During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.13 to 1.15 g/`ml^(−1)` and sulphuric acid of the density of 1.3 g/`ml^(−1)`is 40% by mass and that of the density of 1.15 g/`ml^(−1)` is 20% by mass. The battery holds 3.5 litre of acid and the volume practically remained constant during the 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)rarrPbSO_(4)+2e^(-)`(charging)
`PbO_(2)+4H^(+)+SO_(4)^(2-)+2erarrPnSO_(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