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प्रश्न
Dissociation constant and molar conductance of an acetic acid solution are 1.78 × 10–5 mol L–1 and 48.15 S cm–2 mol–1 respectively. The conductivity of the solution is (considering molar conductance at infinite dilution is 390.5 S cm–2 mol–1)
पर्याय
4.9 × 10–2 S cm–1
4.9 × 102 S cm–1
4.9 × 10–5 S cm–1
4.9 × 105 S cm–1
MCQ
उत्तर
4.9 × 10–5 S cm–1
Explanation:
Kα = 1.78 × 10–5 mol L–1
`Λ_m` = 48.15 S cm–5 mol–1
`Λ_m^(circ)` = 390.5 S cm2 mol–1
`alpha = Λ_m/Λ_m^(circ) = 48.15/390.5` = 0.123
`K_alpha = (Calpha^2)/((1 - alpha)) = (C xx (0.123)^2)/((1 - 0.123)) = (C xx 0.01513)/0.877`
C = `(1.78 xx 10^(-5) xx 0.877)/0.01513` mol L–1 = 103.17 × 10–5
`Λ_m` = `(1000 xx k)/C`
`k = (48.15 xx 103.17 xx 10^(-5))/1000` = 4.9 × 10–5
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