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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 `10^(6)`) of `CaCO_(3)` is:A. 100B. 200C. 10D. 20 |
Answer» Correct Answer - B Temporary hardness is due to `HCO_(3)^(ɵ)` of `Ca^(2+)` and `Mg^(2+)` `Ca (HCO_(3))_(2) + (CaO)/(56 g) rarr (2 CaCO_(3))/((2 xx 100) g) + H_(2) O` `0.56 g CaO -= 2 g CaCO_(3) "in" 10 L H_(2) O` `= 2 g CaCO_(3) "in" 10^(4) mL H_(2) O` `= 200 g CaCO_(3) "in" 10^(6) mL H_(2) O` |
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