The conductivity of 0.001 mol solution of `CH_(3)COOH` is `3.905xx10^(-5)" S.cm"^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `["Given: "lambda^(@)(H^(+))="349.6 S.cm"^(2)."mol"^(-1)` and `lambda^(@)(CH_(3)COO^(-))="40.9 S.cm"^(2)."mol"^(-1)]`

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))

The conductivity of 0.001 "mol" L^(-1) solution of CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1) . Calculate its molar conductivity and degree of dissociation (alpha) . "Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and " lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) "mol"^(-1) (b) Define electrochemical cell. What happens if external potential applied becomes greater than E_(cell)^(@) of electrochemical cell?

The conductivity of 0.001 "mol" L^(-1) solution of CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1) . Calculate its molar conductivity and degree of dissociation (alpha) . "Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and " lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) per mol)

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)

The conductivity of 0.001 mol L^-1 solution of CH_3COOH is 3.905xx10^-5 S cm^-1 . Calculate the molar conductivity and degree of dissociation Give that lambda^@(H^+)=349.65 S cm^2 mol^-1 and lambda^@(CH_2COO^-40.9 S cm^2 mol^-1 .