Correct Answer - C
Equilibrium expression and composition are
`{:(,HA(aq.)hArr,H^(+)(aq.)+,A^(-)(aq.)),("Initial moles",C,0,0),("Equilibrium moles",C-Calpha,Calpha,Calpha):}`
or `C(1-alpha)`
According to the law of chemical equilibrium,
`K_(a)= (C_(H^(+))C_(A^(-)))/(C_(HA))= ((Calpha)(Calpha))/(C(1-alpha))`
`= (Calpha^(2))/(1-alpha)`
Considering `alpha lt lt 1`, we have
`K_(a)=Calpha^(2)`
or `alpha=sqrt((K_(a))/(C ))=sqrt((1.0xx10^(-5))/(0.1))`
`=sqrt(10^(-4))=10^(-2)`
`:.` % Dissociation `= 10^(-2)xx100%`
`=1%`