The hardness of water sample containing 0.002 of water is expressed as

1.	The hardness of water sample containing 0.002 of water is expressed as:
Answer» 20 PPM 200 ppm 2000 ppm 120 ppm Solution :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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