Chemistry · Chemical equations and stoichiometry
of hard water required of lime (CaO) for removing hardness. Hence, the temporary
\( 10 \mathrm{L} \) of hard water required \( 0.56 \mathrm{g} \) of lime (CaO) for removing hardness. Hence, the temporary hardness in ppm (part per million) of \( C a C O_{3} \) is :
- A. 100
- B. 200
- C. 10
- D. 20
Step-by-step solution
0.56 g CaO = 0.01 mol (molar mass 56 g/mol). It removes temporary hardness equivalent to 0.01 mol CaCO3 (mass = 1 g) from 10 L water. Thus, hardness = 1 g / 10 L = 0.1 g/L = 100 mg/L = 100 ppm.
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