Solve the following

Question:

$100 \mathrm{~mL}$ of a water sample contains $0.81 \mathrm{~g}$ of calcium bicarbonate and $0.73 \mathrm{~g}$ of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $\mathrm{CaCO}_{3}$ is:

(molar mass of calcium bicarbonate is $162 \mathrm{~g} \mathrm{~mol}^{-1}$ and magnesium bicarboante is $146 \mathrm{~g} \mathrm{~mol}^{-1}$ )

1. $5,000 \mathrm{ppm}$

2. $1,000 \mathrm{ppm}$

3. $100 \mathrm{ppm}$

4. $10,000 \mathrm{ppm}$

Correct Option:

Solution:

Moles of $\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}=\frac{0.81}{162}=0.005$

Moles of $\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}=\frac{0.73}{146}=0.005$

Hardness in terms of $\mathrm{CaCO}_{3}$ in ppm

$=\frac{(0.005+0.005) \times 100}{100} \times 10^{6}=10^{4} \mathrm{ppm}$