AP EAMCET · Chemistry · Electrochemistry
The molar conductivity of \(0.027 \mathrm{M}\) methanoic acid is \(40.42 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\). The value of dissociation constant of this acid is
\(\left(\right.\) Given \(\lambda_{\mathrm{H}^{+}}^{\circ}=349.6 \mathrm{~S}_{.}, \mathrm{cm}^2 \mathrm{~mol}^{-1}\) and \(\lambda_{\mathrm{HCOO}^{-}}^{\circ}=54.6 \mathrm{~S}\) \(\left.\mathrm{cm}^2 \mathrm{~mol}^{-1}\right)\)
- A \(1.5 \times 10^{-5}\)
- B \(6.0 \times 10^{-5}\)
- C \(4.5 \times 10^{-4}\)
- D \(3.0 \times 10^{-4}\)
Answer & Solution
Correct Answer
(D) \(3.0 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {} \Lambda_{\mathrm{m}}^{\mathrm{o}}(\mathrm{HCOOH})=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{HCOO}^{-}}^0 \\ & =349.6+54.6 \\ & =404.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \Rightarrow…
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