1.

Equivalent conductivity of dilute monovalent acidic solution (M)/(32) is 8.0 mol cm^(2) and at infinite dilution equivalent conductivity is 400 mol cm^(2) then the dissociation constant for acid is

Answer»

`1.25xx10^(-6)`
`6.25xx10^(-4)`
`1.25xx10^(-4)`
`1.25xx10^(-5)`

Solution :Degree of dissociation :
`alpha=(Lamda)/(Lamda_(OO))`
`=(8.00)/(400)=2XX10^(-2)`
`K_(a)=(Calpha^(2))/((1-alpha))~~Calpha^(2)`
`=(1)/(32)xx(2xx10^(-2))^(2)`
`=1.25xx10^(-5)`.


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