If 111 mg of `CaCl_(2)` and 12 mg of `MgSO_(4)` are present in 2L of water, what is its hardness (in gram `CaCO_(3)//p p m)` ?

Options

A. 5
B. 10
C. 15
D. 203

Answer

B

Explanation

`underset(=111xx10^(3)mg)underset(40xx35xx2)(CaCl_(2))-= underset(=100xx10^(3)mg)underset(40+10+48)(CaCO_(3))`
`underset(=120xx10^(3)mg)underset(24+32+64)(MgSO_(4))-=underset(=100xx10^(3)mg)underset(40+12+40)(CaCO_(3))`
`111xx10^(3)mg` of
`CaCl_(2)=100xx10^(3)mg" of "CaCO_(3)`
11.1 mg of `CaCl_(2)`
`=(100xx10^(3)xx11.1)/(111xx10^(3))mg " of "CaCO_(3)`
`=10 mg " of "CaCO_(3)`
Similarly, `120xx10^(3)mg" of " MgSO_(4)`
`=100xx10^(3)mg " of " CaCO_(3)`
`12 mg " of " MgSO_(4)=(100xx10^(3))/(120xx10^(3))xx12mg" of "CaCO_(3)`
`=10mg" of "CaCO_(3)`
In 2 L of water, total weight of
`CaCO_(3) =10+10=20mg`
`therefore ` In 1 L of water, total weight of
`CaCO_(3)=(20)/(2)mg=10mg`
`therefore " In "10^(6)`mg of water, total weight of
`CaCO_(3)=10mg`
So, in `10^(6)` part water, hardness of water in terms of `CaCO_(3)=10`ppm