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A definite amount of reducing agent is oxidised by 20 mL of 1 M KMnO4 in acid medium, then the same amount of reducing agent is oxidised to the same state by how many mL of 1 M KMnO4 in neutral medium itself changing to Mn4+ state?
- a)3 mL
- b)33.3 mL
- c)12 mL
- d)24 mL
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A definite amount of reducing agent is oxidised by 20 mL of 1 M KMnO4...
Given data:
- Volume of 1 M KMnO4 in acid medium = 20 mL
- Volume of 1 M KMnO4 in neutral medium = ?
- State of Mn in neutral medium = Mn4+
Solution:
- In acid medium, KMnO4 is reduced to Mn2+ state.
- In neutral medium, KMnO4 is reduced to Mn4+ state.
Step 1: Calculation in Acid medium
- Number of moles of KMnO4 in 20 mL of 1 M solution = 1 * (20/1000) = 0.02 moles
- Since the same amount of reducing agent is oxidized in both cases, the number of moles of reducing agent oxidized in acid medium = 0.02 moles
Step 2: Calculation in Neutral medium
- In neutral medium, KMnO4 is reduced to Mn4+ state.
- The balanced equation for this reaction is:
5Fe2+ + MnO4^- + 8H+ → 5Fe3+ + Mn4+ + 4H2O
- The molar ratio between Fe2+ and MnO4^- is 5:1
- Number of moles of Fe2+ oxidized to Fe3+ state = 0.02 moles
- Number of moles of MnO4^- required for this oxidation = 0.02/5 = 0.004 moles
Volume of 1 M KMnO4 in neutral medium:
- Volume = (0.004 moles) / (1 mol/L) = 0.004 L = 4 mL
Therefore, the same amount of reducing agent is oxidized to the same state by 4 mL of 1 M KMnO4 in neutral medium changing to Mn4+ state. The correct answer is option B - 4 mL.
- Volume of 1 M KMnO4 in acid medium = 20 mL
- Volume of 1 M KMnO4 in neutral medium = ?
- State of Mn in neutral medium = Mn4+
Solution:
- In acid medium, KMnO4 is reduced to Mn2+ state.
- In neutral medium, KMnO4 is reduced to Mn4+ state.
Step 1: Calculation in Acid medium
- Number of moles of KMnO4 in 20 mL of 1 M solution = 1 * (20/1000) = 0.02 moles
- Since the same amount of reducing agent is oxidized in both cases, the number of moles of reducing agent oxidized in acid medium = 0.02 moles
Step 2: Calculation in Neutral medium
- In neutral medium, KMnO4 is reduced to Mn4+ state.
- The balanced equation for this reaction is:
5Fe2+ + MnO4^- + 8H+ → 5Fe3+ + Mn4+ + 4H2O
- The molar ratio between Fe2+ and MnO4^- is 5:1
- Number of moles of Fe2+ oxidized to Fe3+ state = 0.02 moles
- Number of moles of MnO4^- required for this oxidation = 0.02/5 = 0.004 moles
Volume of 1 M KMnO4 in neutral medium:
- Volume = (0.004 moles) / (1 mol/L) = 0.004 L = 4 mL
Therefore, the same amount of reducing agent is oxidized to the same state by 4 mL of 1 M KMnO4 in neutral medium changing to Mn4+ state. The correct answer is option B - 4 mL.
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Community Answer
A definite amount of reducing agent is oxidised by 20 mL of 1 M KMnO4...
∴ Milliequivalents of KMnO4 in acid medium = Milliequivalents of KMnO4 in neutral medium.
1 × 5 × 20 = 1 × 3 × V
∴ V = 33.3mL
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A definite amount of reducing agent is oxidised by 20 mL of 1 M KMnO4 in acid medium, then the same amount of reducing agent is oxidised to the same state by how many mL of 1 M KMnO4 in neutral medium itself changing to Mn4+ state?a)3 mLb)33.3 mLc)12 mLd)24 mLCorrect answer is option 'B'. Can you explain this answer?
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