One litre of water contains 40.5 mg of Ca(HCO3)2 (molar mass 162) and 73.0 mg of Mg(HCO3)2 (molar mass 146). Clark's process uses Ca(OH)2 (molar mass 74): Ca(HCO3)2 + Ca(OH)2 -> 2CaCO3 + 2H2O and Mg(HCO3)2 + 2Ca(OH)2 -> Mg(OH)2 + 2CaCO3 + 2H2O. The mass of pure Ca(OH)2 needed to soften 1000 L of this water is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq211

One litre of water contains $40.5 mg$ of $Ca (HCO X_{3}) X_{2}$ (molar mass $162$ ) and $73.0 mg$ of $Mg (HCO X_{3}) X_{2}$ (molar mass $146$ ). Clark's process uses $Ca (OH) X_{2}$ (molar mass $74$ ): $Ca (HCO X_{3}) X_{2} + Ca (OH) X_{2} 2 CaCO X_{3} + 2 H X_{2} O$ and $Mg (HCO X_{3}) X_{2} + 2 Ca (OH) X_{2} Mg (OH) X_{2} + 2 CaCO X_{3} + 2 H X_{2} O$ . The mass of pure $Ca (OH) X_{2}$ needed to soften $1000 L$ of this water is:

Per litre:

Ca (HCO X_{3}) X_{2}

= 40.5/162 = 0.250

mmol (needs 0.250 mmol lime) and

Mg (HCO X_{3}) X_{2}

= 73.0/146 = 0.500

mmol (needs 2 x 0.500 = 1.000 mmol lime). Total lime

= 1.250

mmol per litre. For

1000 L

1.250 mol \times 74 = 92.5 g