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Write Brief Answer · Q8

Q.The conductivity of a 0.01M solution of a 1:1 weak electrolyte at 298 K is 1.5×10−4 S cm−11.5\times10^{-4}\ \text{S cm}^{-1}.

i) molar conductivity of the solution
ii) degree of dissociation and the dissociation constant of the weak electrolyte Given that λcationo=248.2 S cm2mol−1\lambda^{o}_{cation} = 248.2\ \text{S cm}^2\text{mol}^{-1} and λaniono=51.8 S cm2mol−1\lambda^{o}_{anion} = 51.8\ \text{S cm}^2\text{mol}^{-1}.
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Step 1. Molar conductivity: Λm=1000 κM=1000×1.5×10−40.01=0.150.01=15\Lambda_m = \dfrac{1000\,\kappa}{M} = \dfrac{1000\times1.5\times10^{-4}}{0.01} = \dfrac{0.15}{0.01} = 15 S cm² mol⁻¹.

Step 2. Limiting molar conductivity, by Kohlrausch's law: Λmo=λcationo+λaniono=248.2+51.8=300\Lambda_m^{o} = \lambda^{o}_{cation}+\lambda^{o}_{anion} = 248.2+51.8 = 300 S cm² mol⁻¹.

Step 3. Degree of dissociation: α=Λm/Λmo=15/300=0.05\alpha = \Lambda_m/\Lambda_m^{o} = 15/300 = 0.05. …

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