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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)`.

Text Solution

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Temporary hardness is due to `HCO_(3)^(ᶱ)` of `Ca^(2+)` and `Mg^(2+)`
`Ca(HCO_(3))_(2)+CaOrarr2CaCO_(3)+H_(2)O`
`56g 2 xx 100g`
`56 g CaO = 200g CaCO_(3)` in `10 L` of `H_(2)O`
`0.28g CaO = (200 xx 0.28)/(56)`
`=1 g CaCO_(3)` in `10 L` of `H_(2)O`
`=1 g CaCO_(3)` in `10 xx 1000mL` of `H_(2)O`
`1g CaCO_(3)` in `10^(4) mL` of `H_(2)O`
`100g CaCO_(3)` in `10^(6) mL` of `H_(2)O = 100 "ppm"`
Hence, temporary hardness of `CaCO_(3) = 100"ppm"` .
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