10L of hard water required `0.56g` of lime `(CaO)` for removing hardness. Hence, temporary hardness in p p m (part per million, `10^6`) of `CaCO_3` is

Options

A. 100
B. 200
C. 10
D. 20

Answer

B

Explanation

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`