Question

What is the degree hardness of a sample of water containing 24 mg of $MgS{O_4}$ (mol. mass 120) per kg of water?
A.10 ppm
B.15 ppm
C.20 ppm
D.25 ppm

Answer 1

The two types of hardness of water is temporary and permanent. The hardness of water depends on the bicarbonates of Ca and Mg which is temporary hardness and the chlorides and sulphate of Ca and Mg occur as permanent hardness. Here degree of hardness calculation in ppm of magnesium sulphate. Magnesium sulphate occurs in permanent hardness in water.

Complete step by step answer:
Water contains permanent and temporary hardness which is dependent on the salt of Magnesium and Calcium.
Here the amount of Mg is given 24 mg.
As we know that 1 kilogram $=$ 1000 gram
As given, 24 mg of $MgS{O_4}$ contains 1 kilogram of water.
Here the degree of hardness is found with the help of calcium carbonate.
Water contains salts of Mg and Ca, so the calcium carbonate and magnesium sulphate contain the same number of moles.
If we take 1 mole of $MgS{O_4}$ , then the mole of $CaC{O_3}$ is 1.
Therefore,
1 mole of $MgS{O_4}$ = 1 mole of $CaC{O_3}$
Here $1{\text{ }}mg{\text{ }}water = 1000 \times 1000$
$24{\text{ }}mg{\text{ }}water = 24 \times 1000{\text{ }}g{\text{ }}ofMgS{O_4}$
We take 120 gm mole mass of Mg, so the equation is as below,
120g of $MgS{O_4}$ = 100 g of $CaC{O_3}$
24g of $MgS{O_4}$ $= 100 \times \dfrac{{24}}{{120}}$
$=$20 ppm

Hence, the correct answer is Option (C) which is 20 ppm.

Note:
The degree of hardness of water depends on the sulphate of magnesium and calcium carbonate. The value is found with the help of mole of magnesium sulphate and calcium carbonate. So the degree of hardness of water is found by the help of salts of Ca and Mg.