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प्रश्न
Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if `Lambda_m^0` for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
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उत्तर
Given: Conductivity (κ) = 7.896 × 10−5 S cm−1
Molar conductivity (M) = 0.00241
`Lambda_m^0` = 390.5 S cm2 mol−1
`∧_m^c = (kappa xx 1000)/"Molarity"`
= `(7.896 xx 10^-5 xx 1000)/0.00241`
= 32.763 S cm2 mol−1
α = `(∧_m^c)/(∧_m^0)`
= `32.763/390.5`
= 0.084
Kα = `(alpha^2 c)/(1 - alpha)`
= `((0.084)^2 xx 0.00241)/(1 - 0.084)`
= `((0.084)^2 xx 0.00241)/(0.916)`
= 1.86 × 10−5
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