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प्रश्न
The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α) Given λ°(H+)= 349.6 S cm2 mol-1 and λ°(CH3COO)= 40.9S cm2mol-1.
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उत्तर
Given:
Conductivity (k) = 3.905 x 10-5 S cm-1
Concentration ofelectrolyte (c) = 0.001 mol L-1
`^^_m=k/cxx1000=39.05S cm^2mol^(-1)`
`^^_m^@=lambda_(CH_3COO^-)^@+lambda_(H+)^@`
= 40.9 + 349.6
= 390.5 S cm2 mol-1
Degree of dissociation = `39.05/390.5=0.1`
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\[\begin{array}{cc}
\end{array}\]\[\begin{bmatrix}
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