Question

A water sample contains $22.2mg$ of $CaCl_{2}$ in $1$ $L$ of sample. The degree of hardness of water in terms of $CaCO_{3}$ equivalent is ------ $ppm.$ $\left[\right.$ Molar mass of $CaCO _{3}=100 g mol ^{-1}$, molar mass of $\left.CaCl _{2}=111 gmol ^{-1}\right]$

Answer 1

ppm is parts per million $\left(10^{6}\right)$. This is equivalent to mg per litre.
$1L$ of water contains $22.2mg$ of $CaCl_{2}$ Now, $111g$ of $CaCl_{2}\equiv 100gCaCO_{3}$
$22.2mg$ of $CaCl_{2}=\frac{22 . 2 \times 100}{111}=20$
It is equivalent to $20mg\text{ of }CaCO_{3}\text{ in }1L\text{ of water }$
$\therefore $ Hardness of water sample $=20ppm$