Question

A sample of hard water contains 1 mg of $CaC{l}_2$ and 1 mg $MgC{l}_2$ per litre. Calculate the hardness of water in terms of $CaC{O}_3$ present in per ${10}^6$ parts of water.

Answer 1

Molecular weight of $CaC{l}_2$ = 111 g
Molecular weight of $CaC{O}_3$ = 100 g
Molecular weight of $MgC{l}_2$ = 95 g
Now, 111 g $CaC{l}_2$ = 100 g $CaC{O}_3$
1 mg $CaC{l}_2$ = $\frac{100}{111}$ mg $CaC{O}_3$
=0.9 mg $CaC{O}_3$
Similarly,
1 mg $MgC{l}_2$ = $\frac{100}{95}$ mg $CaC{O}_3$
= 1.05 mg $CaC{O}_3$
Hardness of $CaC{O}_3$ in ppm,
$\frac{(0.9+1.05)\times {10}^{-3}g\times {10}^{6}mL}{{10}^{3}ml}$ = 1.95 ppm