During the discharge of a lead storage battery the density of `H_(2)SO_(4)` talls from `rho_(1)g//cc` to `rho_(2)g//C`, `H_(2)SO_(4)` of density of `rho_(1)g//C`. C is `X%` by weight and tat of density fo `rho_(2)gc.c` is `Y%` by weight. The battery holds V litre of acid before discharging. Calculate the total charge released at anode of the battery. The reactions occurring during discharging are ltbr. At anode: `Pb+SO_(4)^(2-)toPbSO_(4)+2e^(-)`
At cathode: `PbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)toPbSO_(4)+2H_(2)O`

Explanation

Mass of acid solution before discharge of lead storage battery `(LSB)=(Vxx10^(3)xxrho_(1))g`
`=(1000xVrho_(1))g`
Mass of `H_(2)SO_(4)` before discharge of `LSB=(1000xxVrho_(1)xx(X)/(100))g=(10xxVrho_(1)X)g`
Net reaction during discharging: `Pb+PbO_(2)+2H_(2)SO_(4)toPbSO_(4)+2H_(2)O`
From the reaction, it is evident that the moles of electron exhanged (lost at anode and gai at cathode) is equal to the moles of `H_(2)SO_(4)` consumed or moles of `H_(2)O` produced. Let the moles of `H_(2)SO_(4)` produced be, x then.
Mass of `H_(2)O` produced during discharge of `LSB=(18x)g`
Mass of `H_(2)SO_(4)` consumed during discharge of `LSB=(98x)g`
`therefore` Mass of `H_(2)SO_(4)` after discharge of `LSB=[(10Vrho_(1)X)]-98x]g`
Mass of acid solution after discharge of `LSB=[(1000Vrho_(1))-98x+18x]=[(1000Vrho_(1))-80x]g`
`therefore%` of `H_(2)SO_(4)` after discharge of `LSB=("mass of" H_(2)SO_(4) "after discharge")/("mass of acid solution after discharge")xx100`
`Y=([(1000xxVrho_(1))-98x])/([(1000xxVrho_(1))-80x])xx100`
`x` can be calculated as all other quantities are known.
`therefore` Total charge released at cathode, `Q=nF=xF`