The conductivity of `10^(-3)` mol/L acetic acid at 25 °C is `4.1 xx 10^(-5) cm^(-1)` . Calculate its degree of dissociation, `Lambda_(m)^(@)` if for acetic acid at 25 °C is `390.5 S cm^(2) mol^(-1)`.

Explanation

Apply the following relation:
`Lambda_(m) = (1000 k)/C`
Substituting the values in the above equation, we have
`Lambda_(m) =(1000 xx 4.1 xx 10^(-5))/10^(-3) = 41 S cm^(2) mol^(-1)`
Degree of dissociation is given by the following equation:
`alpha = Lambda_(m)^(C )/Lambda_(m)^(@) = 41/(390.5)= 0.105` or `10.5%`