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Q.

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:

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a

100

b

200

c

10

d

20

answer is B.

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Detailed Solution

Temporary hardness is due to HCO3⊝ of Ca2+ and Mg2+CaHCO32+CaO56g→ 2CaCO3(2×100)g+H2O0.56g CaO≡2g CaCO3 in 10 L H2O =2g CaCO3 in 104mL H2O=200gCaCO3 in 106mL H2O

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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: