If 11.1 mg of `CaCl_(2)` and 12 mg of `MgSO_(4)` are present in 2 L of water, what is its hardness (in gram `CaCO_(3)`/ppm)?

Options

A. 5
B. 10
C. 15
D. 20

Answer

B

Explanation

`underset(40+35.5 times 2)(CaCl_(2)) equiv underset(40+12+48)(CaCO_(3))`
`=111 times 10^(3)mg=100 times 10^(3)mg`
`underset(underset(=120 times 10^(3)mg)(24+32+64))(MgSO_(4)) equiv underset(underset(=100 times 10^(3)mg)(40+12+40))(CaCO_(3))`
`111 times 10^(3)" mg of "CaCl_(2)=100 times 10^(3)" mg of "CaCO_(3)`
`11.1" mg of "CaCl_(2)=(100 times 10^(3) times 11.1)/(111 times 10^(3))mg`
of ` CaCO_(3)=` 10 of `CaCO_(3)`
Similarly, `120 times 10^(3)" mg of "MgSO_(4)`
`=100 times 10^(3)" mg of "CaCO_(3)`
12 mg of `MgSO_(4)=(100 times 10^(3))/(120 times 10^(3)) times 12" mg of "CaCO_(3)`
`" "=10" mg of "CaCO_(3)`
In 2 L of water, total weight of `CaCO_(3)`
`=10+10=20` mg
In 1 L of water, total weight of `CaCO_(3)`
`20/2mg=10mg`
In `10^(6)` mg of water, total weight of `CaCO_(3)=10` mg
So, in `10^(6)` part water, hardness of water in terms of `CaCO_(3)=10` ppm