Sunil Kumar FP
Last Activity: 11 Years ago
conductivity of .001028 mol/l acetic acid =4.95*10^-5Scm^-1
therefore molar conductivity=4.95*10^-5/.001028 =4815.17/1000=48.15Scm^3mol^-1
degree of dissociation,d=48.15/390.5=.1233
now dissociation constane=cd^2/1-d=.001028*.1233^2/1-.1233=1.78*10^-5molL^-1