Calculate the volume of 0.5(M) `H_(2)SO_(4)` solution requried to dissolve 0.5g of copper (II) carbonate.

Explanation

Reaction: `underset("1 mol")(CuCO_(3))+underset("1 mol")(H_(2)SO_(4)) to CuSO_(4)+CO_(2)+H_(2)O`
`therefore`1 mol `CuCO_(3)=63.5+12+48=123.5g`
`therefore` 1 mole of `H_(2)SO_(4)` is required to dissolve 123.5g of `CuCO_(3)`
`therefore` Amount of `H_(2)SO_(4)` required to dissolve 0.5g of `CuCO_(3)`
`=(1xx0.5)/(123.5)=0.004` mol
Now, 0.5 mol of `H_(2)SO_(4)` is present in 1000 mL solution.
`therefore` 0.004 mol `H_(2)SO_(4)` is present in `(1000xx0.004)/(0.5)=8mL` of
given `H_(2)SO_(4)` solution. so, volume of `H_(2)SO_(4)` required=8mL.