Step I. Calculation of mass of `H_(2)SO_(4)`

The chemical equation for the reaction is :

`underset(underset(=123.5g)(63.5+12+48))(CuCO_(3))+underset(underset(=98g)(2+32+64))(H_(2)SO_(4))rarrCuSO_(4)+CO_(2)+H_(2)O`

123.5 of `CuCO_(3)` require `H_(2)SO_(4)=98g`

0.5 g of `CuCO_(3)` require `H_(2)SO_(4)=(98)/(123.5)xx0.5=0.397g`

Step II. Calculation of millilitres of `H_(2)SO_(4)`

Molarity of solution `= ("Mass of "H_(2)SO_(4)//"Molar mass of "H_(2)SO_(4))/("Volume of solution in litres")`

`:.` Volume of solution `= ("Mass of "H_(2)SO_(4)//"Molar mass of "H_(2)SO_(4))/("Molarity of solution")`

`= ((0.397g))/((98"g mol"^(-1))xx("0.5 mol"^(-1)))=0.0081L=0.0081xx1000=8.1" mL"`.