Conductivity of `2xx10^(-3)M` methanoic acid is `8xx10^(-5)" S cm"^(-1)`. Calculate its molar conductivity and degree of dissociation if `Lambda_(m)^(@)` for methanoic acid is `404" S cm"^(2)" mol"^(-1)`.

Explanation

Molar conductivity `Lamda_(m)=(kxx1000)/(C)=(8xx10^(-5)" S cm"^(-1)xx1000)/(2xx10^(-3)" mol L"^(-1))`
`=(8xx10^(-2))/(2xx10^(-3))40" S cm"^(2)" mol"^(-1)`
Degree of dissociation `(Lamda_(m))/(Lamda_(m)^(@))=(40)/(404)=0.099`