At 300 K specific conductivity of ethanol is `4xx10^(-10)mhocm^(-1)`. The ionic conductances of `H^(+),C_(2)H_(5)O^(-)` at his temperature is 300 and 100 `mhocm^(2)" equivalent"^(-1)` respectively. Then the negative logarithm of ionic product of alcohol will be 18.

At 300 K specific conductivity of ethanol is `4xx10^(-10)mhocm^(-1)`.

1.	At 300 K specific conductivity of ethanol is `4xx10^(-10)mhocm^(-1)`. The ionic conductances of `H^(+),C_(2)H_(5)O^(-)` at his temperature is 300 and 100 `mhocm^(2)" equivalent"^(-1)` respectively. Then the negative logarithm of ionic product of alcohol will be 18.
Answer» Correct Answer - T `lamda^(infty)=lamda_(H^(+))^(infty)+lamda_(C_(2)H_(5)O^(-))^(infty)=400` `thereforelamda=kxx(1000)/(C)" "C=[H^(+)]=[.^(-)OC_(2)H_(5)]` `thereforeC=(4xx10^(-10)xx1000)/(400)=10^(-9)M` `thereforeK_("alcohol")=[H^(+)][OC_(2)H_(5)]=(10^(-9))^(2)` `thereforepK_("alcohol")=-log(10^(-18))=18`

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At 300 K specific conductivity of ethanol is `4xx10^(-10)mhocm^(-1)`. The ionic conductances of `H^(+),C_(2)H_(5)O^(-)` at his temperature is 300 and 100 `mhocm^(2)" equivalent"^(-1)` respectively. Then the negative logarithm of ionic product of alcohol will be 18.

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