Question

The hardness of water sample (in terms of equivalents of \[CaC{O_3}\]) containing \[CaS{O_4}{10^{ - 3}}M\] is (Molar mass of \[CaS{O_4}\]= \[136g/mol\])

Answer 1

Hint: The hardness is defined as the water which won't give foam with soap. The hardness of water is mainly due to the dissolved sodium bicarbonates, chlorides, and sulphates salt of calcium and magnesium.

Complete step by step answer:
The molarity of \[CaS{O_4}\] present in hard water is given as \[{10^{ - 3}}mole\]. It is known that the molarity of a solution is nothing but moles of solute per litre of the solution.
Thus, the number of moles present in one litre of calcium sulphate solution= \[{10^{ - 3}}mole\]
Mole is defined as the ratio of mass by molecular weight. Thus, the weight of calcium sulphate can be calculated as,
\[mole = \dfrac{{weight}}{{mol.weight}}\]
\[ {10^{ - 3}}mole = \dfrac{{weight}}{{136g/mol}}\]
\[ {10^{ - 3}} \times 136g = weight\]
\[ \Rightarrow 0.136g = Weight\]
Thus, mass of calcium sulphate present in water is \[0.136g\]
Parts per million (ppm) is one of the terms to describe the concentration of solution. ppm usually denotes very low concentration. Since the moles present in calcium sulphate is low and thus, hardness of water can be determined by using ppm formula.
One ppm is equal to a milligram of solute per litre of water.
Therefore, the mass of calcium sulphate present in water (in milligrams) = \[136mg\]
In the question, it is given that the hardness of water can be expressed in terms of calcium carbonate. Thus,
\[nCaC{O_3} = nCaS{O_4}\]
The molar mass of calcium carbonate is \[100g/mol\]
\[ 100g/mol = 136g/mol(CaS{O_4})\]
\[ \Rightarrow 100mg/L = 136mg(CaS{O_4})\]
Since the mass of calcium sulphate present in water (in milligrams) is \[136mg\]
\[ \Rightarrow 136mg(CaS{O_4}) = 100ppm\]
Thus, \[136mg(CaS{O_4})\] is equal to \[100ppm\] of calcium carbonate since ppm is milligram of solute per litre of water.

Thus, the hardness of the sample in terms of calcium carbonate= \[100ppm\]

Note: Alternate method:
Parts per million can also be calculated by using the formula,
\[ppm = \dfrac{{mass(solute)(g)}}{{mass(solution)(g)}} \times {10^6}\]
In this question, it is given that \[nCaC{O_3} = nCaS{O_4}\]
Thus, calcium carbonate has the same mole of \[{10^{ - 3}}\] as calcium sulphate contains.
The weight (g) of calcium carbonate can be calculated as by multiplying mole with molar mass of the given compound.
\[mass \Rightarrow {10^{ - 3}}mol \times 100g/mol\]
\[mass \Rightarrow 0.1g\]
Thus, ppm of calcium carbonate in hard water can be calculated as,
\[ppm = \dfrac{{0.1g}}{{1000g}} \times {10^6}\]
\[ \Rightarrow 100ppm\]
Thus, the hardness of sample in terms of calcium carbonate= \[100ppm\]