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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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