During the discharge of lead storage battery, density of `H_(2)SO_(4)` fell from `"1.294 to 1.139 g.cm"^(-3)`. Sulphuric acid of density `"1.294g.cm"^(-3)` is `39%` by mass and that of density `"1.139 g.cm"^(-3)` is `20%` by mass. The battery holds 3.5L of acid and the volume remains practically constant during discharge. Calculate the no. of ampere-hours for which battery must have been used.

Explanation

`39%H_(2)SO_(4)` by mass: As the density of the solution is `"1.294 g.cm"^(-3)`, 100 g of it has a volume of `(100)/(1.294)=77.28"cc"`.
This much of volume contains `39g=(39)/(98)=0.398` mol of `H_(2)SO_(4)`. Hence, 3.5L of this solution contains `(0.398xx3.5)/(77.28xx10^(-3))=18.025" mol of "H_(2)SO_(4)`.
`20% H_(2)SO_(4)` by mass : For 100 g of this solution, volume is `(100)/(1.139)=87.8" cc"`. This much of volume conains `20g=(20)/(98)="0.204 mol of "H_(2)SO_(4)`.
Hence, 3.5L of this solution contains
`=(0.204xx3.5)(87.8xx10^(-3))="8.132 mol of "H_(2)SO_(4)`.
During discharging the change in number of mol of `H_(2)SO_(4)=18.025-8.132=9.893` mol.
Discharging of lead - storage battery involves the following reactions :
`{:("Anode: "Pb(s)+SO_(4)^(2-)(aq)rarrPbSO_(4)(s)+2e),(ul("Cathode: "PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(-)(aq)+2erarrPbSO_(4)(s))),("Cell "Pb(s)+PbO(s)+2H_(2)SO_(4)(aq)rarr2PbSO_(4)(s)):}`
Thus, consumption of 2 mol of `H_(2)SO_(4)-=2F` of electricity and that of `"9.893 mol of "H_(2)SO_(4)-=9.893F` of electricity
`9.893F=9.893xx96500C=9.893xx96500A.s`
`=(9.893xx96500)/(3600)A.h=265.18A.h`
So, the no. of ampere-hours that has been used is 265.18.