Question

One litre of a sample of hard water contain $4.44mg$ $Ca{Cl}_2$ and $1.9mg$ of $Mg{Cl}_2$, what is the total hardness in terms of ppm of $Ca{CO}_3$ :

• A. $2$ ppm
• B. $3$ ppm
• C. $4$ ppm
• D. $6$ ppm

Answer 1

The correct option is D $6$ ppm

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

$\therefore 100g$ $CaC{O}_3$ produces $111g$ of $CaC{l}_2$

$\therefore 4.44mg$ $CaC{l}_2$ produces $\frac{4.44\times 100}{111}mg$ $CaC{O}_3$

Similarly, $100g$ $CaC{O}_3$ is required for $95g$ $MgC{l}_2$

$\therefore 1.9mg$ $MgC{l}_2=\frac{1.9\times 100}{95}mg$ $CaC{O}_3$

$=2mg$ $CaC{O}_3$

Total hardness= Hardness due to $CaC{l}_2$+Hardness due to $MgC{l}_2$

=$4+2$

Total hardness=$6$ $ppm$