Question

975 $mL$ of a water sample contains 1 $mgCaC{l}_2$ and 1 $mgMgC{l}_2$. Total hardness of water (in ppm) is

Answer 1

Hardness is measured in terms of ppm of calcium carbonate.
1 mol of $MgC{l}_2\equiv$ 1 mol of $CaC{O}_3$
$95g$ of $MgC{l}_2\equiv 100g$ of $CaC{O}_3$
$1\times {10}^{-3}g$ of $MgC{l}_2\equiv \frac{100}{95}\times {10}^{-3}g$ of $CaC{O}_3$


1 mol of $CaC{l}_2\equiv$ 1 mol of $CaC{O}_3$
$111g$ of $CaC{l}_2\equiv 100g$ of $CaC{O}_3$
$1\times {10}^{-3}g$ of $CaC{l}_2\equiv \frac{100}{111}\times {10}^{-3}g$ of $CaC{O}_3$
${10}^{3}g$ of $H_{2}O$ contains $\frac{100}{111}\times {10}^{-3}g$ of $CaC{O}_3$
So,
Taking density of water to be 1 $gm/mL$
975 $gm$ of water has $(\frac{100}{95}+\frac{100}{111})\times {10}^{-3}gm$ of $CaC{O}_3$
So, ${10}^{6}gm$ of water has $\left(\frac{(\frac{100}{95}+\frac{100}{111})\times {10}^{-3}\times {10}^6}{975}\right)$ gm = 2 ppm