Question

# A sample of hard water contains 96 ppm of SO42âˆ’4 ions, 183 ppm of HCOâˆ’3 ions and the only cation present is Ca2+. X moles of CaO are required to remove the HCOâˆ’3 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 ppmDensity of hard water is approximately 1 g/ml 1L=1000g=1kg 183 ppm of HCO−3 represent 183 mg/L or 3×10−3 moles per 1 kg, So for 1000 kg sample of hard water number of moles of HCO−3 =3 moles We know that ratio of Ca2+ to HCO−3 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+CaO1→CaCO31+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 SO2−4 ions which counter ion is Ca2+ of molecular formula CaSO4 As given: SO2−4=96 ppm=1×10−3 moles 1×10−3 moles of Ca2+=40 ppm All Ca(HCO3)2 so only 40 ppm of Ca2+ ions left in solution.

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