Calculate the volume of 0.5 M `H_2SO_4` required to dissolve 0.5 g of copper (II) carbonate `(CuCO_3)`.

Explanation

`Mw(CuCO_3)=63.5+12+3xx16=123.5gmol^(-1)`
`eW(Cu^(2+)CO_3^(2-))=("Molecular weight")/("Valency")=(123.5)/(2)`
`H_2SO_4-=CuCO_3`
`mEq=mEq`
`NV(mL)-=("weight")/("Equivalent weight")xx10^(3)`
`0.5xx2` (n-factor)`xxV=(0.5xx10^(3))/((123.5)/(2))`
`V=(10^(3))/(123.5)=8.09mL`