Question

One litre of a samle of hard water contains 1 mg of $CaC{l}_2$ and 1 mg of $MgC{l}_2$.Then the total hardness in terms of parts of $CaC{O}_3$ per ${10}^8$ parts of water by mass is:

• A. 1.954 ppm
• B. 1.26
• C. 0.946
• D. none of these

Answer 1

The correct option is A 1.954 ppm

$1$ mole of $CaC{l}_2\equiv 1$ mole of $CaC{O}_3\equiv 1$ mole of $MgC{l}_2$

$\therefore$ $100mg$ $CaC{O}_3$ is produced by $95mg$ $MgC{l}_2$

$\therefore$ $1mg$ of $MgC{l}_2$ gives $\frac{100}{95}mg$ $CaC{O}_3=$ $1.05mg$ $CaC{O}_3$

Similarly, $1mg$ of $CaC{l}_2$ gives $\frac{100}{111}mg$ $CaC{O}_3=$ $0.90mg$ $CaC{O}_3$

$\therefore$ Total $CaC{O}_3$ per litre of water= $1.05+0.90=1.95$ $mg$

Weight of $1000ml$ of water= ${10}^{3}g$=${10}^{6}mg$

$\therefore$ Total hardness in terms of parts of $CaC{O}_3$ per ${10}^6$ parts of water by weight = $\frac{1.95}{{10}^6}\times {10}^6$

=$1.95$ $ppm$

Answer: (A) $1.954$ $ppm$