# 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 densi

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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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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.