3 grms of MgSO4 and 8.1 grms of Ca(HCO3)2 are present in 10 kg of water .The permanenet and temprary hardness are respectively
Answers
Answer:
0.75ppm(0.25+0.5)
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Explanation:
It is given that 10 kg of water contains 3 g of MgSO4.
i.e., 1000 g of water contains 0.3g of MgSO4
therefore, 1 million grams of water (10⁶g ) contains 0.3g × 1000 = 300 g of MgSO4
1 mole of MgSO4 = 1 mole of CaCO3
The molecular weight of MgSO4 = 120g/mol and CaCO3= 100g/mol
120 g of MgSO4 = 100 g ofCaCO3
Therefore, the equivalent amount of CaCO3 for 0.3g of MgSO4 = 0.3 × 100 /120 = 0.25 ppm.(permanent hardness)
Hence, the degree of hardness of water due to MgSO4 is 0.25 ppm.
similarly with calcium bicarbonate
mass of calcium bicarbonate present in 1 kg of water is 0.81g
162 g of Ca(HCO3)2 = 100 g of CaCO3
Therefore, the equivalent amount of CaCO3 for 0.81g of Ca(HCO3)2=0.81*100/162=0.5ppm(temprary hardness)
total hardness is 0.75ppm