(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.

(a) The conductivity of 0.001 mol L^(-1)solution of CH_3COOH is 3.905

1.	(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.
Answer» Solution : (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)