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प्रश्न
The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1. Calculate its degree of dissociation and dissociation constant. Given \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol−1 and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol−1.
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उत्तर
\[\ce{∧^0_{{m}(HCOOH)} = λ^0_{(H^+)} + λ^0_{(HCOO^-)}}\]
= 349.6 + 54.6
= 404.2 S cm2 mol−1
Given: \[\ce{∧_{{m}(HCOOH)}}\] = 46.1 S cm2 mol−1
Degree of dissociation, α = `∧_"m"/(∧_"m"^0)`
= `(46.1 "S cm"^2 "mol"^-1)/(404.2 "S cm"^2 "mol"^-1)`
= 0.114
\[\ce{HCOOH ⇌ HCOO^- + H^+}\]
| Initial concentration | c mol L−1 | 0 | 0 |
| Concentration at equilibrium | c(1 − α) | cα | cα |
∴ Kα = `("c"α . "c"α)/("c"(1 - α))`
= `("c"α^2)/(1 - α)`
= `(0.025 xx (0.114)^2)/(1 - 0.114)`
= 3.67 × 10−4
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