A lead storage cell is discharged which causes the `H_(2)SO_(4)` electrolyte to change from a concentration of `34.6%` by weight (density `1.261gml^(-1)` at `25^(@)C`) to one of `27%` by weight. The original volume of electrolyte is one litre. Calculate the total charge released at anode of the battery. Note that the water is produced by the cell reaction as `H_(2)SO_(4)` is used up. over all reaction is.
`Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(l)to2PbSO_(4)(s)+2H_(2)O(l)`

Explanation

Before the discharge of lead storage battery.
mass of solution `=1000xx1.261=1261g`
mass of `H_(2)SO_(4)=(1261xx34.6)/(100)=436.3g`
mass of water `=1261-436.3=824.3g`
after the discharge of lead storage battery.
Let the mass of `H_(2)O` produce as a result of net reaction during discharge
`(Pb+PbO_(2)+2H_(2)SO_(4)to2PbSO_(4)+2H_(2)O)` is `xxg`
`therefore` moles of `H_(2)O` produced `=(x)/(18)=` moles of `H_(2)SO_(4)` consumed
Mass of `H_(2)SO_(4)` consumed `=(x)/(18)xx98`
Now, mass of solution after discharge `=1261-(98x)/(18)+x`
`%` by the mass of `H_(2)SO_(4)` after discharge `=("mass of " H_(2)SO_(4)" left")/("mass of solution after discharge")xx100=27`
`=(436.3-(98x)/(18))/(1261-(98x)/(18)+x)xx100=27`
`thereforex=22.59g`