(a) The conductivity of 0.001 mol `L^(-1)` solution of `CH_3COOH` is `3.905 xx 10^(-5) S cm^(-1)`. Calculate its molar conductivity and degree of dissociation (`alpha`).
Given :`lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1)` and `lambda^(@)(CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)`
(b) What type of battery is lead storage battery ? Write the overall reaction occurring in lead storage battery.

Explanation

(a) Molar conductivity is given by the relation :
`Lambda_(m) = k/C`
Substituting the values in the above equation, we have
`Lambda_(m) =(3.905 xx 10^(-5)S cm^(-1))/(0.001 mol L^(-1)) =(3.905 xx 10^(-5) S cm^(-1))/(0.001 mol (cm^(3))^(-1))`
or `Lambda_(m) = lambda_(H^(+))^(@) + lambda_(CH_(3)COO^(-))^(@) = 349.6 + 40.9 = 390.5 S cm^(2) mol^(-1)`
`alpha =lambda_(m)/lambda_(0) = (39.05)/(390.5) = 0.1`
Substituting the values in the above equation, we have:
`Pb(s) + PbO_(2)(s) + 2SO_(4)^(2-) (aq) + 4H^(+) (aq) to 2PbSO_(4)(s) + 2H_(2)O` (l)