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The molar conductivity of 0.025 mol L^(-...

The molar conductivity of 0.025 mol `L^(-1)` methanoic acid is 46.1 S `cm^(2)mol^(-1)`. Calculate its degree of dissociation and dissociation constant. Given `lamda^(@)(H^(+))=349.6" S "cm^(2)mol^(-1) and lamda^(@)(HCO O^(-))=54.6" S "cm^(2)mol^(-1)`.

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To solve the problem, we need to calculate the degree of dissociation (α) and the dissociation constant (K) of methanoic acid (HCOOH) given the molar conductivity and the individual ion conductivities. ### Step-by-Step Solution: 1. **Identify Given Data:** - Molar conductivity of methanoic acid (λC) = 46.1 S cm² mol⁻¹ - Concentration (C) = 0.025 mol L⁻¹ - Conductivity of H⁺ ion (λ⁰(H⁺)) = 349.6 S cm² mol⁻¹ ...
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The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Its degree of dissociation (alpha) and dissociation constant. Given lambda^(@)(H^(+))=349.6 S cm^(-1) and lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

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