Question

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

• A. $10\text{ppm}$
• B. $15\text{ppm}$
• C. $20\text{ppm}$
• D. $25\text{ppm}$

Answer 1

The correct option is C $20\text{ppm}$

Degree of hardness is defined as the number of parts of calcium carbonate or equivalent to various calcium and magnesium salts present in a million parts of water by mass. It is expressed in ppm.
$24\text{mg of}MgS{O}_4\text{present in}{10}^3\text{g of water}{10}^6\text{g water will contain}=24000\text{mg of}MgS{O}_4=24\text{g of}MgS{O}_4$
$120\text{g of}MgS{O}_4=100\text{g of}CaC{O}_3$
So, $24\text{g of}MgS{O}_4=\frac{100}{120}\times 24=20\text{g of}CaC{O}_3$
$\text{Hardness of water}=20\text{ppm}$