    Question

# A sample of hard water contains 96 ppm of SO42−4 ions, 183 ppm of HCO−3 ions and the only cation present is Ca2+. X moles of CaO are required to remove the HCO−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.  Suggest Corrections  0      Explore more