Mole ConceptHard
Question
10 L of hard water required 0.56 g of lime (CaO) for removing hardness. Hence, temporary hardness in ppm (part per million 106) of CaCO3 is:
Options
A.100
B.200
C.10
D.20
Solution
Temporary hardness is due to HCO3- of Ca2+ and Mg2+
Ca(HCO3)2 +
+ H2O
mole of CaCO3 = 2
Ca(HCO3)2 +
+ H2Omole of CaCO3 = 2
ppm of CaCO3 = (2/104)×106 =200
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