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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 `∧_m^0` for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
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उत्तर
`∧_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
K = `(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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\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
\ce{\Lambda^{\circ}_{H^+} = 350 S cm^2 mol^{-1}}\\
\ce{\Lambda^{\circ}_{CH_3COO^-} = 50 S cm^2 mol^{-1}}
\end{bmatrix}\]
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