Question

In Question above, the number of ampere hours for which the battery is used containing `1L` of the acid is `16.08x` ampere hour. Calculate the value of `x`.

Answer 1

Correct Answer - 5
From Question above,
Decrease in amount of `H_(2)SO_(4)` as battery yield current `=` Change in molarity `xxMw_(2)xx`Volume of acid in `L`
`=(3xx98xx1L)g`
Overall change `:`

`2F-=2 mol of H_(20SO_(4)` or `1F-=1 mol of H_(2)SO_(4)=98g`
`98g H_(2)SO_(4)` requires `=1F`
Ampere` //` hour `=(3Fxx96500A-s)/(3600 s h^(-1))`
`=80.4 A-h`
`:. 16.08x=80.4`
`x=(80.4)/(16.08)=5`