One litre of a sample of hard water contains 1 mg of CaCl2 and 1 mg of

1.	One litre of a sample of hard water contains 1 mg of CaCl2 and 1 mg of MgCl2 Find out the total hardness in terms of parts of CaCO3 per 106 parts of water by mass, (i) Mol. Mass of CaCl2 = 111
Answer» Now lllg CaCl₂ = 100g of CaCO₃ 1 mg of CaCl₂ = \(\frac{100}{111}\) × 1 mg of CaCO = 0.9 mg of CaCO₃ (ii) Mol. Mass of MgCk = 95 Now 95g of MgCl₂ = 100g of CaCO₃ 1 mg of MgCl₂ = \(\frac{100}{95}\) × 1 mg of CaCO₃ = 1.05 mg of CaCO₃ Thus, 1 litre of hard water contain = 0.90 + 1.05 = 1.95 mg of CaCO₃.

One litre of a sample of hard water contains 1 mg of CaCl2 and 1 mg of MgCl2 Find out the total hardness in terms of parts of CaCO3 per 106 parts of water by mass, (i) Mol. Mass of CaCl2 = 111

Answer»

Now lllg CaCl₂ = 100g of CaCO₃

1 mg of CaCl₂ = \(\frac{100}{111}\) × 1 mg of CaCO = 0.9 mg of CaCO₃

(ii) Mol. Mass of MgCk = 95

Now 95g of MgCl₂ = 100g of CaCO₃

1 mg of MgCl₂ = \(\frac{100}{95}\) × 1 mg of CaCO₃ = 1.05 mg of CaCO₃

Thus, 1 litre of hard water contain = 0.90 + 1.05 = 1.95 mg of CaCO₃.

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One litre of a sample of hard water contains 1 mg of CaCl2 and 1 mg of MgCl2 Find out the total hardness in terms of parts of CaCO3 per 106 parts of water by mass, (i) Mol. Mass of CaCl2 = 111

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