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The conductivity of `0.001028M` acetic acid is `4.95xx10^(-5)Scm^(-1)`. Calculate its dissociation constant if `Lambda_(m)^(0)` for acetic acid is `390.5Scm^(2)mol^(-1)`. |
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Answer» We know, `Lambda_(m)=kxx(1000)/(M)`.....(i) Given : `"k=4.95 x 10"^(-5)"S cm"^(-1),M=0.001028` `therefore"From"(i)Lambda_(m)=4.95xx10^(-5)xx(1000)/(0.001028)` `="48.15 ohm"^(-1)cm^(2)mol^(-1)` Degree of dissociation, |
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