During the discharge of a lead storage battery, the density of sulphuric acid fell from `1.294 g mL^(-1)` to `1.139 g mL^(-)`. Sulphuric acid of density `1.294 g mL^(-1)` is `39%` by weight and that of density `1.139 g mL^(-1)` is `20%` by weight. The battery hold `3.5` litre of acied and discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are:
`Pb+SO_(4)^(2-) rarr PbSO_(4)+2e` (charging)
`PbO_(2)+4H^(+)+SO_(4)^(2-)+2e rarr PbSO_(4)+2H_(2)O` (discharging)

Explanation

The overall hattery reaction is:
`Pb+PbO_(2)+2H_(2)SO_(4) to 2PbSO_(4)+2H_(2)O`
Two moles of electrons are involved for reaction of two moles of `H_(2)SO_(4)`
Eq. weight of `H_(2)SO_(4)=" Molecular weight of "H_(2)SO_(4)=98`
No. of eq. of `H_(2)SO_(4)` present in 3.5 litres of solution of charged battery
`=3500 xx 1.294 xx (39)/(100) xx (98)=18.0235`
No. of equilvalent of `H_(2)SO_(4)` present in 3.5 litres of solution after getting discharged
`3500 xx 1.139 xx (20)/(100) xx (1)/(98) =8.1357`
Number of eq. of `H_2SO_4` lost =18.0235 -8.1357
Moles of electricity produced by the battery = 9.8878F
`= 9.8878 xx 96500" coulomb"`
`=9.8878 xx 96500" amp sec"`
`=(9.8878 xx 96500)/(60 xx 60)"amp hour=265 ampere hour"`