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)

Answer

`265A-h`

Explanation

Discharging reaction `:`

`M_(1)=(10xxdxx%by mass)/(Mw_(2))=(10xx1.294xx39)/(98)=5.149`
`M_(2)=(10xx1.139xx20)/(98)=2.324`
Change in molarity `(M_(2)-M_(1)),`
`5.149-2.324=2.825M`
Decrease in amount of `H_(2)SO_(4)` as battery yields current
`=` Change in molarity `xxMw` of `H_(20SO_(4) xx` Volume of acid
`=2.825xx98xx3.5=969g`
Overall change `:`

`2F-=2 mol of H_(2)SO_(4) or 1F-=1 Mol of H_(2)SO_(4)`
`969 g of H_(2)SO_(4)` is condemned by `=(1)/(98)xx969=9.888F`
Ampere hour `=(9.888xx96500"ampere second")/(3600 "second"// "hour")`
`=265 ` ampere hour