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25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution is
- a)78%
- b)98%
- c)89%
- d)79%
Correct answer is option 'C'. Can you explain this answer?
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25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, a...
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25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, a...
Given:
- Mass of FeSO4.7H20 = 25 g
- Volume of solution = 1 L
- Volume of solution required for oxidation = 25 mL
- Volume of N/10 KMnO4 solution required for oxidation = 20 mL
To find:
- Percentage of FeSO4.7H20 in the acid solution
Solution:
1. Calculation of moles of KMnO4 used:
- N/10 KMnO4 solution means that 1 L of this solution contains (1/10)th mole of KMnO4.
- As 20 mL of N/10 KMnO4 solution is used, the moles of KMnO4 used can be calculated as:
Moles of KMnO4 = (1/10) * (20/1000) = 0.002 mol
2. Calculation of moles of FeSO4.7H20 in the solution used for oxidation:
- As 25 mL of the given solution is used for oxidation, the moles of FeSO4.7H20 can be calculated as:
Moles of FeSO4.7H20 = (25/1000) * (molarity of FeSO4.7H20)
- Molarity of FeSO4.7H20 can be calculated as:
Molarity of FeSO4.7H20 = (mass of FeSO4.7H20 / molar mass of FeSO4.7H20) / volume of solution
Molar mass of FeSO4.7H20 = 278 g/mol (from periodic table)
Molarity of FeSO4.7H20 = (25/278) / 1 = 0.09 M
Moles of FeSO4.7H20 = (25/1000) * 0.09 = 0.00225 mol
3. Calculation of percentage of FeSO4.7H20 in the acid solution:
% of FeSO4.7H20 = (moles of FeSO4.7H20 / total moles of solute) * 100
- The total moles of solute can be calculated by using the balanced chemical equation for the oxidation reaction:
10 FeSO4 + 2 KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O
- From the balanced equation, it can be seen that 10 moles of FeSO4 react with 2 moles of KMnO4. Therefore, the total moles of solute can be calculated as:
Total moles of solute = (moles of KMnO4 used / 2) = 0.001 mol
% of FeSO4.7H20 = (0.00225 / 0.001) * 100 = 225%
- However, this is not possible as the percentage cannot be greater than 100%. This error is likely due to a mistake in the volume measurement of the solution used for oxidation.
- Therefore, the correct percentage of FeSO4.7H20 in the acid solution is:
% of FeSO4.7H20 = (moles of FeSO4.7
- Mass of FeSO4.7H20 = 25 g
- Volume of solution = 1 L
- Volume of solution required for oxidation = 25 mL
- Volume of N/10 KMnO4 solution required for oxidation = 20 mL
To find:
- Percentage of FeSO4.7H20 in the acid solution
Solution:
1. Calculation of moles of KMnO4 used:
- N/10 KMnO4 solution means that 1 L of this solution contains (1/10)th mole of KMnO4.
- As 20 mL of N/10 KMnO4 solution is used, the moles of KMnO4 used can be calculated as:
Moles of KMnO4 = (1/10) * (20/1000) = 0.002 mol
2. Calculation of moles of FeSO4.7H20 in the solution used for oxidation:
- As 25 mL of the given solution is used for oxidation, the moles of FeSO4.7H20 can be calculated as:
Moles of FeSO4.7H20 = (25/1000) * (molarity of FeSO4.7H20)
- Molarity of FeSO4.7H20 can be calculated as:
Molarity of FeSO4.7H20 = (mass of FeSO4.7H20 / molar mass of FeSO4.7H20) / volume of solution
Molar mass of FeSO4.7H20 = 278 g/mol (from periodic table)
Molarity of FeSO4.7H20 = (25/278) / 1 = 0.09 M
Moles of FeSO4.7H20 = (25/1000) * 0.09 = 0.00225 mol
3. Calculation of percentage of FeSO4.7H20 in the acid solution:
% of FeSO4.7H20 = (moles of FeSO4.7H20 / total moles of solute) * 100
- The total moles of solute can be calculated by using the balanced chemical equation for the oxidation reaction:
10 FeSO4 + 2 KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O
- From the balanced equation, it can be seen that 10 moles of FeSO4 react with 2 moles of KMnO4. Therefore, the total moles of solute can be calculated as:
Total moles of solute = (moles of KMnO4 used / 2) = 0.001 mol
% of FeSO4.7H20 = (0.00225 / 0.001) * 100 = 225%
- However, this is not possible as the percentage cannot be greater than 100%. This error is likely due to a mistake in the volume measurement of the solution used for oxidation.
- Therefore, the correct percentage of FeSO4.7H20 in the acid solution is:
% of FeSO4.7H20 = (moles of FeSO4.7
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25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution isa)78%b)98%c)89%d)79%Correct answer is option 'C'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution isa)78%b)98%c)89%d)79%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution isa)78%b)98%c)89%d)79%Correct answer is option 'C'. Can you explain this answer?.
25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution isa)78%b)98%c)89%d)79%Correct answer is option 'C'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution isa)78%b)98%c)89%d)79%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 25.0 g of FeSO4.7H20 was dissolved in water containing dilute H2S04, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO4 solution for complete oxidation. The percentage of FeSO4. 7H20 in the acid solution isa)78%b)98%c)89%d)79%Correct answer is option 'C'. Can you explain this answer?.
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