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A sample of hard water contains 96 ppm of SO424 ions, 183 ppm of HCO3 ions and the only cation present is Ca2+. X moles of CaO are required to remove the HCO3 ions from 1000 kg of this sample of water. When the same amount of CaO is added to the water sample, the concentration of residual Ca2+ ions (in ppm) becomes Y. (Assume CaCO3to be completely insoluble in water)
Calculate the value of X and Y

A
3.0 mol, 80 ppm
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B
1.5 mol, 80 ppm
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C
1.5 mol, 40 ppm
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D
3.0 mol, 40 ppm
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Solution

The correct option is C 1.5 mol, 40 ppm
Density of hard water is approximately 1 g/ml
1L=1000g=1kg
183 ppm of HCO3 represent 183 mg/L or 3×103 moles per 1 kg, So for 1000 kg sample of hard water number of moles of HCO3
=3 moles
We know that ratio of Ca2+ to HCO3 is 1:2 in Ca(HCO3)2 , number of moles of Ca2+=1.5 moles.

Intially number of moles of Ca(HCO3)2=1.5
Complete balanced reaction to remove Ca(HCO3)2
Ca(HCO3)21+CaO1CaCO31+H2O+CO2

One mole of Ca(HCO3)2 replaced by one mole of CaO so 1.5 mole of CaOrequired to remove 1.5 mole of Ca(HCO3)2.
And solution contains SO24 ions which counter ion is Ca2+ of molecular formula CaSO4
As given:
SO24=96 ppm=1×103 moles
1×103 moles of Ca2+=40 ppm
All Ca(HCO3)2 so only 40 ppm of Ca2+ ions left in solution.






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