Question

The hardness of water sample (in terms of equivalents of $CaC{{O}_{3}}$) containing ${{10}^{-3}}$ M $CaS{{O}_{4}}$ is:
[Molar mass of $CaS{{O}_{4}}$ = 136 $gmo{{l}^{-1}}$$gmo{{l}^{-1}}$].

A. 100ppm
B. 50ppm
C. 10ppm
D. 90ppm

Answer 1

Hint: Hard water is a mixture of magnesium and calcium together with chloride, bicarbonate, sulphate, etc. Molecular weight of calcium carbonate is 100gm/mol and here the mass of water is 1000g. To find the degree of hardness in ppm we have a formula,
DOH (in terms of $CaC{{O}_{3}}$ )= $\dfrac{mass\,of\,CaC{{O}_{3}}}{mass\,of\,{{H}_{2}}O}\times {{10}^{6}}$

Complete step by step solution:
From your chemistry lessons you have learned about the hardness of water and what is hard water and its constituents. Hard water is the mixture of magnesium and calcium together with sulphate, bicarbonate, chloride, etc. Carbonate which is made up of 1 carbon and 3 atoms of oxygen and they together combine to form carbonate molecules $(C{{O}_{3}})$.
-Here carbonate is an anion which has negative charge on it so for the stability it gets associated with a positively charged ion or we can say cation. The ion can be metals or hydrogen like calcium, magnesium, potassium, etc. And in this way calcium carbonate $(CaC{{O}_{3}})$and magnesium carbonate are formed.
-The hardness of water is usually expressed in ppm and measured in terms of calcium carbonate because the molecular weight of $CaC{{O}_{3}}$ is 100g/mol and it will be easier to calculate the values in respect to 100 as comparison with the molecular weight of other agents that cause hardness . And to calculate the hardness it requires one standard value which is used as a constant to equate with and that makes it easy to calculate the hardness of water in terms of ppm.
-In the question the amount of $CaS{{O}_{4}}$ is given as ${{10}^{-3}}$ M
-So, ${{10}^{-3}}$ molar $CaS{{O}_{4}}$ = ${{10}^{-3}}$ moles of $CaS{{O}_{4}}$ present in 1 litre.
-Here we have to find the hardness of water in terms of $CaC{{O}_{3}}$,
-Therefore moles of $CaC{{O}_{3}}$= moles of $CaS{{O}_{4}}$
Molecular weight of calcium carbonate = 100 g/mol
Mass of water = 100g
-So, the degree of hardness in term of $CaC{{O}_{3}}$ in ppm is,
-DOH(in terms of $CaC{{O}_{3}}$)=$\dfrac{mass\,of\,CaC{{O}_{3}}}{mass\,of\,{{H}_{2}}O}\times {{10}^{6}}$
-DOH (in terms of $CaC{{O}_{3}}$)=$\dfrac{moles\,of\,CaC{{O}_{3}}\times molecular\,weight\,of\,CaC{{O}_{3}}\,}{mass\,of\,{{H}_{2}}O}$
-Therefore, DOH (in terms of $CaC{{O}_{3}}$) =$\dfrac{{{10}^{-3}}\times 100}{1000}\times {{10}^{6}}=100ppm$

Thus the correct option will be (A).

Note: The hardness of water is not only due to calcium but magnesium and other multivalent cations also present to some extent in the hard water. Calcium carbonates are essentially present in the form of limestone and chalk. When we tell that hardness is in terms of calcium carbonate then it is calculated as if magnesium or other cations were present there as calcium.