Question

What is the degree of hardness ( $\text{in ppm}$) of a sample of water containing $24\text{mg of}MgS{O}_4$ (molecular mass $=120$) per $\text{kg}$ of water.

Answer 1

$\text{24 mg of}MgS{O}_4\text{present in}{10}^3\text{g}$ of water.
${10}^6\text{g water will contain}=24000\text{mg of}MgS{O}_4=24\text{g of}MgS{O}_4$
$1\text{mole of}MgS{O}_4\equiv 1\text{mole of}CaC{O}_3$
$120\text{g of}MgS{O}_4\equiv 100\text{g of}CaC{O}_3$
So,
$24\text{g of}MgS{O}_4=\frac{100}{120}\times 24=20\text{g of}CaC{O}_3$
$\text{Hardness of water =20 ppm}$