A sample of hard water contains 1 mg `CaCl_2` and 1 mg `MgCl_2` per litre. Calculate the hardness of water in terms of `CaCO_3` present in per `10^(6)` parts of water.
(a). 2.5 ppm
(b). 1.95 ppm
(c). 2.15 ppm
(d). 195 ppm

Explanation

Mw of `CaCl_2=11.0g`
`mw=CaCO_3=100g`
Mw of `MgCl_295.0g`
`CaCl_2+Na_2CO_2toCaCO_3+2NaCl`
`MgCl_2+Na_2CO_3toMgCO_3+2NaCl`
(i). `111.0gCaCl_2-=100gCaCO_3`
1 mg `CaCl_2-=(100)/(111)mgCaCO_3=0.9mgCaCO_3`
(ii). `95.0g MgCl_2-=100gCaCO_3`
1 mg `MgCl_2-=(100)/(95)mgCaCO_3=1.05mgCaCO_3`
Hardness of `CaCO_3` ppm
`-=((0.9+1.05)xx10^(-3)gxx10^(6)mL)/(10^(3)mL)`
`1.95 ppm`