10 L of hard water required 5.6 g of lime for removing hardness. Hence temporary hardness in ppm of CaCO₃ is:

$\begin{array}{l}\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2+\mathrm{CaO}⟶{\mathrm{CaCO}}_3+{\mathrm{H}}_2\mathrm{O}\\ 162\mathrm{g}-56\mathrm{g} \mathrm{CaO}\\ 16.2\mathrm{g}-5.6\mathrm{g} \mathrm{CaO}\\ 162\mathrm{gCa}{\left({\mathrm{HCO}}_3\right)}_2=100\mathrm{g}{\mathrm{CaCO}}_3\\ 16.2\mathrm{gCa}{\left({\mathrm{HCO}}_3\right)}_2=10\mathrm{g}{\mathrm{CaCO}}_3\end{array}$
$\begin{array}{l}\mathrm{Hardness} =\frac{ \mathrm{Mass} \mathrm{of} {\mathrm{CaCO}}_3}{\mathrm{Mass} \mathrm{of} \mathrm{water}}\times {10}^6\\ =\frac{10}{{10}^4}\times {10}^6=1000\mathrm{ppm}\end{array}$