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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) |
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Answer» Correct Answer - `265 Amp.hr` 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. |
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