Calculate the degree of dissociation `(alpha)` of acetic acid if its molar conductivity `(^^_(m))` is 39.05 `S cm^(2) mol^(-1)`
Given `lamda^(@) (H^(+)) = 349.6 cm^(2) mol^(-1) and lamda^(2) (CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)`

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 xx 10^(-5)" S "cm^(-1) . Given . lambda_(H^(+))^(@) = 349.6" S " cm^(2)" mol"^(-1), lamda_(CH_(3)COO^(-))^(@)= 40.9" S " cm^(2) " mol "^(-1) (b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer.

Calculate degree of dissocation of 0.02 M acetic acid at 298K. Given that Lamda_(m) (CH_(3)COOH)= 17.37 S cm^(2) mol^(-1), lamda_(m)^(@) (H^(+)) = 345.8 S cm^(2) mol^(-1), lamda_(m)^(@) (CH_(3)COO^(-)) = 40.2S cm^(2) mol^(-1)

Calculate degree of dissociation of 0.02 M acetic acid at 298 K given that mho_(m)(CH_(3)COOH)=17.37 cm^(2) mol^(-1),lambda_(m)^(@)(H+)=345.8 S cm^(2) mol^(-1), lambda_(m)^(@)(CH_(3)COO^(-))=40.2 Scm^(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?