The specific conductivity of 0.1 M solution of a weak acid at 298 K is 0.00159 `"ohm^(-1) cm^(-1)` . Calculate the degree of dissociation of the acid if its molar conductance at infinite dilution is 350.5 `"ohm"^(-1) cm^(2) mol^(-1)` .

Explanation

Specific conductivity, `k= 0.00159 ohm^(-1) cm^(-1)`
`C= 0.1M = 0.1mol L^(-1)=1 0^(-1) mol cm^(-3)`
Molar conductivity, `Lamda_(m)^(@) = (k)/(C )= (0.00159 ohm^(-1) cm^(-1))/(10^(-4) cm^(-3) mol) = 15.9 ohm^(-1) cm^(2) mol^(-1)`
Molar conductance at infinite dilution `=350.5 ohm^(-1) cm^(2) mol^(-1)`
Degree of dissociation `alpha= (Lamda_(m))/(Lamda_(m)^(@))= (15.9 ohm^(-1) cm^(2) mol^(-1))/(350.5 ohm^(-1) cm^(2) mol^(-1))= 0.0454`
`therefore` Degree of dissociation = 4.54%