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15,000 litres of hard water was passed through a zeolite softener. The exhausted zeolite required 120 litres of NaCl having 30 g/litre of NaCl. Calculate the hardness of water.
written 3.6 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 3.6 years ago
engineering chemistry
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written 3.6 years ago by gravatar for teamques10 teamques10 ★ 68k

1L NaCl contain = 30 g

120 L NaCl contain = 120 × 30 g

Thus, the amount of substance (W) = 3600 g

 

Equivalent mass of Na = 23

Equivalent mass of Cl = 35.5

Hence, equivalent mass of NaCl (E) = 58.5

 

Equivalent of CaCO3 = (W × 50)/E

= (3600× 50)/58.5

= 3076.9 g of CaCO3

which means

15000 L of water = 3076.9 g of CaCO3

                           = 3076.9 × 1000 g of CaCO3

 

Hence, 1 L water = 3076.9 × 1000 / 15000

= 205.12 ppm CaCO3

 

Hardness of Water Sample = 205 ppm
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