`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:

Options

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

Answer

B

Explanation

Temporary hardness is due to `HCO_(3)^(-)" of "Ca^(2+) and Mg^(2+)`
`Ca(HCO_(3))_(2)+underset(56g)(CaO)rarr underset((2xx100g))(2CaCO_(3))+H_(2)O`
`0.56g CaO-=2g CaCO_(3)" in "10 L H_(2)O`
`=2g CaCO_(3)" in "10^(4)mL H_(2)O`
`=200g CaCO_(3)" in "10^(6)mL H_(2)O`
`=200" ppm of "CaCO_(3)`