PrepRung
Sign inSign up free

JEE Main 2017 · Chemistry · s-Block, p-Block Elements and Hydrogen · hard

Temporary hardness of a water sample is due to 81 ppm of Ca(HCO3)2 (M=162) and 73 ppm of Mg(HCO3)2 (M=146) (1 ppm =1 mg L^-1). In Clarke's process…

Single correct option

Temporary hardness of a water sample is due to 81 ppm81\ \mathrm{ppm} of Ca(HCO3)2\mathrm{Ca(HCO_3)_2} (M=162M=162) and 73 ppm73\ \mathrm{ppm} of Mg(HCO3)2\mathrm{Mg(HCO_3)_2} (M=146M=146) (1 ppm =1 mg L−1=1\ \mathrm{mg\,L^{-1}}). In Clarke's process, in which CaCO3\mathrm{CaCO_3} and Mg(OH)2\mathrm{Mg(OH)_2} are precipitated, the mass of pure Ca(OH)2\mathrm{Ca(OH)_2} (M=74M=74) needed to soften 2000 L2000\ \mathrm{L} of the water is
  1. A. 148 g148\ \mathrm{g}
  2. B. 222 g222\ \mathrm{g}
  3. C. 333 g333\ \mathrm{g}
  4. D. 111 g111\ \mathrm{g}
Show answer and solution

Answer: (B) 222 g222\ \mathrm{g}

Ca(HCO3)2=81/162=0.50\mathrm{Ca(HCO_3)_2}=81/162=0.50 mmol/L; Mg(HCO3)2=73/146=0.50\mathrm{Mg(HCO_3)_2}=73/146=0.50 mmol/L. Ca(HCO3)2+Ca(OH)2→2CaCO3+2H2O\mathrm{Ca(HCO_3)_2+Ca(OH)_2\to2CaCO_3+2H_2O} needs 11 lime per Ca salt. Mg(HCO3)2+2Ca(OH)2→Mg(OH)2+2CaCO3+2H2O\mathrm{Mg(HCO_3)_2+2Ca(OH)_2\to Mg(OH)_2+2CaCO_3+2H_2O} needs 22 lime per Mg salt. Lime =0.50+2(0.50)=1.5=0.50+2(0.50)=1.5 mmol/L =111=111 mg/L. For 20002000 L: 222222 g. (148148 g comes from taking 1:1 for magnesium; 111111 g is for 10001000 L.)
Report a mistake

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake