(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 x 10-5 S cm!Given 2 =349.6S cm’ mol-'; %ch.coo = 40.98 cm’ mol'

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if co

1.	(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 x 10-5 S cm!Given 2 =349.6S cm’ mol-'; %ch.coo = 40.98 cm’ mol'
Answer» K =8.0×10^-5 s cm^-1M = 0.0024Molar conductance = (k× 1000)/m= (8.0×10^-5×1000)/0.0024= 33.33 s cm^2 mol^-1Now degree of DISSOCIATION = MOLAR conductance/sum of conductance of H+ and CH3COO-= 33.33/390.5= 0.0853