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)

Explanation

Weight of solution before discharge `= 3500 xx 1.294`
`= 4529g`
Weight of `H_(2)SO_(4)` before discharge `= (39)/(100) xx 4529`
`= 1766.31 g`
Weight of solution after discharge `= 3500 xx 1.139`
`= 3986.5 g`
Weight of `H_(2)SO_(4)` after discharge `= (20)/(100) xx 3986.5`
`= 797.3 g`
Loss in mass of `H_(2)SO_(4)` during discharge
`= 1766.31 - 797.3 = 969.01 g`
Now from first law of electrolysis,
`W = (Q xx E)/(96500)`
`969.01 = (Q xx 98)/(96500)`
`Q = 954178.21` coulomb
Ampere-hour `= ("Coulomb")/(3600) = (954178.21)/(3600)`
= 265.04 ampere-hour