cm² mol^-1 

Degree of dissociation = 0.525

The correct answer is option 3.

Molar Conductivity = 210 S cm² mol^-1 

Degree of dissociation = 0.525

Concentration is 2.5 x 10^-4 M
K = 5.25 × 10^-5 Scm^-1

\(\Lambda ^c_m = \frac{1000 × \kappa}{concentration }\)

\(⇒ \Lambda ^c_m = \frac{1000 × 5.2 × 10^{-5}}{2.5 × 10^{-4}} S \text{ }cm^2 \text{ }mol^{-1}\)

\(⇒ \Lambda ^c_m = \frac{ 5.2 × 10^2}{2.5} S \text{ }cm^2 \text{ }mol^{-1}\)

\(⇒ \Lambda ^c_m = \frac{ 525}{2.5} S \text{ }cm^2 \text{ }mol^{-1}\)

\(⇒ \Lambda ^c_m =210 S \text{ }cm^2 \text{ }mol^{-1}\)

Molar conductivity at infinite dilution,

\(\Lambda ^0_m = \lambda ^0_{H^+} + \lambda ^0_{HCOO^-}\)

\(⇒ \Lambda ^0_m = 349.5 + 50.5 = 400 S\text{ }cm^2\text{ }mol^{-1}\)

Degree of dissociation,

\(\alpha = \frac{\Lambda ^c_m}{\Lambda ^0_m}\)

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