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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»


Solution :From QUESTION above,
Decrease in amount of `H_(2)SO_(4)` as batteryyield 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`


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