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If a sample of hard water contain 68 ppm of $CaS{O_4}$, then the hardness of the sample of water is:(A) 68(B) 100(C) 200(D) 50

Last updated date: 20th Jun 2024
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Hint: ppm stands for parts per million. We can say that if the sample of water contains the calcium or magnesium salt whose weight (in ppm) is equal to its molecular weight (in g/mol), then the hardness of water is 100%.

Complete step by step solution:
We are given the concentration of calcium sulphate in ppm in the solution. Here, we need to find the degree of hardness of the solution.
- If calcium ions are present in the water, then the water is called hard water. The water which does not contain calcium or magnesium salts is called soft water.
- Let’s find the molecular weight of calcium sulphate first.
Molecular weight of $CaS{O_4}$ = Atomic weight of Ca + Atomic weight of S + 4 (Atomic weight of O)
Molecular weight of $CaS{O_4}$ = 40 + 32 + 4(16) = 136$gmmo{l^{ - 1}}$
- Now, we can say that if the sample contains 136 ppm of calcium sulphate, then the degree of hardness of the water sample would be 100%. So, here it is given that there is 68 ppm of calcium present in the solution. So, the hardness of the water will be $\dfrac{{68}}{{136}} \times 100\% = 50\%$ .
Thus, we obtained that the hardness of the sample of water that contains 68 ppm of calcium sulphate is 50%.

So, the correct answer to this question is (D).

Note: There are two types of hardness of water; permanent and temporary. Permanent hardness is present if water contains calcium and magnesium metal’s chloride or carbonates. If calcium or magnesium hydrogen carbonates are present, then the harness is called temporary hardness.