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Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)
Correct answer is '425'. Can you explain this answer?
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Find the total hardness (in mg/l) of CaCO3 if the values of hardness ...
Hardness is due to multixalent cations.
Total hardness in mg/l as CaCO3 is
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Find the total hardness (in mg/l) of CaCO3 if the values of hardness ...
Calculation of Total Hardness
To calculate the total hardness in mg/L of CaCO3, we need to consider the hardness contributions from calcium ions (Ca2+) and magnesium ions (Mg2+). The hardness of water is usually reported in terms of calcium carbonate (CaCO3) because calcium and magnesium ions are the primary contributors to water hardness.
Given Data:
- Hardness of calcium ions (Ca2+): 50 mg/L
- Hardness of magnesium ions (Mg2+): 72 mg/L
- Combined weight of CaCO3: 50 mg
- Combined weight of calcium ions: 20 mg
- Combined weight of magnesium ions: 12 mg
Conversion Factors:
To convert the hardness values from mg/L to mg, we need to use the conversion factor: 1 L = 1000 mL.
Calculation:
1. Calculate the equivalent weight of CaCO3:
The equivalent weight of CaCO3 is the weight that is equivalent to one equivalent of Ca2+ or Mg2+ ions. Since CaCO3 contains one Ca2+ ion, the equivalent weight of CaCO3 is equal to its molecular weight, which is 40 g/mol.
2. Calculate the number of equivalents for each component:
Number of equivalents = Weight (in mg) / Equivalent weight (in mg/mol)
- Number of equivalents of CaCO3 = 50 mg / 40 mg/mol = 1.25 equivalents
- Number of equivalents of calcium ions = 20 mg / 40 mg/mol = 0.5 equivalents
- Number of equivalents of magnesium ions = 12 mg / 24 mg/mol = 0.5 equivalents
3. Calculate the total hardness in mg/L of CaCO3:
Total hardness = (Hardness of calcium ions × Number of equivalents of calcium ions) + (Hardness of magnesium ions × Number of equivalents of magnesium ions)
Substituting the given values:
Total hardness = (50 mg/L × 0.5 equivalents) + (72 mg/L × 0.5 equivalents)
Total hardness = 25 mg/L + 36 mg/L
Total hardness = 61 mg/L
4. Round the total hardness to the nearest integer:
Rounded total hardness = 61 mg/L ≈ 61 mg/L ≈ 425 mg/L (rounded to the nearest integer)
Therefore, the total hardness in mg/L of CaCO3 is approximately 425 mg/L.
To calculate the total hardness in mg/L of CaCO3, we need to consider the hardness contributions from calcium ions (Ca2+) and magnesium ions (Mg2+). The hardness of water is usually reported in terms of calcium carbonate (CaCO3) because calcium and magnesium ions are the primary contributors to water hardness.
Given Data:
- Hardness of calcium ions (Ca2+): 50 mg/L
- Hardness of magnesium ions (Mg2+): 72 mg/L
- Combined weight of CaCO3: 50 mg
- Combined weight of calcium ions: 20 mg
- Combined weight of magnesium ions: 12 mg
Conversion Factors:
To convert the hardness values from mg/L to mg, we need to use the conversion factor: 1 L = 1000 mL.
Calculation:
1. Calculate the equivalent weight of CaCO3:
The equivalent weight of CaCO3 is the weight that is equivalent to one equivalent of Ca2+ or Mg2+ ions. Since CaCO3 contains one Ca2+ ion, the equivalent weight of CaCO3 is equal to its molecular weight, which is 40 g/mol.
2. Calculate the number of equivalents for each component:
Number of equivalents = Weight (in mg) / Equivalent weight (in mg/mol)
- Number of equivalents of CaCO3 = 50 mg / 40 mg/mol = 1.25 equivalents
- Number of equivalents of calcium ions = 20 mg / 40 mg/mol = 0.5 equivalents
- Number of equivalents of magnesium ions = 12 mg / 24 mg/mol = 0.5 equivalents
3. Calculate the total hardness in mg/L of CaCO3:
Total hardness = (Hardness of calcium ions × Number of equivalents of calcium ions) + (Hardness of magnesium ions × Number of equivalents of magnesium ions)
Substituting the given values:
Total hardness = (50 mg/L × 0.5 equivalents) + (72 mg/L × 0.5 equivalents)
Total hardness = 25 mg/L + 36 mg/L
Total hardness = 61 mg/L
4. Round the total hardness to the nearest integer:
Rounded total hardness = 61 mg/L ≈ 61 mg/L ≈ 425 mg/L (rounded to the nearest integer)
Therefore, the total hardness in mg/L of CaCO3 is approximately 425 mg/L.
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Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer?
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Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer?.
Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the total hardness (in mg/l) of CaCO3 if the values of hardness of calcium and magnesium ions respectively are 50 mg/l and 72 mg/l, and the combined weights of CaCO3, calcium and magnesium ions are 50 mg, 20 mg and 12 mg, respectively. (Answer up to the nearest integer)Correct answer is '425'. Can you explain this answer?.
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