The molar conductance of acetic acid at infinite dilution is `390.7 S cm^(2) mol^(-1)` . Calculate the molar conductance of 0.01 M acetic acid solution , given that the dissociation constant of a acetic acid is `1.8 xx 10^(-5)` .

Explanation

`underset(c(1-alpha))(CH_(3)COOH)hArrunderset(calpha)(CH_(3)COO^(-))+underset(calpha)(H^(+))`
`K_(eq)=calpha^(2)" or "alpha=sqrt((K_(eq))/(c))=(wedge_(m)^(c))/(wedge_(m)^(oo))`
`therefore(wedge_(m)^(c))/(390.7)=sqrt((1.8xx10^(-5))/(0.01))=sqrt(18xx10^(-4))=4.243xx10^(-2)` or `wedge_(m)^(c)=16.57Omega^(-1)cm^(2)`.