How many millilitres of 0.5 M `H_(2)SO_(4)` are needed to dissolve 0.5 g of copper (II) Carbonate ?

Explanation

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"`.