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The conductivity of "0.001028 mol.L"^(-1...

The conductivity of `"0.001028 mol.L"^(-1)` acetic acid is `4.95xx10^(-5)"S.cm"^(-1)`. Calculate its dissociation constant if `Lambda_(m)^(@)` for acetic acid is `"390.5 S.cm"^(2)."mol"^(-1)`.

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`Lambda_(m)=1000xx(kappa)/(C)=(1000xx4.96xx10^(-5))/(0.001028)="48.15 S.cm"^(3)."mol"^(-1)`
`therefore" "alpha=(Lambda_(m))/(Lambda_(m)^(@))=("48.15 S.cm"^(2)."mol"^(-1))/("390.5 S.cm"^(2)."mol"^(-1))=0.1233,`
`therefore" "kappa=(C alpha^(2))/((1-alpha))=("0.001028 mol.L"^(-1)xx(0.1233)^(2))/(1-0.1233)`
`=1.78xx10^(-5)" mol.L"^(-1)`
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