The conductivity of 0.001 mol `L^(-1)` solution of `CH_(3)COOH` is `3.905xx10^(-5)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))`

Answer

`Lambda_(m) = 39.05 S cm^(2) mol^(-1) alpha = 0.1`

Explanation

`Lambda_(m) = (k xx 1000)/(M) = (3.905 xx 10^(-5) xx 1000)/(0.001)`
`= 39.05 S cm^(2) mol^(-1)`
`CH_(3) COOH hArr CH_(3) COO^(-) + H^(+)`
`Lambda_(m) = lambda_(CH_(3)COO^(-))^(@) + lambda_(H^(+))^(@)`
`= 40.9 + 349.6 = 390 .5 S cm^(2) mol^(-1)`
Degree of dissociation ,
`alpha = (Lambda_(m))/(Lambda_(m)^(@)) = (39.05)/(390.5) = 0.1`