What is the degree of hardness of a sample of water containing 10.8 mg of MgSO₄ (mol. mass 120) per kg of water?

$10.8\mathrm{mg}\text{ of }{\mathrm{MgSO}}_4\text{ is present in }{10}^3\mathrm{g}\text{ of water. }$

One kg of water contains 10.8 mg of MgSO4. One kg of water has 1000 g of water.

1 millions grams of water is 10⁶ gm of water which contains 10.8 g of MgSO4.

$$\begin{gathered} 1 \mathrm{mole} {\mathrm{MgSO}}_4=1 \mathrm{mol} \mathrm{of} {\mathrm{CaCO}}_3 \\ 120 \mathrm{g} {\mathrm{MgSO}}_4=100 \mathrm{g} {\mathrm{CaCO}}_3 \\ 10.8 \mathrm{g} {\mathrm{MgSO}}_4=\frac{10.8\times 100}{120}=9 \end{gathered}$$
Hardness of water = 9 ppm