Question

A particular water sample has 136 ppm $CaS{O_4}$ what fraction of the water must be evaporated from the sample before solid $CaS{O_4}$ begins to deposit. (${K_{sp}}$ of $CaS{O_{_4}} = 9 \times {10^{ - 6}}$)
A.20%
B.46%
C.67%
D.92%

Answer 1

We have to remember that the 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 answer:
We are given the concentration of calcium sulphate in ppm in the solution. Here, we need to find the fraction of the water.
If calcium ions are present within the water, then the water is named as hard water. The water which doesn’t contain calcium or magnesium salts is called soft water.
Let’s find the relative molecular weight of calcium sulphate is,
Relative molecular weight of $CaS{O_4}$ = Atomic weight of Ca + atomic weight of S +4(atomic weight O)
Molecular weight of calcium sulphate= 40+32+4(16) = 136
Now, we can say that if the sample contains 136 ppm of calcium sulphate, then it must be 100% hard water.
136 ppm of $CaS{O_4}$ means 136 mole of $CaS{O_4}$ ,
${K_{SP}}$= ${S^2} = 9 \times {10^{ - 6}}$
$S = 3 \times {10^{ - 3}}$
That is $3 \times 10^-3$ moles can be present. Hence one mole can be present in $\dfrac{{1000}}{3}$ litre. So, 2/3 of the volume of solution should evaporate.
Therefore, the solution to this present question is option C that is 67%.

Note:
We have to know that there are two types of hardness of water; permanent and temporary. Permanent hardness is present if water contains calcium and magnesium metals chloride or carbonates. If calcium or magnesium hydrogen carbonates are present, then the hardness is called temporary hardness.