One litre sample of water contains 0.9 mg `CaCl_(2)` and 0.9 mg of `MgCl_(2)`. Find the total hardness in terms of parts permillion of `CaCO_(3)`.
Text Solution
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`111 g CaCl_(2)-=100 g CaCO_(3)` `therefore " " 0.9 mg CaCl_(2) -=(100)/(111)xx0.9 mg CaCO_(3)` `-=0.81 mg " of" CaCO_(3)` ` 95 g MgCl -=100 g CaCO_(3)` ` therefore " " 0.9 mg MgCl_(2) -=(100)/(95)xx0.9 mg CaCO_(3)` `-=0.94 mg " of " CaCO_(3)` Thus, 1 litre water to `10^(6)`. mg water contains (0.81+0.94 = 1.75 mg) i.e., hardness of water is 1.75 ppm.
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