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One litre of sample of hard water contai...

One litre of sample of hard water contains 0.8 mg of `CaCl_(2)` and `"0.8 mg of MgCl"_(2)`. Find the total hardness in terms of parts of `CaCO_(3)`, per `10^(6)` parts of water by mass.

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To find the total hardness of hard water in terms of parts of calcium carbonate (CaCO₃) per 10^6 parts of water by mass, we will follow these steps: ### Step 1: Determine the molecular mass of CaCl₂ and MgCl₂ - **Calcium Chloride (CaCl₂)**: - Atomic mass of Calcium (Ca) = 40 g/mol - Atomic mass of Chlorine (Cl) = 35 g/mol - Molecular mass of CaCl₂ = 40 + (2 × 35) = 40 + 70 = 110 g/mol ...
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A water is said to be soft water if it produces sufficient foam with the soap and water that does not produce foam with soap is known as hard water. Hardness has been classified into two types (i)Temporary hardness (ii) Permanent hardness. Temporary hardness is due to presence of calcium and magnesium bicarbonate. It is simply removed by boiling as Ca(HCO_(3))_(2)overset(Delta)rarr CaCO_(3)darr+CO_(2)uarr+H_(2)O Mg(HCO_(3))_(2)overset(Delta)rarr MgCO_(3)darr+CO_(2)uarr+H_(2)O temporary hardness can also be removed by addition of slaked lime, Ca(OH)_(2) Ca(HCO_(3))_(2)+Ca(OH)_(2) to 2CaCO_(3)darr+2H_(2)O permanent hardsness is due to presencce of sulphates and chlorides of Ca,Mg,etc. It is removed by washing soda as CaCl_(2)+Na_(2)CO_(3) to CaCO_(3)darr+2NaCl CaSO(4)+Na_(2)CO_(3)to CaCO_(3)darr+Na_(2)SO_(4) Permanent hardness also removed by ion exchange resin process as 2RH+Ca^(2+)toR_(2)Ca+2H^(+) 2ROH+SO_(4)^(2-) to R_(2)SO_(4)+2OH^(-) The degree of hardness of water is measured in terms of PPm of CaCO_(3) 100 PPm means 100 g of CaCO_(3) is present in 10^(6) g of H_(2)O . If any other water sample which contain 120 PPm of MgSO_(4) , hardness in terms of CaCO_(3) is equal to =100 PPm. One litre of a sample of hard water (d=1 g/mL) cotains 136 mg of CaSO_(4) and 190 mg of MgCl_(2) . What is the total hardness of water in terms of CaCO_(3) ?

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