Question

How many millilitres of $0.5M{H}_{2}S{O}_4$ are needed to dissolve $0.5g$ of copper (II) carbonate?

Answer 1

The balanced chemical reaction is
$CuC{O}_3+H_{2}S{O}_4⟶CuS{O}_4+H_{2}O+C{O}_2$
millimol of $CuC{O}_3=\frac{0.5\times 1000}{123.5}=4.048$
$\because$ Millimol of $H_{2}S{O}_4$ required $=4.048$
Millimol $=$ Molarity $\times$ Volume (in mL)
$\Rightarrow$ Volume $=\frac{4.048}{0.50}=8.096mL$