During discharge of a lead stronge cell the density of sulphuric acid in the cell:-

Answer

`265 Amp.hr`

Explanation

The overall battery reaction is
`Pb +PbO_(2) +2H_(2)SO_(4) = 2PbSO_(4) +2H_(2)O`
`:.` two moles of electrons are involved for the reaction of two moles of `H_(2)SO_(4)`.
`:.` eq.wt.of `H_(2)SO_(4) =mol.wt.of H_(2)SO_(4) = 98`
no.of eq.of `H_(2)SO_(4)` present in `3.5 L` of solution of a charged battery
`= (39)/(98) xx (1.294)/(100) xx 3500 = 18.0235`
No. of equivalents of `H_(2)SO_(4)` present in `3.5L` of solution after getting discharged
`=(20)/(98) xx (1.139)/(100) xx 3500 = 8.1357`
Number of eq. of `H_(2)SO_(4)` lost `(H_(2)SO_(4) =18.0235 -8.1357 = 9.8878)`
`:.` moles of electric charge produced by the battery `=9.8878F`
`= 9.8878 xx 96500` coulombs
`=9.8878 xx 96500` amp-seconds
`= (9.8878 xx 96500)/(60 xx 60)` amp-hours `= 265` amp-hours.