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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)? |
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Answer» 5 `=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 |
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