Mumbai University > First Year Engineering > Sem1 > Applied Chemistry 1

Marks: 4M

Year: May 2015

50g i.e.50 X $10^3$ mg of NaCl → 1 litre

∴ 200 litre of solution contains → 200 X 50 X $10^3$ = $10^7$ mg

$CaCO_3$ equivalent = $10^7 X \frac{10^2}{2 X 58.5} = \frac{10^9}{117} = 8.55 X 10^{-3} X 10^9 \\ = 8.55 X 10^6 \ \ mg \ of \ \ CaCO3 \ \ equivalent \\ $

Now, 1 litre of water → 50 ppm

∴ x → 8.55 X $10^6$

∴ $x = \frac{8.55 X 10^6}{50}$ = 0.171 X 106 litres of water

∴ 171 X $10^3$ litres of water can be softened by zeolite softener.