JEE Main 2019 · Chemistry · s-Block, p-Block Elements and Hydrogen · hard

Hard water contains 81 mg of Ca(HCO3)2 (M =162) and 73 mg of Mg(HCO3)2 (M =146) per litre. Using Clark's method, with Ca(OH)2 (M =74) precipitating…

Single correct option

Hard water contains

81\ \mathrm{mg}

\mathrm{Ca(HCO_3)_2}

=162

) and

73\ \mathrm{mg}

\mathrm{Mg(HCO_3)_2}

=146

) per litre. Using Clark's method, with

\mathrm{Ca(OH)_2}

=74

) precipitating

\mathrm{CaCO_3}

and

\mathrm{Mg(OH)_2}

, the mass of slaked lime needed to soften

1000\ \mathrm{L}

of this water is

Show answer and solution

Answer: (B) $111\ \mathrm{g}$

Per litre:

\mathrm{Ca(HCO_3)_2} = 0.5

mmol and

\mathrm{Mg(HCO_3)_2} = 0.5

mmol.

\mathrm{Ca(HCO_3)_2 + Ca(OH)_2 \to 2CaCO_3 + 2H_2O}

needs

1

lime per formula unit, so

0.5

mmol.

\mathrm{Mg(HCO_3)_2 + 2Ca(OH)_2 \to Mg(OH)_2 + 2CaCO_3 + 2H_2O}

needs

2

lime, so

1.0

mmol. Total lime

=1.5

mmol/L

=1.5\times74 = 111

mg/L, hence

111\ \mathrm{g}

per

1000\ \mathrm{L}

. Counting only one lime per salt gives

74\ \mathrm{g}

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake