Question:
The hardness of a water sample containing $10^{-3} \mathrm{M} \mathrm{MgSO}_{4}$ expressed as $\mathrm{CaCO}_{3}$ equivalents (in ppm) is___________. (molar mass of $\mathrm{MgSO}_{4}$ is $120.37 \mathrm{~g} / \mathrm{mol}$ )
Solution:
(100.00)
$10^{-3}$ molar $\mathrm{MgSO}_{4} \equiv 10^{-3}$ moles of $\mathrm{MgSO}_{4}$ present in $1 \mathrm{~L}$ solutions.
$10^{-3} \mathrm{M} \mathrm{MgSO}_{4} \equiv 10^{-3} \mathrm{M} \mathrm{CaCO}_{3}$
$10^{-3} \mathrm{M} \mathrm{CaCO}_{3}=10^{-3} \times 100 \mathrm{~g} \mathrm{CaCO}_{3}$ in $1 \mathrm{~L}$ water
$\mathrm{ppm}_{\text {(in term of } \mathrm{CaCO}_{3} \text { ) }}=\frac{10^{-3} \times 100}{1000} \times 10^{6}$
$\mathrm{ppm}_{\text {(in term of } \mathrm{CaCO}_{3} \text { ) }}=100 \mathrm{ppm}$