Water contains 48.6 mg/L of Ca(HCO3)2 (molar mass 162) and 72 mg/L of MgSO4 (molar mass 120). It is softened by pure lime, Ca(OH)2 (molar mass 74), which reacts 1 : 1 with each salt: Ca(HCO3)2 + Ca(OH)2 -> 2CaCO3 + 2H2O and MgSO4 + Ca(OH)2 -> Mg(OH)2 + CaSO4. Find the mass of lime in grams needed for 5000 litres (nearest integer).

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JEE Advanced levelNumerical answerDifficulty 3 of 3eq105

Water contains $48.6 mg/L$ of $Ca (HCO X_{3}) X_{2}$ (molar mass 162) and $72 mg/L$ of $MgSO X_{4}$ (molar mass 120). It is softened by pure lime, $Ca (OH) X_{2}$ (molar mass 74), which reacts 1 : 1 with each salt: $Ca (HCO X_{3}) X_{2} + Ca (OH) X_{2} 2 CaCO X_{3} + 2 H X_{2} O$ and $MgSO X_{4} + Ca (OH) X_{2} Mg (OH) X_{2} + CaSO X_{4}$ . Find the mass of lime in grams needed for $5000$ litres (nearest integer).

Numerical answer type, answer in g, tolerance 1.

Answer: 333 g

Ca(HCO3)2 = 48.6/162 = 0.30 mmol/L; MgSO4 = 72/120 = 0.60 mmol/L. Lime = 0.30 + 0.60 = 0.90 mmol/L = 66.60 mg/L. For 5000 L: 66.60 x 5000 = 333.0 g, i.e. 333 g.

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