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

Options

A. `2.5 p p m`
B. `1.95 p p m`
C. `2.15 p p m`
D. `195 p p m`

Answer

B

Explanation

(b) Molecular weight of `CaCl_(2)=111.0g`
`MW` of `CaCO_(3)=100g`
`MW` of `MgCl_(2)=95.0 g`
`CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl`
`MgCl_(2)+Na_(2)CO_(3)toMgCO_(3)+2NaCl`
`(i) 111.0g CaCl_(2)=100 g CaCO_(3)`
`1 mg CaCl_(2)=100/111 mg CaCO_(3)=0.9 mg CaCO_(3)`
(ii) `95.0 g MgCl_(2)=100 g CaCO_(3)`
`1 mg MgCl_(2)-=100/95 mg CaCO_(3)=1.05 mg CaCO_(3)`
Hardness of `CaVO_(3) p p m`
`=((0.9+1.05)xx10^(-3)gxx10^(16)mL)/(10^(3)mL)`
`=1.95 p p m`