The conductivity of 0.001 `" mol "L^(-1)` solution of `CH_(3)COOH` is `4.95xx10^(-5)" S "cm^(-1)`. Calculate its molar conductance and degree of dissociation (alpha).`"Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)`

Explanation

`k=4.95xx10^(-5)" S "cm^(-1)`
`C=0.001" mol "L^(-1)=((0.001" mol"))/((10^(3)"cm"^(3)))=10^(-6)" mol "cm^(-3)`
Molar conductance at given concentration
`Lambda_(m)^(c)=(k)/(C )=((4.95xx10^(-5)" S "cm^(-1)))/((10^(-6)" mol "cm^(-3)))=49.5" S "cm^(2)mol^(-1)`.
Molar conductance at infinite dilution. `(Lambda^(@))`
`Lambda_(m)^(@)=(lambda_(CH_(3)COO^(-))^(@)+lambda_(H^(+))^(@))=(40.9+349.6)" S "cm^(2)mol^(-1)`
`=390.5" S "cm^(2)mol^(-1)`
Degree of dissociation `(alpha)=(Lambda_(m)^(c))/(Lambda_(m^(@)))=((49.5" S "cm^(2)mol^(-1)))/((390.5" S "cm^(2)mol^(-1)))=0.127`