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10 L of hard water requires 0.28 g of li...

`10 L` of hard water requires `0.28 g` of lime `(CaO)` for removing hardness. Calculate the temporary hardness in ppm of `CaCO_(3)`.

A

100

B

200

C

10

D

20

Text Solution

Verified by Experts

The correct Answer is:
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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