Each litre of a water sample contains 48.6 mg Ca(HCO3)2, 33.3 mg CaCl2 and 31.2 mg MgSO4. Washing soda treatment removes only the permanent hardness, each mole of CaCl2 or MgSO4 using one mole of Na2CO3. Using CaCl2=111, MgSO4=120, Na2CO3=106, find the mass of Na2CO3 (in g, nearest integer) needed for 5000 L of the water.

JEE Advanced levelNumerical answerDifficulty 3 of 3eq216

Each litre of a water sample contains $48.6 mg$ $Ca (HCO X_{3}) X_{2}$ , $33.3 mg$ $CaCl X_{2}$ and $31.2 mg$ $MgSO X_{4}$ . Washing soda treatment removes only the permanent hardness, each mole of $CaCl X_{2}$ or $MgSO X_{4}$ using one mole of $Na X_{2} CO X_{3}$ . Using $CaCl X_{2} = 111$ , $MgSO X_{4} = 120$ , $Na X_{2} CO X_{3} = 106$ , find the mass of $Na X_{2} CO X_{3}$ (in $g$ , nearest integer) needed for $5000 L$ of the water.

Numerical answer type, answer in g, tolerance 1.

The bicarbonate does not use soda. Per litre:

CaCl X_{2}

= 33.3/111 = 0.300

mmol and

MgSO X_{4}

= 31.2/120 = 0.260

mmol, total

0.560

mmol. For

5000 L

2800

mmol

= 2.80

mol. Mass

= 2.80 \times 106 = 296.8 g