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If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ?
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If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht i...
Calculation of Hardness of Water
Step 1: Calculate the equivalent weight of CaCO3.
The equivalent weight of CaCO3 is 50 g/mol.
Step 2: Calculate the number of moles of CaCl2 and MgSO4.
Number of moles of CaCl2 = 11.1 mg / (40.08 g/mol) = 0.0002779 mol
Number of moles of MgSO4 = 12 mg / (120.37 g/mol) = 0.0000997 mol
Step 3: Calculate the total concentration of Ca2+ and Mg2+ ions.
Concentration of Ca2+ ions = (0.0002779 mol / 2 L) * 2 = 0.0002779 M
Concentration of Mg2+ ions = (0.0000997 mol / 2 L) * 2 = 0.0000997 M
Step 4: Calculate the total hardness of water.
Total hardness of water = (Concentration of Ca2+ ions * 2.5) + (Concentration of Mg2+ ions * 4.1)
Total hardness of water = (0.0002779 M * 2.5) + (0.0000997 M * 4.1) = 0.0009105
Step 5: Convert the total hardness of water to gm CaCO3 / ppm.
1 ppm = 1 mg/L
1 gm CaCO3 / L = 1000 ppm
Therefore, hardness of water = (0.0009105 * 1000) / 50 = 18.21 ppm = 18.21 mg/L
Answer: The hardness of water is 18.21 mg/L or 18.21 gm CaCO3 / ppm.
Step 1: Calculate the equivalent weight of CaCO3.
The equivalent weight of CaCO3 is 50 g/mol.
Step 2: Calculate the number of moles of CaCl2 and MgSO4.
Number of moles of CaCl2 = 11.1 mg / (40.08 g/mol) = 0.0002779 mol
Number of moles of MgSO4 = 12 mg / (120.37 g/mol) = 0.0000997 mol
Step 3: Calculate the total concentration of Ca2+ and Mg2+ ions.
Concentration of Ca2+ ions = (0.0002779 mol / 2 L) * 2 = 0.0002779 M
Concentration of Mg2+ ions = (0.0000997 mol / 2 L) * 2 = 0.0000997 M
Step 4: Calculate the total hardness of water.
Total hardness of water = (Concentration of Ca2+ ions * 2.5) + (Concentration of Mg2+ ions * 4.1)
Total hardness of water = (0.0002779 M * 2.5) + (0.0000997 M * 4.1) = 0.0009105
Step 5: Convert the total hardness of water to gm CaCO3 / ppm.
1 ppm = 1 mg/L
1 gm CaCO3 / L = 1000 ppm
Therefore, hardness of water = (0.0009105 * 1000) / 50 = 18.21 ppm = 18.21 mg/L
Answer: The hardness of water is 18.21 mg/L or 18.21 gm CaCO3 / ppm.
Community Answer
If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht i...
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If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ?
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If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ?.
If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If 11.1 mg of cacl2 and 12mg of mgso4 are present in 2l of water wht is it's hardness (in gm caco3/ppm) ?.
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