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Question

In a 103 L sample of hard water, CaSO4 & MgSO4 are present. If elevation in boiling point is 0.00208 C, then calculate the hardness of water in terms of ppm of CaCO3. (Kb for H2O=0.52 Kkgmol1)

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Solution

We know,
Elevation in boiling point, ΔTb=iKb(m1+m2)
0.00208=2×0.52[n1103+n2103]
where n1 and n2 are the moles of CaSO4 and MgSO4 .
n1+n2=2.082×0.52
n1+n2=2
n factor of CaSO4 and MgSO4 are 2
Equivalent of CaSO4+MgSO4=2×2=4
Hardness of water is always expressed in ppm of CaCO3
So, equivalent of CaCO3=4
Mass. of CaCO3=(4×50) g=200 g
Again, mass of water = 103 L = 103 kg = 106 g
Hence, ppm =200106×106=200

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