The molar conductivity of 0.025 M methanoic | vedclass

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Question

(B) The molar conductivity at infinite dilution for methanoic acid $(HCOOH)$ is calculated using Kohlrausch's law: $\Lambda_m^0(HCOOH) = \lambda^0(H^+

Vedclass · Accepted Answer

$0.114$

Vedclass · Answer

(B) The molar conductivity at infinite dilution for methanoic acid $(HCOOH)$ is calculated using Kohlrausch's law: $\Lambda_m^0(HCOOH) = \lambda^0(H^+) + \lambda^0(HCOO^-)$.Substituting the given values: $\Lambda_m^0(HCOOH) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$.The degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at a given concentration $(\Lambda_m)$ to the molar conductivity at infinite dilution $(\Lambda_m^0)$: $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$.$\alpha = \frac{46.1}{404.2} \approx 0.114$.

The molar conductivity of $0.025 \ M$ methanoic acid is $46.1 \ S \ cm^2 \ mol^{-1}$. Calculate its degree of dissociation. Given: $\lambda^0(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^0(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.

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