1.

The hardness of water sample containing 0.002 of water is expressed as :A. 20 ppmB. 200 ppmC. 2000 ppmD. 120 ppm

Answer» Correct Answer - B
Amount of `MgSO_(4)=0.002xx120xx1000`
`=120` mg
Now 120 mg `MgSO_(4)-=100 mg CaCO_(3)`
`therefore` 240 mg `MgSO_(4)-=200 mg CaCO_(3)`
1 L of water contains =200 mg of `CaCO_(3)`
or `10^(6)` mg `H_(2)O` contains =200 mg of `CaCO_(3)`
`therefore ` Degree of hardness =200 ppm.


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