• # question_answer 100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $CaC{{O}_{3}}$ is: (molar mass of calcium bicarbonate is $162\,g\,\,mo{{l}^{-\,1}}$  and magnesium bicarbonate is $146\text{ }g\text{ }mo{{l}^{-1}}$) A) 100 ppm                        B) 1,000 ppm C) 5,000 ppm                     D) 10,000 ppm

 Ppm of $CaC{{O}_{3}}=?$ Degree of hardness $=\frac{weight\,\,of\,hardness\,\,cau\sin g\,\,salt}{Mw}\times 100$ 1 ppm=1 part $CaC{{O}_{3}}$ eq in ${{10}^{6}}$parts water $ppm\,\,of\,\,CaC{{O}_{3}}=\,\frac{\left( \frac{0.73}{146}+\frac{0.81}{162} \right)\times 100}{100}\times {{10}^{6}}$$={{10}^{4}}ppm$ $=10,000\,\,ppm$