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.
Answer & explanation
Correct answer: option 3
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,
\(\alpha = \frac{\Lambda ^c_m}{\Lambda ^0_m}\)
\(⇒ \alpha = \frac{210}{400} = 0.525\)