A 10L water sample has $40.5mg$ of $Ca{\left(HC{O}_3\right)}_2$, $73 mg$ of $Mg{\left(HC{O}_3\right)}_2$,$30 mg$ of $MgS{O}_4$. Calculate the degree of hardness of water in terms of $CaC{O}_3$.

$\begin{array}{l}D.O.H\\ \left[Ca{\left(HC{O}_3\right)}_2\right]\end{array}$$=\frac{40.5\times {10}^{-6}\times {10}^6\times {10}^2}{162\times 10}=0.25\times {10}^1=2.5$

$D.O.H\left[Mg{\left(HC{O}_3\right)}_2\right]=\frac{73\times {10}^{-6}\times {10}^6\times {10}^2}{146\times 10}=0.5\times {10}^1=5$

$D.O.H\left[MgS{O}_4\right]=\frac{30\times {10}^{6120}}{146\times 10}=2.5$

Total hardness = 2.5+5+2.5 =10ppm